Large deviations asymptotics and the spectral theory of multiplicatively regular Markov pro

尝试新事物体现勇敢的英语作文Embracing the Unknown: The Courage to Try New Things.In the vast expanse of life, the unknown looms large, a mysterious and often intimidating territory. It is the realm of the uncharted, the unexplored, and the untested.It is here, in this realm, that the true essence of courage is tested. The courage to step into the unknown, to embrace change, and to try new things is a bravery that defines our character and shapes our destiny.The fear of the unknown is a primal emotion, hardwired into our DNA. It is the reason why we hesitate before taking a leap of faith, why we cling to familiar routines and avoid deviation. But it is also the reason why we grow, why we evolve, and why we push the boundaries of what is possible. The courage to try new things is the antidote to this fear, the key that unlocks the door to personal growth and transformation.Trying new things means venturing into territorieswhere there are no maps, no guides, and no guarantees of success. It means taking risks, making mistakes, andpossibly encountering failure. But it is also the path to discovery, to learning, and to the expansion of our horizons. It is the way we find new passions, new interests, and new purposes.The courage to try new things is not a matter of being fearless, but of being willing to face our fears. It is about stepping out of our comfort zones, even when it feels safer to stay put. It is about embracing the unknown, not shying away from it. It is about finding the balance between caution and audacity, between risk and reward.This courage is not something that we are born with; it is something that we cultivate over time. It grows through practice, through repeated exposure to new experiences, and through the gradual erosion of our fears. It isstrengthened by setbacks and failures, because they teachus resilience and the importance of perseverance.Trying new things can be exciting, terrifying, rewarding, and challenging all at the same time. It is a journey that takes us into the heart of our being, revealing our strengths, weaknesses, and the true essence of our character. It is a journey that transforms us, making us more resilient, more open-minded, and more empathetic.In the end, the courage to try new things is not just about achieving success or avoiding failure. It is about living life to the fullest, about growing as individuals, and about embracing the opportunities and challenges that come our way. It is about living bravely, about stepping into the unknown with confidence and hope, and aboutfinding joy and fulfillment in the journey of life.As we embark on this journey, let us remember that courage is not the absence of fear, but the willingness to face it. Let us remember that trying new things is not just about achieving goals or overcoming obstacles, but about discovering who we are, what we are capable of, and what we can contribute to the world. Let us embrace the unknown,step into the void, and find the courage to try new things. Because in the end, it is this courage that will define us, that will shape our destiny, and that will make all the difference.。

2020年12月13日托福阅读答案解析12月13日托福阅读词汇题：Obviously=clearlyWidespread=commonDense=thickThus=consequentlyresultantShallow=smalldepthexerciseProfound=very strongEmergence=riseTactic=strategyAdjacent to=near toParallel=match12月13日托福阅读第一篇题材划分：生物类主要内容：板块运动能够改变生物多样性，提到生物区的划分，少于百分之二十的物种相似度就是不同的区越多说明那里的多样性越高。

比如板块分开的时候，多样性增加，反之亦然。

一个山脉能够把原本的湿润风给挡了，就变成沙漠不适合生长了。

或者一个障碍的形成能够把本来的一个物种分成两个，一南一北，等到在合并的时候，发现北部的能够到南部生活，但南部的很少到北部生活。

相似TPO练习推荐TPO31- Speciation in Geographically Isolated Populations相关背景知识：Speciation is the evolutionary process by which new biological species arise. The biologist Orator F. Cook wasthe first to coin the term 'speciation' for the splitting of lineages or "cladogenesis," as opposed to "anagenesis" or "phyletic evolution" occurring within lineages. Charles Darwin was the first to describe the role of naturalselection in speciation.There is research comparing the intensity of sexual selection in different clades with their number of species.There are four geographic modes of speciation in nature, based on the extent to which speciating populations are isolated from one another： allopatric, peripatric, parapatric, and sympatric. Speciation may also be induced artificially, through animal husbandry, agriculture, or laboratory experiments. Whether genetic drift is a minor or major contributor to speciation is the subject matter of much ongoing discussion.All forms of natural speciation have taken place over the course of evolution; however, debate persists as to the relative importance of each mechanism in driving biodiversity.One example of natural speciation is the diversity of the three-spined stickleback, a marine fish that, after the lastglacial period, has undergone speciation into new freshwater colonies in isolated lakes and streams. Over an estimated 10,000 generations, the sticklebacks show structural differences that are greater than those seen betweendifferent genera of fish including variations in fins, changes in the number or size of their bony plates, variable jaw structure, and color differences.During allopatric speciation, a population splits into two geographically isolated populations (for example, by habitat fragmentation due to geographical change such as mountain formation). The isolated populations then undergo genotypic and/or phenotypic divergence as： (a) they become subjected to dissimilar selective pressures; (b) they independently undergo genetic drift; (c) different mutations arise in the two populations. When the populations come back into contact, they have evolved such that they are reproductively isolated and are no longer capable of exchanging genes. Island genetics is the term associated with the tendency of small, isolated genetic pools to produce unusual traits. Examples include insular dwarfism and the radical changes among certain famous island chains, for example on Komodo. The Galápagos Islands are particularly famous for their influence on Charles Darwin. During his five weeks there he heard that Galáp agos tortoises could be identified by island, and noticed that finches differed from one island to another, but it was only nine months later that he reflected that such facts could show that species were changeable. When he returned to England, his speculation on evolution deepened after experts informed him that these were separate species, not just varieties, and famously that other。

⾦融市场与机的构Madura第九版题库ch11Chapter 11Stock Valuation and Risk1. The common price-earnings valuation method applied the ______ price-earnings ratio to ________earnings per share in order to value the firm’s stock.A) firm’s; industryB) firm’s; firm’sC) average industry; industryD) average industry; firm’sANSWER: D2. A firm is expected to generate earnings of $2.22 per share next year. The mean ratio of share price toexpected earnings of competitors in the same industry is 15. Based on this information, the valuation of the firm’s shares based on the price-earnings (PE) method isA) $2.22.B) $6.76.C) $33.30.D) none of the aboveANSWER: C3. The PE method to stock valuation may result in an inaccurate valuation for a firm if errors are madein forecasting the firm’s future earnings or in choosing the industry composite used to derive the PE ratio.A) TrueB) FalseANSWER: A4. Bolwork Inc. is expected to pay a dividend of $5 per share next year. Bolwork’s dividends areexpected to grow by 3 percent annually. The required rate of return for Bolwork stock is 15 percent.Based on the dividend discount model, a fair value for Bolwork stock is $_______ per share.A) 33.33B) 166.67C) 41.67D) 60.00ANSWER: C5. Protsky Inc. just paid a divid end of $2.20 per share. The dividend growth rate for Protsky’s dividendsis 3 percent per year. If the required rate of return on Protsky stock is 12 percent, the stock should be valued at $_______ pershare according to the dividend discount model.A) 24.44B) 25.182 Stock Valuation and RiskC) 18.88D) 75.53ANSWER: B6. The limitations of the dividend discount model are more pronounced when valuing stocksA) that pay most of their earnings as dividends.B) that retain most of their earnings.C) that have a long history of dividends.D) that have constant earnings growth.ANSWER: B7. Hancock Inc. retains most of its earnings. The company currently has earnings per share of $11.Hancock expects its earnings to grow at a constant rate of 2 percent per year. Furthermore, theaverage PE ratio of all other firms in Hancock’s industry is 12. Hancock is expected to pay dividends per share of $3.50 during each of the next three years. If investors require a 10 percent rate of return on Hancock stock, a fair price for Hancock stock today is $________.A) 113.95B) 111.32C) 105.25D) none of the aboveANSWER: A8. When evaluating stock performance, ______ measures variability that is systematically related tomarket returns; ______ measures total variabili ty of a stock’s returns.A) beta; standard deviationB) standard deviation; betaC) intercept; betaD) beta; error termANSWER: A9. The ___________ is commonly used as a proxy for the risk-free rate in the Capital Asset PricingModel.A) Treasury bond rateB) prime rateC) discount rateD) federal funds rateANSWER: A10. Arbitrage pricing theory (APT) suggests that a stock’s price is influenced only by a stock’s beta.A) TrueB) FalseANSWER: BStock Valuation and Risk 3 11. Stock prices of U.S. firms primarily involved in exporting are likely to be ________ affected by aweak dollar and __________ affected by a strong dollar.A) favorably; adverselyB) adversely; adverselyC) favorably; favorablyD) adversely; favorablyANSWER: A12. A weak dollar may enhance the value of a U.S. firm whose sales are dependent on the U.S. economy.A) TrueB) FalseANSWER: A13. The January effect refers to the __________ pressure on ______ stocks in January of every year.A) downward; largeB) upward; largeC) downward; smallD) upward; smallANSWER: D14. The expected acquisition of a firm typically results in ____________ in the target’s stock price.A) an increaseB) a decreaseC) no changeD) none of the aboveANSWER: A15. The _______ index can be used to measure risk-adjusted performance of a stock while controlling forthe stock’s volatility.A) SharpeB) TreynorC) arbitrageD) marginANSWER: A16. The _______ index can be used to measure risk-adjusted performance of a stock while controlling for the stock’s beta.A) SharpeB) TreynorC) arbitrageD) marginANSWER: B4 Stock Valuation and Risk17. Stock price volatility increased during the credit crisis.A) TrueB) FalseANSWER: A18. The Sharpe Index measures theA) average return on a stock.B) variability of stock returns per unit of returnC) stock’s beta adjusted for risk.D) excess return above the risk-free rate per unit of risk.ANSWER: D19. A stock’s average return is 11 percent. The average risk-free rate is 9 percent. The stock’s beta is 1and its standard deviation of returns is 10 percent. What is the Sharpe Index?A) .05B) .5C) .1D) .02E) .2ANSWER: E20. A stock’s average return is 10 percent. The average risk-free rate is 7 percent. The standarddeviation of the stock’s return is 4 percent, and the stock’s beta is 1.5. What is the Treynor Index for the stock?A) .03B) .75C) 1.33D) .02E) 50ANSWER: D21. If security prices fully reflect all market-related information (such as historical price patterns) but do not fully reflect all other public information, security markets areA) weak-form efficient.B) semi-strong form efficient.C) strong form efficient.D) B and CE) none of the aboveANSWER: A22. If security markets are semi-strong form efficient, investors cannot solely use ______ to earn excess returns.A) previous price movementsB) insider informationStock Valuation and Risk 5C) publicly available informationD) A and CANSWER: D23. The ______ is commonly used to determine what a stock’s price should have been.A) Capital Asset Pricing ModelB) Treynor IndexC) Sharpe IndexD) B and CANSWER: A24. A stock’s beta is estimated to be 1.3. The risk-free rate is 5 percent, and the market return is expected to be 9 percent. What is the expected return on the stock based on the CAPM?A) 5.2 percentB) 11.7 percentC) 16.7 percentD) 4 percentE) 10.2 percentANSWER: E25. According to the text, other things being equal, stock prices of U.S. firms primarily involved inexporting could be ______ affected by a weak dollar. Stock prices of U.S. importing firms could be ______ affected by a weak dollar.A) adversely; favorablyB) favorably; adverselyC) favorably; favorablyD) adversely; adverselyANSWER: B26. The demand by foreign investors for the stock of a U.S. firm sold on a U.S. exchange may be higherwhen the dollar is expected to ______, other things being equal. (Assume the firm’s operations are unaffected by the value of the dollar.)A) strengthenB) weakenC) stabilizeD) B and CANSWER: A27. A higher beta of an asset reflectsA) lower risk.B) lower covariance between the asset’s returns and market returns.C) higher covariance between the asset’s returns and the market returns.D) none of the above6 Stock Valuation and RiskANSWER: C28. The “January effect” refers to a largeA) rise in the price of small stocks in January.B) decline in the price of small stocks in January.C) decline in the price of large stocks in January.D) rise in the price of large stocks in January.ANSWER: A29. Technical analysis relies on the use of ______ to make investment decisions.A) interest ratesB) inflationary expectationsC) industry conditionsD) recent stock price trendsANSWER: D30. The arbitrage pricing theory (APT) differs from the capital asset pricing model (CAPM) in that it suggests that stock pricesA) are influenced only by the market itself.B) can be influenced by a set of factors in addition to the market.C) are not influenced at all by the market.D) cannot be influenced at all by the industry factors.ANSWER: B31. According to the capital asset pricing model, the required return by investors on a security isA) inversely related with the risk-free rate.B) inversely related with the firm’s beta.C) inversely related with the market return.D) none of the aboveANSWER: D32. Boris stock has an average return of 15 percent. Its beta is 1.5. Its standard deviation of returns is 25 percent. The average risk-free rate is 6 percent. The Sharpe index for Boris stock isA)0.35.B)0.36.C)0.45.D)0.28.E)none of the aboveANSWER: B33. Morgan stock has an average return of 15 percent, a beta of 2.5, and a standard deviation of returns of20 percent. The Treynor index of Morgan stock isA)0.04.B)0.05.Stock Valuation and Risk 7C)0.35.D)0.03.E)none of the above34. Zilo stock has an average return of 15 percent, a beta of 2.5, and a standard deviation of returns of 20percent. The Sharpe index of Zilo stock isA)0.36.B)0.35.C)0.28.D)0.45.E)none of the aboveANSWER: B35. Sorvino Co. is expected to offer a dividend of $3.2 per share per year forever. The required rate ofreturn on Sorvino stock is 13 percent. Thus, the price of a share of Sorvino stock, according to the dividend discount model, is $_________.A) 4.06B) 4.16C)40.63D)24.62E)none of the aboveANSWER: D36. Kudrow stock just paid a dividend of $4.76 per share and plans to pay a dividend of $5 per share nextyear, which is expected to increase by 3 percent per year subsequently. The required rate of return is15 percent. The value of Kudrow stock, according to the dividend discount model, is $__________.A)39.67B)41.67C)33.33D)31.73E)none of the aboveANSWER: B37. LeBlanc Inc. currently has earnings of $10 per share, and investors expect that the earnings per sharewill grow by 3 percent per year. Furthermore, the mean PE ratio of all other firms in the sameindustry as LeBlanc Inc. is 15. LeBlanc is expected to pay a dividend of $3 per share over the next four years, and an investor in LeBlanc requires a return of 12 percent. What is the forecasted stock price of LeBlanc in four years, using the adjusted dividend discount model?A)$150.00B)$163.91C)$45.00D)$168.83E)none of the above8 Stock Valuation and Risk38. Tarzak Inc. has earnings of $10 per share, and investors expect that the earnings per share will growby 3 percent per year. Furthermore, the mean PE ratio of all other firms in the same industry asTarzac is 15. Tarzac is expected to pay a dividend of $3 per share over the next four years, and an investor in Tarzac requires a return of 12 percent. The estimated stock price of Tarzak today should be __________ using the adjusted dividend discount model.A)$116.41B)$104.91C)$161.15D)none of the aboveANSWER: A39. The standard deviation of a stock’s returns is used to measure a stock’sA)volatility.B)beta.C)Treynor Index.D)risk-free rate.ANSWER: A40. The formula for a stock portfolio’s volatility does not contain theA)weight (proportional investment) assigned to each stock.B)variance (standard deviation squared) of returns of each stock.C) correlation coefficients between returns of each stock.D) risk-free rate.ANSWER: D41. If the returns of two stocks are perfectly correlated, thenA) their betas should each equal 1.0.B) the sum of their betas should equal 1.0.C) their correlation coefficient should equal 1.0.D) their portfolio standard deviation should equal 1.0.ANSWER: C42. A stock’s beta can be measured from the estimate of the using regression analysis.A) interceptB) market returnC) risk-free rateD) slope coefficientANSWER: D43. A beta of 1.1 means that for a given 1 percent change in the value of the market, theis expected to change by 1.1 percent in the same direction.A)risk-free rateB)stock’s valueStock Valuation and Risk 9C)s tock’s standard deviationD)correlation coefficientANSWER: B44. Stock X has a lower beta than Stock Y. The market return for next month is expected to be either–1 percent, +1 percent, or +2 percent with an equal probability of each scenario. The probability distribution of Stock X returns for next month isA)the same as that of Stock Y.B)more dispersed than that of Stock Y.C)less dispersed than that of Stock Y.D)zero.ANSWER: C45. The beta of a stock portfolio is equal to a weighted average of theA)betas of stocks in the portfolio.B)betas of stocks in the portfolio, plus their correlation coefficients.C)standard deviations of stocks in the portfolio.D)correlation coefficients between stocks in the portfolio.ANSWER: A46. Value at risk estimates the a particular investment for a specified confidence level.A)beta ofB)risk-free rate ofC)largest expected loss toD)standard deviation ofANSWER: C47. A stock has a standard deviation of daily returns of 1 percent. It wants to determine the lowerboundary of its probability distribution of returns, based on 1.65 standard deviations from theexpected outcome. The stock’s expected daily return is .2 percent. The lower boundary isA)–1.45 percent.B)–1.85 percent.C)0 percent.D)–1.65 percent.ANSWER: A48. A stock has a standard deviation of daily returns of 3 percent. It wants to determine the lower boundary of its probability distribution of returns, based on 1.65 standard deviations from the expected outcome. The stock’s expected daily return is .1 percent. The lower boundary isA)–1.65 percent.B)–3.00 percent.C)–4.85 percent.D)–5.05 percent.10 Stock Valuation and RiskANSWER: C49. Which of the following is not commonly used as the estimate of a stock’s volatility?A)the estimate of its standard deviation of returns over a recent periodB)the trend of historical standard deviations of returns over recent periodsC)the implied volatility derived from an option pricing modelD)the estimate of its option premium derived from an option pricing modelANSWER: D50. The credit crisis only affected the stock performance of stocks in the U.S.A) TrueB) FalseANSWER: B51. When new information suggests that a firm will experience lower cash flows than previously anticipated or lower risk, investors will revalue the corresponding stock downward.A) TrueB) FalseANSWER: B52. A relatively simple method of valuing a stock is to apply the mean price-earnings (PE) ratio of all publicly traded competitors in the respective industry to the firm’s expected earnings for the year.ANSWER: A53. While the previous year’s earnings are often used as a base for forecast ing future earnings, the recent year’s earnings do not always provide an accurate forecast of the future.A) TrueB) FalseANSWER: A54. If investo rs agree on a firm’s forecasted earnings, they will derive the same value for that firm using the PE method to value the firm’s stock.A) TrueB) FalseANSWER: B55. The dividend discount model states that the price of a stock should reflect the present value of the stock’s future dividends.A) TrueB) FalseStock Valuation and Risk 11 ANSWER: A56. The dividend discount model can be adapted to assess the value of any firm, even those that retain most or all of their earnings.A) TrueB) FalseANSWER: A57. For firms that do not pay dividends, a more suitable valuation may be the free cash flow model.A) TrueB) FalseANSWER: A58. The capital asset pricing model (CAPM) is based on the premise that the only important risk of a firm is unsystematic risk.A) TrueB) FalseANSWER: B59. The prime rate is commonly used as a proxy for the risk-free rate in the capital asset pricing model60. A stock with a beta of 2.3 means that for every 1 percent change in the market overall, the stock tends to change by 2.3 percent in the same direction.A) TrueB) FalseANSWER: A61. Stocks that have relatively little trading are normally subject to less price volatility.A) TrueB) FalseANSWER: B62. A firm’s stock price is affected not only by macroeconomic and market conditions but also by firm specific conditions.A) TrueB) FalseANSWER: A12 Stock Valuation and Risk63. Stock repurchases are commonly viewed as an unfavorable signal about the firm.A) TrueB) FalseANSWER: B64. The main source of uncertainty in computing the return of a stock is the dividend to be received next year.A) TrueB) FalseANSWER: B65. A stock portfolio has more volatility when its individual stock returns are uncorrelated.A) TrueB) FalseANSWER: B66. Beta serves as a measure of risk because it can be used to derive a probability distribution of return67. The value-at-risk method is intended to warn investors about the potential maximum loss that couldoccur.A) TrueB) FalseANSWER: A68. Regarding the value-at-risk method, the same methods used to derive the maximum expected loss ofone stock can be applied to derive the maximum expected loss of a stock portfolio for a givenconfidence level.A) TrueB) FalseANSWER: A69. Portfolio managers who monitor systematic risk rather than total risk are more concerned about stockvolatility than about beta.A) TrueB) FalseANSWER: BStock Valuation and Risk 13 70. Regarding the implied standard deviation, by plugging in the actual option premium paid by investorsfor a specific stock in the option-pricing model, it is possible to derive the anticipated volatility level.A) TrueB) FalseANSWER: A71. One way to forecast a portfolio’s beta is to first forecast the betas of the individual stocks in theportfolio and then sum the individual forecasted betas, weighted by the market value of each stock.A) TrueB) FalseANSWER: B72. If beta is thought to be the appropriate measure of risk, a stock’s risk-adjusted returns should bedetermined by the Sharpe index.ANSWER: B73. The Treynor index is similar to the Sharpe index, except that is uses beta rather than standarddeviation to measure the stock’s risk.A) TrueB) FalseANSWER: A74. Fabrizio, Inc. is expected to generate earnings of $1.50 per share this year. If the mean ratio of shareprice to expected earnings of competitors in the same industry is 20, then the stock price per share is $_________.A)13.33B) 3.00C)20.00D)30.00E)none of the aboveANSWER: D75. Which of the following is not a reason the PE ratio method may result in an inaccurate valuation for afirm?A)potential errors in the forecast of the firm’s betaB)potential errors in the forecast of the firm’s future earningsC)potential errors in the choice of the industry composite used to derive the PE ratioD)All of the above are reasons the PE ratio method may result in an inaccurate valuation for a firm.ANSWER: A14 Stock Valuation and Risk76. Sorvino Co. is expected to offer a dividend of $3.2 per share per year forever. The required rate ofreturn on Sorvino stock is 13 percent. Thus, the price of a share of Sorvino stock, according to the dividend discount model, is $_________.A) 4.06B) 4.16C)24.62D)40.63E)none of the aboveANSWER: Csubsequently. The required rate of return is 15 percent. The value of Kudrow stock, according to the dividend discount model, is $__________.A)39.67B)33.33C)31.73D)41.67E)none of the aboveANSWER: D78. The limitations of the dividend discount model are most pronounced for a firm thatA)has a high beta.B)has high expected future earnings.C)distributes most of its earnings as dividends.D)retains all of its earnings.E)none of the aboveANSWER: D79. Which of the following is incorrect regarding the capital asset pricing model (CAPM)?A)It is sometimes used to estimate the required rate of return for any firm with publicly traded stock.B)It is based on the premise that the only important risk of a firm is systematic risk.C)It is concerned with unsystematic risk.D)All of the above are true.ANSWER: C80. The _______________ is not a factor used in the capital asset pricing model (CAPM) to derive thereturn of an asset.A)prevailing risk-free rateB)dividend growth rateC)market returnD)covariance between the asset’s returns and market returnsE)All of the above are factors used in the CAPM.NSWER: BStock Valuation and Risk 15 81. Schwimmer Corp. has a beta of 1.5. The prevailing risk-free rate is 5 percent and the annual marketreturn in recent years has been 11 percent. Based on this information, the required rate of return on Schwimmer Corp. stockB) 6.5C)16.5D)14.0E)none of the aboveANSWER: D82. Which of the following is not a type of factor that drives stock prices, according to your text?A)economic factorsB)market-related factorsC)firm-specific factorsD)All of the above are factors that affect stock prices.ANSWER: D83. ______________ is (are) not a market-related factor(s) that affect(s) stock prices.A)Interest ratesB)Noise tradingC)TrendsD)January effectE)All of the above are market-related factors that affect stock prices.ANSWER: A84. _____________ is (are) not a firm-specific factor(s) that affect(s) stock prices.A)Exchange ratesB)Dividend policy changesC)Stock offerings and repurchasesD)Earnings surprisesE)All of the above are firm-specific factors that affect stock prices.ANSWER: A85. The ____________ is not a measure of a st ock’s risk.A)stock’s price volatilityB)stock’s returnC)stock’s betaD)value-at-risk methodE)All of the above are measures of a stock’s risk.ANSWER: B86. If the standard deviation of a stock’s returns over the last 12 quarters i s 4 percent, and if there is no16 Stock Valuation and RiskB)68; 8C)95; 8D)95; 6E)none of the above ANSWER: A。

英文：Engineering ToleranceIntroductionA solid is defined by its surface boundaries. Designers typically specify a component’s nominal dimensions such that it fulfils its requirements. In reality, components cannot be made repeatedly to nominal dimensions, due to surface irregularities and the intrinsic surface roughness. Some variability in dimensions must be allowed to ensure manufacture is possible. However, the variability permitted must not be so great that the performance of the assembled parts is impaired. The allowed variability on the individual component dimensions is called the tolerance.The term tolerance applies not only to the acceptable range of component dimensions produced by manufacturing techniques, but also to the output of machines or processes. For example , the power produced by a given type of internal combustion engine varies from one engine to another. In practice, the variability is usually found to be modeled by a frequency distribution curve, for example the normal distribution (also called the Gaussian distribution).One of the tasks of the designer is to specify a dimension on a component and the allowable variability on this value that will give acceptable performance.Component TolerancesControl of dimensions is necessary in order to ensure assembly and interchangeability of components. Tolerances are specified on critical dimensions that affect clearances and interferences fits. One method of specifying tolerances is to state the nominal dimension followed by the permissible variation, so a dimension could be stated as 40.000mm ± 0.003mm.This means that the dimension should be machined so that it is between 39.997mm and 40.003mm.Where thevariation can vary either side of the nominal dimension, the tolerance is called a bilateral tolerance. For a unilateral tolerance, one tolerance is zero, e.g. 40+0.006 .0.000Most organizations have general tolerances that apply to dimensions when an explicit dimension is not specified on a drawing. For machined dimensions a general tolerance may be ±0.5mm. So a dimension specified as 15.0mm may range between 14.5mm and 15.5mm. Other general tolerances can be applied to features such as angles, drilled and punched holes, castings，forgings, weld beads and fillets.When specifying a tolerance for a component, reference can be made to previous drawings or general engineering practice. Tolerances are typically specified in bands as defined in British or ISO standards.Standard Fits for Holes and ShaftsA standard engineering ask is to determine tolerances for a cylindrical component, e.g. a shaft, fitting or rotating inside a corresponding cylindrical component or hole. The tightness of fit will depend on the application. For example, a gear located onto a shaft would require a “tight” interference fit, where the diameter of the shaft is actually slightly greater than the inside diameter of the gear hub in order to be able to transmit the desired torque. Alternatively, the diameter of a journal bearing must be greater than the diameter of the shaft to allow rotation. Given that it is not economically possible to manufacture components to exact dimensions, some variability in sizes of both the shaft and hole dimension must be specified. However, the range of variability should not be so large that the operation of the assembly is impaired. Rather than having an infinite variety of tolerance dimensions that could be specified, national and international standards have been produced defining bands of tolerances. To turn this information into actual dimensions corresponding tables exist，defining the tolerance levels for the size of dimension under consideration.Size：a number expressing in a particular unit the numerical value of a dimension.Actual size：the size of a part as obtained by measurement.Limits of size：the maximum and minimum sizes permitted for a feature.Maximum limit of size the greater of the two limits of size.Minimum limit of size：the smaller of the two limits of size.Basic size：the size by reference to which the limits of size are fixed.Deviation：the algebraic difference between a size and the corresponding basic size.Actual deviation：the algebraic difference between the actual size and the corresponding basic size.Upper deviation：the algebraic difference between the maximum limit of size and the corresponding basic size.Lower deviation：the algebraic difference between the minimum limit of size and the corresponding basic size.Tolerance：the difference between the maximum limit of size and the minimum limit of size.Shaft：the term used by convention to designate all external features of a part.Hole：the term used by convention to designate all internal features of a part.Heat Treatment of MetalThe generally accepted definition for heat treating metals and metal alloys is “heating and cooling a solid metal or alloy in a way so as to obtain specific conditions and I or properties.”Heating for the sole purpose of hot working(as in forging operations) is excluded from this definition．Likewise，the types of heat treatment that are sometimes used for products such as glass or plastics are also excluded from coverage by this definition．Transformation CurvesThe basis for heat treatment is the time-temperature-transformation curves or TTT curves where，in a single diagram all the three parameters are plotted．Becauseof the shape of the curves，they are also sometimes called C-curves or S-curves．To plot TTT curves，the particular steel is held at a given temperature and the structure is examined at predetermined intervals to record the amount of transformation taken place．It is known that the eutectoid steel (T80) under equilibrium conditions contains，all austenite above 723℃，whereas below，it is pearlite．To form pearlite，the carbon atoms should diffuse to form cementite．The diffusion being a rate process，would require sufficient time for complete transformation of austenite to pearlite ．From different samples，it is possible to note the amount of the transformation taking place at any temperature．These points are then plotted on a graph with time and temperature as the axes．Classification of Heat Treating ProcessesIn some instances，heat treatment procedures are clear cut in terms of technique and application．whereas in other instances，descriptions or simple explanations are insufficient because the same technique frequently may be used to obtain different objectives ．For example, stress relieving and tempering are often accomplished with the same equipment and by use of identical time and temperature cycles．The objectives，however，are different for the two processes ．The following descriptions of the principal heat treating processes are generally arranged according to their interrelationships．Normalizing consists of heating a ferrous alloy to a suitable temperature (usually 50°F to 100 °F or 28 ℃to 56℃) above its specific upper transformation temperature. This is followed by cooling in still air to at least some temperature well below its transformation temperature range．For low-carbon steels．the resulting structure and properties are the same as those achieved by full annealing ；for most ferrous alloys, normalizing and annealing are not synonymous.Normalizing usually is used as a conditioning treatment, notably for refining the grain of steels that have been subjected to high temperatures for forging or other hot working operations．The normalizing process usually is succeededby another heat treating operation such as austenitizing for hardening, annealing，or tempering.Annealing is a generic term denoting a heat treatment that consists of heating to and holding at a suitable temperature followed by cooling at a suitable rate．It is used primarily to soften metallic materials，but also to simultaneously produce desired changes in other properties or in microstructure．The purpose of such changes may be，but is not confined to, improvement of machinability, facilitation of cold work ( known as in-process annealing)，improvement of mechanical or electrical properties, or to increase dimensional stability．When applied solely to relieve stresses, it commonly is called stress-relief annealing, synonymous with stress relieving．When the term “anneali ng is applied to ferrous alloys without qualification, full annealing is implied．This is achieved by heating above the alloy’s transformation temperature，then applying a cooling cycle which provides maximum softness．This cycle may vary widely, depending on composition and characteristics of the specific alloy．Quenching is the rapid cooling of a steel or alloy from the austenitizing temperature by immersing the workpiece in a liquid or gaseous medium．Quenching media commonly used include water，5% brine，5% caustic in an aqueous solution，oil，polymer solutions，or gas(usually air or nitrogen)．Selection of a quenching medium depends largely on the hardenability of the material and the mass of the material being treated(principally section thickness)．The cooling capabilities ofthe above-listed quenching media vary greatly．In selecting a quenching medium, it is best to avoid a solution that has more cooling power than is needed to achieve the results，thus minimizing the possibility of cracking and warp of the parts being treated．Modifications of the term quenching include direct quenching，fog quenching，hot quenching，interrupted quenching selective quenching，spray quenching, and time quenching．Tempering ．In heat treating of ferrous alloys ，tempering consists of reheating the austenitized and quench-hardened steel or iron to some preselected temperature that is below the lower transformation temperature (generally below 1300°F or 705℃) ．Tempering offers a means of obtaining various combinations of mechanical properties．Tempering temperatures used for hardened steels are often no higher than 300°F (150℃)．The term “tempering”should not be confused with either process annealing or stress relieving．Even though time and temperature cycles for the three processes may be the same，the conditions of the materials being processed and the objectives may be different．Stress Relieving．Like tempering, stress relieving is always done by heating to some temperature below the lower transformation temperature for steels and irons ．For nonferrous metals，the temperature may vary from slightly above room temperature to several hundred degrees，depending on the alloy and the amount of stress relief that is desired．The primary purpose of stress relieving is to relieve stresses that have been imparted to the workpiece from such processes as forming, rolling，machining or welding．The usual procedure is to heat workpieces to the pre-established temperature long enough to reduce the residual stresses (this is a time-and temperature-dependent operation) to an acceptable level；this is followed by cooling at a relatively slow rate to avoid creation of new stresses．Introduction to CAD/CAMThroughout the history of our industrial society, many inventions have been patented and whole new technologies have evolved. Perhaps the single development that has impacted manufacturing more quickly and significantly than any previous technology is the digital computer. Computers are being used increasingly for both design and detailing of engineering components in the drawing office.Computer-aided design (CAD) is defined as the application of computers and graphics software to aid or enhance the product design from conceptualizationto documentation. CAD is most commonly associated with the use of an interactive computer graphics system, referred to as a CAD system. Computer-aided design systems are powerful tools and are used in the mechanical design and geometric modeling of products and components.There are several good reasons for using a CAD system to support the engineering design function：⑴To increase the productivity⑵To improve the quality of the design⑶To uniform design standards⑷To create a manufacturing data base⑸To eliminate inaccuracies caused by hand-copying of drawingsand inconsistency between drawingsComputer-aided manufacturing (CAM) is defined as the effective use of computer technology in manufacturing planning and control. CAM is most closely associated with functions in manufacturing engineering, such as process and production planning, machining, scheduling, management, quality control, and numerical control (NC) part programming. Computer-aided design and computer-aided manufacturing are often combined into CAD/CAM systems.This combination allows the transfer of information from the design stage into the stage of planning for the manufacturing of a product, without the need to reenter the data on part geometry manually. The database developed during CAD is stored; then it is processed further, by CAM, into the necessary data and instructions for operating an controlling production machinery, material-handling equipment, and automated testing and inspection for product quality.Rationale for CAD/CAMThe rationale for CAD/CAM is similar to that used to justify any technology-based improvement in manufacturing. It grows out of a need to continually improve productivity, quality and competitiveness. There are also other reasons why a company might make a conversion from manual processes toCAD/CAM：⑴Increased productivity⑵Better quality⑶Better communication⑷Common database with manufacturing⑸Reduced prototype construction costs⑹Faster response to customersCAD/CAM HardwareThe hardware part of a CAD/CAM system consists of the following components：(1) one or more design workstations,(2) digital computer, (3) plotters, printers and other output devices, and (4) storage devices. In addition, the CAD/CAM system would have a communication interface to permit transmission of data to and from other computer systems, thus enabling some of the benefits of computer integration.The workstation is the interface between computer and user in the CAD system. The design of the CAD workstation and its available features have an important influence on the convenience, productivity, and quality of the user’s output. The workstation must include a graphics display terminal and a set of user input devices. CAD/CAM applications require a digital computer with a high-speed control processing unit (CPU). It contains the main memory and logic/arithmetic section for the system. The most widely used secondary storage medium in CAD/CAM is the hard disk, floppy diskette, or a combination of both.Input devices are generally used to transfer information from a human or storage medium to a computer where “CAD functions” are carried out. There are two basic approaches to input an existing drawing:model the object on a drawing or digitize the drawing. The standard output device for CAD/CAM is a CRT display. There are two major types of CRT displays: random-scan-line-drawing displays and raster-scan displays. In addition to CRT, there are also plasma paneldisplays and liquid-crystal displays.CAD/CAM SoftwareSoftware allows the human user to turn a hardware configuration into a powerful design and manufacturing system. CAD/CAM software falls into two broad categories, 2-D and 3-D, based on the number of dimensions visible in the finished geometry. CAD packages that represent objects in two dimensions are called 2-D software. Early systems were limited to 2-D. This was a serious shortcoming because 2-Drepresentations of 3-Dobjects is inherently confusing. Equally problem has been the inability of manufacturing personnel to properly read and interpret complicated 2-D representations of objects. 3-D software permits the parts to be viewed with the three-dimensional planes-height, width, and depth-visible. The trend in CAD/CAM is toward 3-D representation of graphic images. Such representations approximate the actual shape and appearance of the object to be produced; therefore, they are easier to read and understand.Applications of CAD/CAMThe emergence of CAD/CAM has had a major impact on manufacturing, by standardizing product development and by reducing design effort, tryout, and prototype work; it has made possible significantly reduced costs and improved productivity.Numerical ControlNumerical control (NC) is a form of programmable automation in which the processing equipment is controlled by means of numbers，letters，and other symbols．The numbers，letters，and symbols are coded in an appropriate format to define a program of instructions for a particular workpart or job. Theinstructions are provided by either of the two binary coded decimal systems: the Electronic Industries Association (EIA) code, or the American Standard Code for Information Interchange (ASCII). ASCII-coded machine control units will not accept EIA coded instructions and vice versa. Increasingly, however, control units are being made to accept instructions in either code. Automation operation by NC is readily adaptable to the operation of all metalworking machines. Lathes, milling machines, drill presses, boring machines, grinding machines, turret punches, flame or wire-cutting and welding machines, and even pipe benders are available with numerical controls.Basic Components of NCA numerical control system consists of the following three basic components：(1) Program instructions(2) Machine control unit(3) Processing equipmentThe program instructions are the detailed step by step commands that direct the processing equipment In its most common form，the commands refer to positions of a machine tool spindle with respect to the worktable on which the part is fixed．More advanced instructions include selection of spindle speeds，cutting tools，and other functions.The machine control unit (MCU) consists of the electronics and control hardware that reads and interprets the program of instructions and convert it into mechanical actions of the machine tool or other processing equipment ．The processing equipment is the component that performs metal process．In the most common example of numerical control ，it is used to perform machining operations. The processing equipment consists of the worktable and spindle as well as the motors and controls needed to drive them．Types of NCThere are two basic types of numerical control systems：point to point and contouring ．Point to point control system, also called positioning, are simpler than contouring control system．Its primary purpose is to move a tool or workpiece from one programmed point to another. Usually the machine function，such as a drilling operation，is also activated at each point by command from the NC Program．Point to point systems are suitable for hole machining operations such as drilling, countersinking，counterboring，reaming，boring and tapping. Hole punching machines，spotwelding machines，and assembly machines also use point to point NC systems．Contouring system，also known as the continuous path system，positioning and cutting operations are both along controlled paths but at different velocities．Because the tool cuts as it travels along a prescribed path ，accurate control and synchronization of velocities and movements are important．The contouring system is used on lathes，milling machines，grinders，welding machinery，and machining centers．Movement along the path，or interpolation, occurs incrementally，by one of several basic methods ．There are a number of interpolation schemes that have been developed to deal with the various problems that are encountered in generating a smooth continuous path with a contouring type NC system．They include linear interpolation, circular interpolation, helical interpolation, parabolic interpolation and cubic interpolation. In all interpolations，the path controlled is that of the center of rotation of the tool．Compensation for different tools，different diameter tools，or tools wear during machining，can be made in the NC program．Programming for NCA program for numerical control consists of a sequence of directions that causes an NC machine to carry out a certain operation ，machining being the most commonly used process ．Programming for NC may be done by aninternal programming department，on the shop floor，or purchased from an outside source．Also，programming may be done manually or with computer assistance．The program contains instructions and commands．Geometric instructions pertain to relative movements between the tool and the workpiece. Processing instructions pertain to spindle speeds，feeds，tools，and so on．Travel instructions pertain to the type of interpolation and slow or rapid movements of the tool or worktable．Switching commands pertain to on/off position for coolant supplies，spindle rotation，direction of spindle rotation tool changes，workpiece feeding，clamping，and so on. The first NC programming language was developed by MIT developmental work on NC programming systems in the late 1950s and called APT(Automatically Programmed Tools).DNC and CNCThe development of numerical control was a significant achievement in batch and job shop manufacturing，from both a technological and a commercial viewpoint．There have been two enhancements and extensions of NC technology，including：(1) Direct numerical control(2) Computer numerical controlDirect numerical control can be defined as a manufacturing system in which a number of machines are controlled by a computer through direct connection and in real time．The tape reader is omitted in DNC，thus relieving the system of its least reliable component．Instead of using the tape reader，the part program is transmitted to the machine tool directly from the computer memory．In principle，one computer can be used to control more than 100 separate machines．(One commercial DNC system during the l970s boasted a control capability of up to 256 machine tools.) The DNC computer is designed to provide instructions to each machine tool on demand ．When the machine needs control commands ，they are communicated to it immediately．Since the introduction of DNC ，there have been dramatic advances in computer technology．The physical size and cost of a digital computer has been significantly reduced at the same time that its computational capabilities have been substantially increased．In numerical control，the result of these advances has been that the large hard-wired MCUs of conventional NC have been replaced by control units based on the digital computer．Initially，minicomputers were utilized in the early 1970s ．As further miniaturization occurred in computers ，minicomputers were replaced by today’s microcomputers．Computer numerical control is an NC system using dedicated microcomputer as the machine control unit ．Because a digital computer is used in both CNC and DNC，it is appropriate to distinguish between the two types of system．There are three principal differences：(1) DNC computers distribute instructional data to，and collect data from， a large number of machines．CNC computers control only one machine，or a small number of machines．(2) DNC computers occupy a location that is typically remote from the machines under their control. CNC computer are located very near their machine tools.(3) DNC software is developed not only to control individual pieces of production equipment, but also to serve as part of a management information system in the manufacturing sector of the firm. CNC software is developed to augment the capabilities of a particular machine tool.中文翻译：工程公差引言固体由其表面边界确定界限。

CHAPTER 07CAPITAL ASSET PRICING AND ARBITRAGE PRICINGTHEORY1. The required rate of return on a stock is related to the required rate of return on thestock market via beta. Assuming the beta of Google remains constant, the increase in the risk of the market will increase the required rate of return on the market, and thus increase the required rate of return on Google.2. An example of this scenario would be an investment in the SMB and HML. As of yet,there are no vehicles (index funds or ETFs) to directly invest in SMB and HML. While they may prove superior to the single index model, they are not yet practical, even for professional investors.3. The APT may exist without the CAPM, but not the other way. Thus, statement a ispossible, but not b. The reason being, that the APT accepts the principle of risk and return, which is central to CAPM, without making any assumptions regardingindividual investors and their portfolios. These assumptions are necessary to CAPM.4. E(r P ) = r f + β[E(r M ) – r f ]20% = 5% + β(15% – 5%) ⇒ β = 15/10 = 1.55. If the beta of the security doubles, then so will its risk premium. The current riskpremium for the stock is: (13% - 7%) = 6%, so the new risk premium would be 12%, and the new discount rate for the security would be: 12% + 7% = 19%If the stock pays a constant dividend in perpetuity, then we know from the original data that the dividend (D) must satisfy the equation for a perpetuity:Price = Dividend/Discount rate 40 = D/0.13 ⇒ D = 40 ⨯ 0.13 = $5.20 At the new discount rate of 19%, the stock would be worth: $5.20/0.19 = $27.37The increase in stock risk has lowered the value of the stock by 31.58%.6. The cash flows for the project comprise a 10-year annuity of $10 million per year plus anadditional payment in the tenth year of $10 million (so that the total payment in the tenth year is $20 million). The appropriate discount rate for the project is:r f + β[E(r M ) – r f ] = 9% + 1.7(19% – 9%) = 26% Using this discount rate:NPV = –20 + +∑=101t t26.1101026.110= –20 + [10 ⨯ Annuity factor (26%, 10 years)] + [10 ⨯ PV factor (26%, 10 years)] = 15.64The internal rate of return on the project is 49.55%. The highest value that beta can take before the hurdle rate exceeds the IRR is determined by:49.55% = 9% + β(19% – 9%) ⇒ β = 40.55/10 = 4.055 7. a. False. β = 0 implies E(r) = r f , not zero.b. False. Investors require a risk premium for bearing systematic (i.e., market orundiversifiable) risk.c. False. You should invest 0.75 of your portfolio in the market portfolio, and theremainder in T-bills. Then: βP = (0.75 ⨯ 1) + (0.25 ⨯ 0) = 0.758.a. The beta is the sensitivity of the stock's return to the market return. Call theaggressive stock A and the defensive stock D . Then beta is the change in the stock return per unit change in the market return. We compute each stock's beta by calculating the difference in its return across the two scenarios divided by the difference in market return.00.2205322A =--=β70.0205145.3D =--=βb. With the two scenarios equal likely, the expected rate of return is an average ofthe two possible outcomes: E(r A ) = 0.5 ⨯ (2% + 32%) = 17%E(r B ) = 0.5 ⨯ (3.5% + 14%) = 8.75%c. The SML is determined by the following: T-bill rate = 8% with a beta equal tozero, beta for the market is 1.0, and the expected rate of return for the market is:0.5 ⨯ (20% + 5%) = 12.5%See the following graph.812.5%S M LThe equation for the security market line is: E(r) = 8% + β(12.5% – 8%) d. The aggressive stock has a fair expected rate of return of:E(r A ) = 8% + 2.0(12.5% – 8%) = 17%The security analyst’s estimate of the expected rate of return is also 17%.Thus the alpha for the aggressive stock is zero. Similarly, the required return for the defensive stock is:E(r D ) = 8% + 0.7(12.5% – 8%) = 11.15%The security analyst’s estimate of the expected return for D is only 8.75%, and hence:αD = actual expected return – required return predicted by CAPM= 8.75% – 11.15% = –2.4%The points for each stock are plotted on the graph above.e. The hurdle rate is determined by the project beta (i.e., 0.7), not by the firm’sbeta. The correct discount rate is therefore 11.15%, the fair rate of return on stock D.9. Not possible. Portfolio A has a higher beta than Portfolio B, but the expected returnfor Portfolio A is lower.10. Possible. If the CAPM is valid, the expected rate of return compensates only forsystematic (market) risk as measured by beta, rather than the standard deviation, which includes nonsystematic risk. Thus, Portfolio A's lower expected rate of return can be paired with a higher standard deviation, as long as Portfolio A's beta is lower than that of Portfolio B.11. Not possible. The reward-to-variability ratio for Portfolio A is better than that of themarket, which is not possible according to the CAPM, since the CAPM predicts that the market portfolio is the most efficient portfolio. Using the numbers supplied:S A =5.0121016=- S M =33.0241018=-These figures imply that Portfolio A provides a better risk-reward tradeoff than the market portfolio.12. Not possible. Portfolio A clearly dominates the market portfolio. It has a lowerstandard deviation with a higher expected return.13. Not possible. Given these data, the SML is: E(r) = 10% + β(18% – 10%)A portfolio with beta of 1.5 should have an expected return of: E(r) = 10% + 1.5 ⨯ (18% – 10%) = 22%The expected return for Portfolio A is 16% so that Portfolio A plots below the SML (i.e., has an alpha of –6%), and hence is an overpriced portfolio. This is inconsistent with the CAPM.14. Not possible. The SML is the same as in Problem 12. Here, the required expectedreturn for Portfolio A is: 10% + (0.9 ⨯ 8%) = 17.2%This is still higher than 16%. Portfolio A is overpriced, with alpha equal to: –1.2%15. Possible. Portfolio A's ratio of risk premium to standard deviation is less attractivethan the market's. This situation is consistent with the CAPM. The market portfolio should provide the highest reward-to-variability ratio.16.a.b.As a first pass we note that large standard deviation of the beta estimates. None of the subperiod estimates deviate from the overall period estimate by more than two standard deviations. That is, the t-statistic of the deviation from the overall period is not significant for any of the subperiod beta estimates. Looking beyond the aforementioned observation, the differences can be attributed to different alpha values during the subperiods. The case of Toyota is most revealing: The alpha estimate for the first two years is positive and for the last two years negative (both large). Following a good performance in the "normal" years prior to the crisis, Toyota surprised investors with a negative performance, beyond what could be expected from the index. This suggests that a beta of around 0.5 is more reliable. The shift of the intercepts from positive to negative when the index moved to largely negative returns, explains why the line is steeper when estimated for the overall period. Draw a line in the positive quadrant for the index with a slope of 0.5 and positive intercept. Then draw a line with similar slope in the negative quadrant of the index with a negative intercept. You can see that a line that reconciles the observations for both quadrants will be steeper. The same logic explains part of the behavior of subperiod betas for Ford and GM.17. Since the stock's beta is equal to 1.0, its expected rate of return should be equal to thatof the market, that is, 18%. E(r) =01P P P D -+0.18 =100100P 91-+⇒ P 1 = $10918. If beta is zero, the cash flow should be discounted at the risk-free rate, 8%:PV = $1,000/0.08 = $12,500If, however, beta is actually equal to 1, the investment should yield 18%, and the price paid for the firm should be:PV = $1,000/0.18 = $5,555.56The difference ($6944.44) is the amount you will overpay if you erroneously assume that beta is zero rather than 1.ing the SML: 6% = 8% + β(18% – 8%) ⇒β = –2/10 = –0.220.r1 = 19%; r2 = 16%; β1 = 1.5; β2 = 1.0a.In order to determine which investor was a better selector of individual stockswe look at the abnormal return, which is the ex-post alpha; that is, the abnormalreturn is the difference between the actual return and that predicted by the SML.Without information about the parameters of this equation (i.e., the risk-free rateand the market rate of return) we cannot determine which investment adviser isthe better selector of individual stocks.b.If r f = 6% and r M = 14%, then (using alpha for the abnormal return):α1 = 19% – [6% + 1.5(14% – 6%)] = 19% – 18% = 1%α2 = 16% – [6% + 1.0(14% – 6%)] = 16% – 14% = 2%Here, the second investment adviser has the larger abnormal return and thusappears to be the better selector of individual stocks. By making betterpredictions, the second adviser appears to have tilted his portfolio toward under-priced stocks.c.If r f = 3% and r M = 15%, then:α1 =19% – [3% + 1.5(15% – 3%)] = 19% – 21% = –2%α2 = 16% – [3%+ 1.0(15% – 3%)] = 16% – 15% = 1%Here, not only does the second investment adviser appear to be a better stockselector, but the first adviser's selections appear valueless (or worse).21.a.Since the market portfolio, by definition, has a beta of 1.0, its expected rate ofreturn is 12%.b.β = 0 means the stock has no systematic risk. Hence, the portfolio's expectedrate of return is the risk-free rate, 4%.ing the SML, the fair rate of return for a stock with β= –0.5 is:E(r) = 4% + (–0.5)(12% – 4%) = 0.0%The expected rate of return, using the expected price and dividend for next year: E(r) = ($44/$40) – 1 = 0.10 = 10%Because the expected return exceeds the fair return, the stock must be under-priced.22.The data can be summarized as follows:ing the SML, the expected rate of return for any portfolio P is:E(r P) = r f + β[E(r M) – r f ]Substituting for portfolios A and B:E(r A) = 6% + 0.8 ⨯ (12% – 6%) = 10.8%E(r B) = 6% + 1.5 ⨯ (12% – 6%) = 15.0%Hence, Portfolio A is desirable and Portfolio B is not.b.The slope of the CAL supported by a portfolio P is given by:S =P fP σr)E(r-Computing this slope for each of the three alternative portfolios, we have:S (S&P 500) = 6/20S (A) = 5/10S (B) = 8/31Hence, portfolio A would be a good substitute for the S&P 500.23.Since the beta for Portfolio F is zero, the expected return for Portfolio F equals therisk-free rate.For Portfolio A, the ratio of risk premium to beta is: (10% - 4%)/1 = 6%The ratio for Portfolio E is higher: (9% - 4%)/(2/3) = 7.5%This implies that an arbitrage opportunity exists. For instance, you can create aPortfolio G with beta equal to 1.0 (the same as the beta for Portfolio A) by taking a long position in Portfolio E and a short position in Portfolio F (that is, borrowing at the risk-free rate and investing the proceeds in Portfolio E). For the beta of G to equal 1.0, theproportion (w) of funds invested in E must be: 3/2 = 1.5The expected return of G is then:E(r G) = [(-0.50) ⨯ 4%] + (1.5 ⨯ 9%) = 11.5%βG = 1.5 ⨯ (2/3) = 1.0Comparing Portfolio G to Portfolio A, G has the same beta and a higher expected return.Now, consider Portfolio H, which is a short position in Portfolio A with the proceedsinvested in Portfolio G:βH = 1βG + (-1)βA = (1 ⨯ 1) + [(-1) ⨯ 1] = 0E(r H) = (1 ⨯ r G) + [(-1) ⨯ r A] = (1 ⨯ 11.5%) + [(- 1) ⨯ 10%] = 1.5%The result is a zero investment portfolio (all proceeds from the short sale of Portfolio Aare invested in Portfolio G) with zero risk (because β = 0 and the portfolios are welldiversified), and a positive return of 1.5%. Portfolio H is an arbitrage portfolio.24.Substituting the portfolio returns and betas in the expected return-beta relationship, weobtain two equations in the unknowns, the risk-free rate (r f ) and the factor return (F):14.0% = r f + 1 ⨯ (F – r f )14.8% = r f + 1.1 ⨯ (F – r f )From the first equation we find that F = 14%. Substituting this value for F into the second equation, we get:14.8% = r f + 1.1 ⨯ (14% – r f ) ⇒ r f = 6%25.a.Shorting equal amounts of the 10 negative-alpha stocks and investing the proceedsequally in the 10 positive-alpha stocks eliminates the market exposure and creates azero-investment portfolio. Using equation 7.5, and denoting the market factor as R M,the expected dollar return is [noting that the expectation of residual risk (e) inequation 7.8 is zero]:$1,000,000 ⨯ [0.03 + (1.0 ⨯ R M)] – $1,000,000 ⨯ [(–0.03) + (1.0 ⨯ R M)]= $1,000,000 ⨯ 0.06 = $60,000The sensitivity of the payoff of this portfolio to the market factor is zero because theexposures of the positive alpha and negative alpha stocks cancel out. (Notice thatthe terms involving R M sum to zero.) Thus, the systematic component of total riskalso is zero. The variance of the analyst's profit is not zero, however, since thisportfolio is not well diversified.For n = 20 stocks (i.e., long 10 stocks and short 10 stocks) the investor will have a$100,000 position (either long or short) in each stock. Net market exposure is zero,but firm-specific risk has not been fully diversified. The variance of dollar returnsfrom the positions in the 20 firms is:20 ⨯ [(100,000 ⨯ 0.30)2] = 18,000,000,000The standard deviation of dollar returns is $134,164.b.If n = 50 stocks (i.e., 25 long and 25 short), $40,000 is placed in each position,and the variance of dollar returns is:50 ⨯ [(40,000 ⨯ 0.30)2] = 7,200,000,000The standard deviation of dollar returns is $84,853.Similarly, if n = 100 stocks (i.e., 50 long and 50 short), $20,000 is placed ineach position, and the variance of dollar returns is:100 ⨯ [(20,000 ⨯ 0.30)2] = 3,600,000,000The standard deviation of dollar returns is $60,000.Notice that when the number of stocks increases by a factor of 5 (from 20 to 100),standard deviation falls by a factor of 5= 2.236, from $134,164 to $60,000. 26.Any pattern of returns can be "explained" if we are free to choose an indefinitely largenumber of explanatory factors. If a theory of asset pricing is to have value, it mustexplain returns using a reasonably limited number of explanatory variables (i.e.,systematic factors).27.The APT factors must correlate with major sources of uncertainty, i.e., sources ofuncertainty that are of concern to many investors. Researchers should investigatefactors that correlate with uncertainty in consumption and investment opportunities.GDP, the inflation rate and interest rates are among the factors that can be expected to determine risk premiums. In particular, industrial production (IP) is a good indicator of changes in the business cycle. Thus, IP is a candidate for a factor that is highlycorrelated with uncertainties related to investment and consumption opportunities in the economy.28.The revised estimate of the expected rate of return of the stock would be the oldestimate plus the sum of the unexpected changes in the factors times the sensitivitycoefficients, as follows:Revised estimate = 14% + [(1 ⨯ 1) + (0.4 ⨯ 1)] = 15.4%29.Equation 7.11 applies here:E(r P) = r f + βP1[E(r1) - r f] + βP2[E(r2) – r f]We need to find the risk premium for these two factors:γ1 = [E(r1) - r f] andγ2 = [E(r2) - r f]To find these values, we solve the following two equations with two unknowns: 40% = 7% + 1.8γ1 + 2.1γ210% = 7% + 2.0γ1 + (-0.5)γ2The solutions are: γ1 = 4.47% and γ2 = 11.86%Thus, the expected return-beta relationship is:E(r P) = 7% + 4.47βP1 + 11.86βP230.The first two factors (the return on a broad-based index and the level of interest rates)are most promising with respect to the likely impa ct on Jennifer’s firm’s cost of capital.These are both macro factors (as opposed to firm-specific factors) that can not bediversified away; consequently, we would expect that there is a risk premiumassociated with these factors. On the other hand, the risk of changes in the price ofhogs, while important to some firms and industries, is likely to be diversifiable, andtherefore is not a promising factor in terms of its impact on the firm’s cost of capital.31.Since the risk free rate is not given, we assume a risk free rate of 0%. The APT required(i.e., equilibrium) rate of return on the stock based on Rf and the factor betas is:Required E(r) = 0 + (1 x 6) + (0.5 x 2) + (0.75 x 4) = 10%According to the equation for the return on the stock, the actually expected return onthe stock is 6 % (because the expected surprises on all factors are zero by definition).Because the actually expected return based on risk is less than the equilibrium return,we conclude that the stock is overpriced.CFA 1a, c and dCFA 2a.E(r X) = 5% + 0.8(14% – 5%) = 12.2%αX = 14% – 12.2% = 1.8%E(r Y) = 5% + 1.5(14% – 5%) = 18.5%αY = 17% – 18.5% = –1.5%b.(i)For an investor who wants to add this stock to a well-diversified equityportfolio, Kay should recommend Stock X because of its positivealpha, while Stock Y has a negative alpha. In graphical terms, StockX’s expected return/risk profile plots above the SML, while Stock Y’sprofile plots below the SML. Also, depending on the individual riskpreferences of Kay’s clients, Stock X’s lower beta may have abeneficial impact on overall portfolio risk.(ii)For an investor who wants to hold this stock as a single-stock portfolio,Kay should recommend Stock Y, because it has higher forecastedreturn and lower standard deviation than S tock X. Stock Y’s Sharperatio is:(0.17 – 0.05)/0.25 = 0.48Stock X’s Sharpe ratio is only:(0.14 – 0.05)/0.36 = 0.25The market index has an even more attractive Sharpe ratio:(0.14 – 0.05)/0.15 = 0.60However, given the choice between Stock X and Y, Y is superior.When a stock is held in isolation, standard deviation is the relevantrisk measure. For assets held in isolation, beta as a measure of risk isirrelevant. Although holding a single asset in isolation is not typicallya recommended investment strategy, some investors may hold what isessentially a single-asset portfolio (e.g., the stock of their employercompany). For such investors, the relevance of standard deviationversus beta is an important issue.CFA 3a.McKay should borrow funds and i nvest those funds proportionally in Murray’sexisting portfolio (i.e., buy more risky assets on margin). In addition toincreased expected return, the alternative portfolio on the capital market line(CML) will also have increased variability (risk), which is caused by the higherproportion of risky assets in the total portfolio.b.McKay should substitute low beta stocks for high beta stocks in order to reducethe overall beta of York’s portfolio. By reducing the overall portfolio beta,McKay will reduce the systematic risk of the portfolio and therefore theportfolio’s volatility relative to the market. The security market line (SML)suggests such action (moving down the SML), even though reducing beta mayresult in a slight loss of portfolio efficiency unless full diversification ismaintained. York’s primary objective, however, is not to maintain efficiencybut to reduce risk exposure; reducing portfolio beta meets that objective.Because York does not permit borrowing or lending, McKay cannot reduce riskby selling equities and using the proceeds to buy risk free assets (i.e., by lendingpart of the portfolio).CFA 4c.“Both the CAPM and APT require a mean-variance efficient market portfolio.”This statement is incorrect. The CAPM requires the mean-variance efficientportfolio, but APT does not.d.“The CAPM assumes that one specific factor explains security returns but APTdoes not.” This statement is c orrect.CFA 5aCFA 6dCFA 7d You need to know the risk-free rate.CFA 8d You need to know the risk-free rate.CFA 9Under the CAPM, the only risk that investors are compensated for bearing is the riskthat cannot be diversified away (i.e., systematic risk). Because systematic risk(measured by beta) is equal to 1.0 for each of the two portfolios, an investor wouldexpect the same rate of return from each portfolio. Moreover, since both portfolios are well diversified, it does not matter whether the specific risk of the individual securities is high or low. The firm-specific risk has been diversified away from both portfolios. CFA 10b r f = 8% and E(r M) = 16%E(r X) = r f + βX[E(r M) – r f] = 8% + 1.0(16% - 8%) = 16%E(r Y) = r f + βY[E(r M) – r f] = 8% + 0.25(16% - 8%) = 10%Therefore, there is an arbitrage opportunity.CFA 11cCFA 12dCFA 13cInvestors will take on as large a position as possible only if the mis-pricingopportunity is an arbitrage. Otherwise, considerations of risk anddiversification will limit the position they attempt to take in the mis-pricedsecurity.CFA 14d。

1Calculation of Areas Under the Normal CurveWe want to calculate the probability that a random event occurs, given that the probability distribution function is Normal with mean μ and standard deviation σ. We may numerically calculate the probabilityWe can do this as follows: The probability of exceeding a certain value x in a normal curve is denoted by Q(x). We, first, normalize the values of x by putting:σμ-←x xThen, the area under the normal curve to the right of x is:⎪⎩⎪⎨⎧<---=>=0)()(102/10)()()(x if x N x Z x if x if x N x Z x Qwhere)2/exp(21)(2x x Z -=π()pxt t t t t t b b b b b x N +=⋅⋅⋅⋅++++=11]})({[)(54321andp =+0.23164192b1=+0.319381530 b2=-0.356563782 b3=+1.781477937 b4=-1.821255978 b5=+1.330274429The probability that a given x is between to values a and b is given, simply,by )()()(b Q a Q b x a p -=≤≤where a < b.I. The Calculation of the Probabilities for a Normal Distribution0) The following are the constants for calculating the area under the Normal curve:p =+0.2316419 b1=+0.319381530 b2=-0.356563782 b3=+1.781477937 b4=-1.821255978 b5=+1.330274429 π2=k1) Normalize: a) offset = μ b) scale = σ2) Normalize the dataNb=(b-μ)/σ Na=(a-μ)/σNalpha=(alpha-μ)/ σ3) Calculate the probability of exceeding "b"fQ=Q(Nb)4) Calculate the probability of not reaching "b”P=1-fQ5) Calculate the probability of exceeding "a" and "b"Qa=Q(Na) Qb=Q(Nb)6) Calculate the probability of being between "a" and "b"a_b=Qa-Qb7) To calculate the prob. of a "point", define a very narrow band (0.1 sd's)alpha1=Nalpha-.05alpha2=Nalpha+.05Q1=Q(alpha1)Q2=Q(alpha2)P_alpha=Q1-Q2Function Q calculates the area under the normal curve by determining whether the sought for area is to the right or left of the mean value.a) If to the right, area is calculated from a direct numerical approximationb) If exactly on the middle p=1/2c) If to the left, calculate the probability from the symmetry of the curveIt uses two other functions:i) Z()ii) N()function Q(arg)if arg>0return Z(arg)*N(arg)endifif arg=0return 0.5endifif arg<03return 1-Z(-arg)*N(-arg)endiffunction Z(arg)return k*e(-arg*arg/2)function N(arg)t=1/(1+p*arg)return *(b1+t*(b2+t*(b3+t*(b4+t*b5))))45II. The Generation of Normally Distributed NumbersAn approximation to normally distributed random numbers y may be found from asequence of uniform random numbers [System/360 Scientific Subroutine Package] using the formula:12/2/1k k x y ki i ∑=-=wherea) x i is a uniformly distributed random numberb) 10≤<i xy approaches true normality asymptotically as k approaches infinity. To reduce execution time, make k=12.Then:66121-⎪⎪⎭⎫ ⎝⎛=∑=i i x yFor a given set μ and σ:μσ+=y y '7III. The 2χ Goodness_of_Fit TestThe theoretical distribution may be approximated using Chebyshev’s polynomials of degree 4. Then:443322102ννννχc c c c c ++++≈For 0.005 confidence level, the coefficients (c 0, ..., c 4) are:For 0.01 confidence level, the coefficients are:For 0.05 confidence level, the coefficients are:If the 2χ statistic, calculated from ∑=-=ri ii i E E O 122)(χ exceeds the theoretical 2χ, calculated from the polynomials, then the probability that the distribution is NOT normal equals the stipulated confidence level. In other words:8l c c c c c E E O Z P ri ii i =⎪⎪⎭⎫ ⎝⎛++++>-⌝∑=4433221012)(|νννν where P(Z) is the probability that the distribution is gaussian, r is the number of intervals and l is the confidence level.9Sampling Distribution of MeansStatistical MethodologyWe want to answer the following. Q : For any given algorithm A(i), what is the probability that we find a certain minimum value (denoted by κ) for any )(x γ given that A(i) is iterated G times?Since one of our premises is that the )(x γ be selected randomly from Ξ we do not know, a priori, anything about the probability distribution function of the κ’s. To answer Q we rely on the following known theorems from statistical theory.T1) Any sampling distribution of means (sdom) is distributed normally for a large enough sample size n .Remark: This is true, theoretically, as ∞→n . However, it is considered that any n >20 is satisfactory. We have chosen n =36.T2) In a normal distribution (with mean X μ and standard deviation X σ) approximately 1/10 of the observations lie in the intervals: X μ–5X σ to X μ-1.29X σ; X μ-1.29X σ to X μ–0.85X σ; X μ-0.85X σ to X μ–0.53X σ; X μ-0.53X σ to X μ–0.26X σ; X μ-0.26X σ to X μ and the symmetrical X μ to 0.26X σ, etc.Remark: These deciles divide, therefore, the area under the normal curve in 10 unequally spaced intervals. The expected number of observed events in each interval will, however, be equal.T3) The relation between the population distribution’s parameters [which we denote with μ (the mean) and σ(the standard deviation)] and the sdom’s parameters (which we denote with X μ and X σ) is given by X μμ= and X n σσ⋅=. Remark: In our case X σσ6=.T4) The proportion of any distribution found within k standard deviations of the mean is, at least, 1-1/k 2.Remark: Chebyshev’s bound generality makes it quite a loose one. Tighter bounds are achievable but they may depend on the characteristics of the distribution under study. We selected k = 4, which guarantees that our observations will occur with probability = 0.9375.T5) For a set of r intervals, a number of O i observed events in the i -th interval, a number of expected E i events in the i -th interval, p distribution parameters and 1--=p r ν degrees of freedom, the following equation holds.05.0)(4433221012=⎪⎪⎭⎫ ⎝⎛++++>-∑=ννννc c c c c E E O Z P ri ii i 98829512.10+≈c06290867.21+≈c 06021040.02-≈c00205163.03+≈c00002637.04-≈c(1)where ≡)(Z P probability that the distribution is normal.10Remarks: The summation on the left of (4) is the 2χ statistic; the polynomial to the right of the inequality sign (call it )(νT ) is a least squares Chebyshev polynomial approximation to the theoretical 2χ for a 95% confidence level. In our case, 7=ν for which ≈)(νT 14.0671. Furthermore, if we choose the deciles as above, we know that i E i ∀=10/η, where η is a sample of size n . A further condition normally imposed on this goodness-of-fit test is that a minimum number of observations θ (usually between 3 and 5) be required in each interval. Thus, (4) is replaced by()05.0)(4433221012=⎥⎥⎦⎤⎢⎢⎣⎡∀<∨⎪⎪⎭⎫ ⎝⎛++++>-∑=i O c c c c c E E O Z P i r i i i i θνννν(2) Making 5=θ and using the parameters’ values described above, equation (5) finally takes the following form.()95.05&0671.1410/)10/()5/(21011012=⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡≥⎪⎪⎪⎪⎪⎭⎫ ⎝⎛≤+-∑∑==ii i i i O O O Z P ηηη (3)We assume that we are exploring a set of algorithms. These are to be characterized by a set of parameters (from 1 to n ) and a method, which is, actually, a meta-heuristic (from 1 to m ).Algorithm for the Determination of the Distribution’s Parameters (of a meta -heuristic) 1. 1α← (determine the parameter set) 2. m ethod)the (determ ine ;1β←3. 1←i (count the number of samples)4. 1←j (count the elements of a sample)5. A function is selected randomly from the suite.6. Experiment βαE is performed with this function and a) the best value and b) the number of satisfied constraints are stored.7. 1+←j j8. If j ≤36, go to step 5 (a sample size of 36 guarantees normality).9. The average ∑=j)x (j f N 1i x of the best fitness’ values is calculated.10. 1+←i i11. If i ≤ 50, go to step 412. According to the central limit theorem, the i x distribute normally. We, therefore, define 10 intervals which are expected to hold 1/10 of the samples assuming a normal distribution: i.e., the intervals are standardized. If the samples are indeed normally distributed the following 2 conditions should hold.a) At least 5 observations should be found in each of the 10 intervals (which explains why we test for 50 in step 11).11b) The values of a 2χ goodness of fit test should be complied with (which we demand to be in the 95% confidence level).We, therefore, check for conditions (a) and (b) above. If they have not been reached, go to step 4.13. Once we are assured (with probability = 0.95) that the i x ’s are normally distributed, we calculate the mean X μ and standard deviation X σ of the sampling distribution of the measured mean values of the best fitnesses for this experiment. Moreover, we may calculate the mean μand the standard deviation σ of the distribution of the best values (rather than the means) from Xσ6σ;X μμ==. Notice that, therefore, we characterize the statistical behavior of experiment αβE quantitatively .14. 1ββ+←. If β< m , go to step 3.15. 1αα+←. If α< n , go to step 2.Program for the Sampling Distribution of MeansWe start by assuming the following data:Next, we define samples of size 3. For example: (3,3,3); (3,6,3); (4,6,10), etc. There are 1000 possible samples of this size.121. Directly from the table we calculate the mean and standard deviation, which turn out to be4.500 and 0.793 respectively. These are the parameters (μ) and (σ) of the population.2. Then we calculate the sampling distribution of means by generating all possible samples[(3,3,3);(3,3,4);(3,3,5);...;(10,9,10);(10,10,10)], the average (or mean) for every sample (3, 3.333,3.667,...,9.667, 10) and the resulting mean and standard deviation of the averages (means). This yields: x μ=4.500 and x σ = 0.458. From the CLT we know that x σ3σ⋅=; and that σ0.4581.732⋅≈0.793≈.3. Now we sample the population as per the algorithm above, to get 0.418x μ=, 0.442x σ= and 0.765σ=.In the figure above we show the intervals under the Normal curve, the number of observations in each interval and the percentage of the total number (81) for which 2χ= 3.814.4. Finally, we calculate the estimators ∑==N 1i i x N 1x and ∑=--=N 1i 2)x i (x 1N 1s , getting 3.666X = and 0.763s =.13。

测绘工程专业英语翻译各位读友大家好，此文档由网络收集而来，欢迎您下载，谢谢篇一：测绘工程专业英语课文翻译Unit 9 Basic Statistical Analysis of Random Errors （随机误差的统计学基本分析）Random errors are those variables that remain after mistakes are detected and eliminated and all systematic errors have been removed or corrected from the measured 后，并且所有系统误差被从测量值中移除或修正后，保留下的那些变量variable变量、变化n.）They are beyond the control of the the random errors are errors the occurrence of which does not follow a deterministic pattern.确定性的模式pattern而发生的误差）In mathematical statistics, they areconsidered as stochastic variables, and despite their irregular behavior, the study of random errors in any well-conducted measuring process or experiment has indicated that random errors follow the following empirical rules:mathematical statistics中，它们被当成随机变量stochastic variable，尽管它们的行为无规律，在任一正确的well-conducted原意为品行端正的，这里指测量实验和活动是无误的测量活动和实验中，对的随机误差的研究显示indicate随机误差遵循以下经验法则empirical⑴A random error will not exceed a certain amount.（随即误差不会超过一个确定的值）⑵Positive and negative random errors may occur at the same frequency.（正负误差出现的频率相同）⑶Errors that are small in magnitude are more likely to occur than those that are larger in magnitude.比数值大的误差出现可能性大be likely to 可能）⑷The mean of random errors tends to zero as the sample size tends to infinite.随机误差的平均值趋近于0）In mathematical statistics, random errors follow statistical behavioral laws such as the laws of 行为behavioral行为的规律，如概率法则）A characteristic theoretical pattern of error distribution occurs upon analysis of a large number of repeated measurements of a quantity, which conform to normal or Gaussian distribution.观测分析analysisn.中的误差分布的一个特征理论模式，遵照conform to遵照正态或高斯分布）在对一个量进行大量重复观测分析后，得到一个误差分布的理论特征——正态或高斯分布The plot of error sizes versus probabilities would approach a smooth curve of the characteristic bell-shape.与……相对概率的关系图，接近一条光滑的特有的characteristic特有的钟形曲线。