FIRST CROSSING PROBABILITY OF TWO RANDOM PROCESSES

【转】⼏种基本的数值采样⽅法（Sampling method）原⽂戳这⾥：/2011/10/10384.html在数值计算中，采样⽅法(sampling method)就是⼀种⽣成⼀系列符合某个特定概率分布(例如⾼斯分布)的随机数的算法。

最简单的采样⽅法就是⽣成符合均匀分布的采样，这在随机数是怎么⽣成的⾥⾯已经提到过，通常其他的⾮均匀分布都是通过⼀个或⼏个均匀分布通过转换⽽得来。

采样⽅法最早是在美国50年代的曼哈顿计划中的蒙特卡洛模拟⽅法 (Monte-Carlo simulations )中发明的，经过近年来的研究已经有很多⾼效的算法，⽐如马尔科夫链蒙特卡洛采样⽅法(Markov Chain Monto Carlo)，吉布斯采样(Gibbs sampling)，切⽚采样(Slice Sampling)等等，其中最基本的采样⽅法包括：转换采样⽅法( transform sampling)如果我们想得到⼀个符合累计分布函数F的变量X，⾸先通过均匀分布得到⼀个[0,1]之间的变量U，然后通过变换，这样得到的X就是符合F的分布。

指数分布和科西分布都可以通过这种逆变换的⽅式得到，这种⽅法虽然简单，但是逆函数不是都存在的，所以它的应⽤范围有了很⼤的限制。

拒绝性采样Rejection Sampling当我们需要的分布f(x)不是简单的基本分布时，从中采样就变得很困难，这时候可以采取拒绝性采样的⽅法。

选择⼀个简单的容易采样的分布g(x)和⼀个常数M使得。

函数⼜被称为对⽐函数。

拒绝性采样的算法：1) 从均匀分布U(0,1)中⽣成⼀个数u2) 从分布g(x)中⽣成⼀个数x3) 判断是否符合，如果是，接受x；如果不是，重复第⼀步。

Matlab 5 中使⽤的⽣成⾼斯分布的算法，极坐标形式的Box–Muller算法就是⼀种拒绝是采样算法。

Box-Muller算法是通过[0,1]之间的均匀分布和单位圆来⽣成⾼斯分布的⼀种算法。

Physics8.421Center for Ultracold Atoms Massachusetts Institute of TechnologySpring2006Lecture NotesFebruary13,2006ContentsPreface iii 1The Two-State System:Resonance11.1Introduction (1)1.2Resonance Studies and Q.E.D (2)1.2.1The language of resonance:a classical damped system (3)1.3Magnetic Resonance:Classical Spin in Time-varying B-Field (5)1.3.1The classical motion of spins in a static magneticfield (5)1.3.2Rotating coordinate transformation (6)1.3.3Larmor’s theorem (6)1.4Motion in a Rotating Magnetic Field (7)1.4.1Exact resonance (7)1.4.2Off-resonance behavior (8)1.5Adiabatic Rapid Passage:Landau-Zener Crossing (9)1.5.1Rotating frame argument (10)1.5.2Quantum treatment-Landau Zener (11)1.6Resonance Of Quantized Spin (12)1.6.1Expected projection of quantized spin (12)1.6.2Resonance of quantized spin12 (13)1.6.2.1The Rabi transition probability (13)1.6.2.2The Hamiltonian of a quantized spin12 (14)1.7Quantum Mechanical Solution for Resonance in a Two-State System (15)1.7.1Interaction representation (15)1.7.2Two-state problem (16)1.8Density Matrix (18)1.8.1General results (18)1.8.2Density matrix for two level system (20)1.8.3Phenomological treatment of relaxation:Bloch equations (21)1.8.4Introduction:Electrons,Protons,and Nuclei (23)iii CONTENTSPrefaceThe original incarnation of these notes was developed to accompany the lectures in the MIT graduate courses in atomic physics.AMO I was created in the late1960s as a one-term in-troductory course to prepare graduate students for research in atomic physics in the Physics Department.Over the years Dan Kleppner and David Pritchard changed the contents of the course to reflect new directions of research,though the basic concepts remained as a constant thread.With the growth of interest in atom cooling and quantum gases,a second one-term course,AMO II,was designed by me a few years ago and presented with AMO I in alternating years.AMO I is still taught in the traditional way.These lecture notes were created by”culling the best of Dan and Dave”and putting them into Latex form.As part of the Joint Harvard/MIT Center for Ultracold Atoms summer school in Atomic Physics, John Doyle got involved and improved the notes.However,they are still work in progress.Wolfgang Ketterleiiiiv CONTENTSChapter1The Two-State System:Resonance1.1IntroductionThe cornerstone of major areas of contemporary Atomic,Molecular and Optical Physics (AMO Physics)is the study of atomic and molecular systemsthrough their resonant in-teraction with applied oscillating electromagneticfields.The thrust of these studies has evolved continuously since Rabi performed thefirst resonance experiments in1938.In the decade following World War II the edifice of quantum electrodynamics was constructed largely in response to resonance measurements of unprecedented accuracy on the properties of the electron and thefine and hyperfine structure of simple atoms.At the same time,nu-clear magnetic resonance and electron paramagnetic resonance were developed and quickly became essential research tools for chemists and solid state physicists.Molecular beam magnetic and electric resonance studies yielded a wealth of information on the properties of nuclei and molecules,and provided invaluable data for the nuclear physicist and physical chemist.With the advent of lasers and laser spectroscopy these studies evolved into the creation of new species,such as Rydberg atoms,to studies of matter in ultra intensefields, to fundamental studies in the symmetries of physics,to new types of metrology,and to the manipulation of matter with laser light,notably the creation of atomic quantumfluids.Resonance techniques may be used not only to learn about the structure of a system, but also to prepare it in a precisely controlled way.Because of these two facets,resonance studies have lead physicists through a fundamental change in attitude-from the passive study of atoms to the active control of their internal quantum state and their interactions with the radiationfield.The chief technical legacy of the early work on resonance spectroscopy is the family of lasers which have sprung up like the brooms of the sorcerer’s apprentice.The scientific applications of these devices have been prodigious.They caused the resurrection of physical optics-now freshly christened quantum optics-and turned it into one of the liveliestfields in physics.They have had a similar impact on atomic and molecular spectroscopy.In ad-dition they have led to new families of physical studies such as single particle spectroscopy, multiphoton excitation,cavity quantum electrodynamics,and laser cooling and trapping.This chapter is about the interactions of a two-state system with a sinusoidally oscil-latingfield whose frequency is close to the natural resonance frequency of the system.The term“two-level”system is sometimes used,but this is less accurate than the term two-state, because the levels could be degenerate,comprising several states.)However,its misusage is so widespread that we adopt it anyway-The oscillatingfield will be treated classically,and12MIT8.421Notes,Spring2006Figure1.1.Spectral profile of the H a line of atomic hydrogen by conventional absorption ponents1)and2)arise from thefine structure splitting.The possibility that a third line lines at position3)was suggested to indicate that the Dirac theory might need to be revised. (From“The Spectrum of Atomic Hydrogen”-Advances.G.W.Series ed.,World Scientific,1988). the linewidth of both states will be taken as zero until near the end of the chapter where relaxation will be treated phenomenologically.The organization of the material is historical because this happens to be also a logical order of presentation.The classical driven oscillator is discussedfirst,then the magnetic resonance of a classical spin,and then a quantized spin.The density matrix is introduced last and used to treat systems with damping-this is a useful prelude to the application of resonance ideas at optical frequencies and to the many real systems which have damping.1.2Resonance Studies and Q.E.D.One characteristic of atomic resonance is that the results,if you can obtain them at all,are generally of very high accuracy,so high that the information is qualitatively different from other types.The hydrogenfine structure is a good example.In the late1930s there was extensive investigation of the Balmer series of hydrogen, (n>2→n=2).The Dirac Theory was thought to be in good shape,but some doubts were arising.Careful study of the Balmer-alpha line(n=3→n=2)showed that the line shape might not be given accurately by the Dirac Theory.Pasternack,in1939,suggested that the2s2S1/2and2p2P1/2states were not degenerate, but that the energy of the2s state was greater than the Dirac value by∼.04cm−1(or, in frequency,∼1,200MHz).However,there was no rush to throw out the Dirac theory on suchflimsy evidence.In1947,Lamb found a splitting between the2S1/2and2P1/2levels using a resonance method.The experiment is one of the classics of physics.Although his veryfirst observa-tion was relatively crude,it was nevertheless accurate to one percent.He foundS H=1h E(2S1/2)−E(2P1/2) =1050(10)MHz(1.1)The inadequacy of the Dirac theory was inescapably demonstrated.1.2.Resonance Studies and Q.E.D.3The magnetic moment of the electron offers another example.In 1925,Uhlenbeck and Goudsmit suggested that the electron has intrinsic spin angular momentum S =1/2(in units of ¯h )and magnetic momentµe =e ¯h 2m=µB (1.2)where µB is the Bohr magneton.The evidence was based on studies of the multiplicity of atomic lines (in particular,the Zeeman structure).The proposal was revolutionary,but the accuracy of the prediction that µe =µB was poor,essentially one significant figure.According to the Dirac theory,µe =µB ,exactly.However,our present understanding is µe µB−1=1.1596521884(43)×10−3(experiment,U.of Washington)(1.3)This result is in good agreement with theory,the limiting factor in the comparison being possible doubts about the value of the fine structure constant.The Lamb shift and the departure of µe from µB resulted in the award of the 1955Nobel prize to Lamb and Kusch,and provided the experimental basis for the theory of quantum electrodynamics for which Feynman,Schwinger and Tomonaga received the Nobel Prize in 1965.1.2.1The language of resonance:a classical damped systemBecause the terminology of classical resonance,as well as many of its features,are car-ried over into quantum mechanics,we start by reviewing an elementary resonant system.Consider a harmonic oscillator composed of a series RLC circuit.The charge obeys¨q +γ˙q +ω20q =0(1.4)where γ=R/L,ω20=1/LC .Assuming that the system is underdamped (i.e.γ2<4ω20),the solution for q is a linear combination of exp −γ2 exp ±iω t (1.5)where ω =ω0 1−γ2/4ω20.If ω0 γ,which is often the case,we have ω ≡ω0.The energy in the circuit is W =12C q 2+12L ˙q 2=W 0e −γt (1.6)where W 0=W (t =0).The decay time of the stored energy is τ=1γ.If the circuit is driven by a voltage E 0e iωt ,the steady state solution is q o e iωt whereq 0=E 02ω0L 1(ω0−ω+iγ/2).(1.7)(We have made the usual resonance approximation:ω2o −ω2≈2ω0(ω0−ω).)The average power delivered to the circuit isP =12E 20R 11+ ω−ω0γ/2 2(1.8)4MIT8.421Notes,Spring2006Figure1.2.Sketch of a Lorentzian curve,the universal response curve for damped oscillators and for many atomic systems.The width of the curve(full width at half maximum)is∆ω=γ,where γis the decay constant.The time constant for decay isτ=γ.In the presence of noise(right),the frequency precision with which the center can be located,δω,depends on the signal-to-noise ratio, S/N:δω=∆ω/(S/N).The plot of P vsω(Fig.1.2)is a universal resonance curve often called a“Lorentzian curve”.The full width at half maximum(“FWHM”)is∆ω=γ.The quality factor of the oscillator isQ=ω0∆ω(1.9)Note that the decay time of the free oscillator and the linewidth of the driven oscillator obeyτ∆ω=1(1.10) This can be regarded as an uncertainty relation.Assuming that energy and frequency are related by E=¯hωthen the uncertainty in energy is∆E=¯h∆ωandτ∆E=¯h(1.11) It is important to realize that the Uncertainty Principle merely characterizes the spread of individual measurements.Ultimate precision depends on the experimenter’s skill:the Uncertainty Principle essentially sets the scale of difficulty for his or her efforts.The precision of a resonance measurement is determined by how well one can“split”the resonance line.This depends on the signal to noise ratio(S/N).(see Fig.1.2)As a rule of thumb,the uncertaintyδωin the location of the center of the line isδω=∆ωS/N(1.12)In principle,one can makeδωarbitrarily small by acquiring enough data to achieve the required statistical accuracy.In practice,systematic errors eventually limit the precision. Splitting a line by a factor of104is a formidable task which has only been achieved a few1.3.Magnetic Resonance:Classical Spin in Time-varying B-Field5 times,most notably in the measurement of the Lamb shift.A factor of103,however,is not uncommon,and102is child’s play.1.3Magnetic Resonance:Classical Spin in Time-varying B-Field1.3.1The classical motion of spins in a static magneticfieldNote:angular momentum will always be expressed in a form such as¯h J,where the vector J is dimensionless.The interaction energy and equation of motion of a classical spin in a static magnetic field are given byW=− µ·B,(1.13)F=−∇W=∇( µ·B),(1.14)torque= µ×B.(1.15) In a uniformfield,F=0.The torque equation(d¯h/dt=torque)givesd¯h J= µ×B.(1.16)dtSince µ=γ¯h J,we haved J=γJ×B=−γB×J.(1.17)dtTo see that the motion of J is pure precession about B,imagine that B is alongˆz and that the spin,J,is tipped at an angleθfrom this axis,and then rotated at an angleφ(t) from theˆx axis(ie.,θandφare the conventionally chosen angles in spherical coordinates). The torque,−γB×J,has no component along J(that is,alongˆr),nor alongθ(because the J−B plane containsθ),hence−γB×J=−γB|J|sinθˆφ.This implies that J maintains constant magnitude and constant tipping angleθ.Since theφ-component of J is d J/dt =|J|sinθ(dφ/dt)it is clear thatφ(t)=−γBt.This solution shows that the moment precesses with angular velocityΩL=−γB(1.18) whereΩL is called the Larmor Frequency.For electrons,γe/2π=2.8MHz/gauss,for protonsγp/2π=4.2kHz/gauss.Note that Planck’s constant does not appear in the equation of motion:the motion is classical.6MIT8.421Notes,Spring20061.3.2Rotating coordinate transformationA second way tofind the motion is to look at the problem in a rotating coordinate system. If some vector A rotates with angular velocityΩ,thend A=Ω×A.(1.19)dtIf the rate of change of the vector in a system rotating atΩis(d A/dt)rot,then the rate of change in an inertial system is the motion in plus the motion of the rotating coordinate system.d A dt= d A dt rot+Ω×A.(1.20)inertThe operator prescription for transforming from an inertial to a rotating system is thusd dt= d dt inert−Ω×.(1.21)rotApplying this to Eq.1.17givesd J dt=γJ×B−Ω×J=γJ×(B+Ω/γ).(1.22)rotIf we letB eff=B+Ω/γ,(1.23) Eq.1.22becomesd J dt=γJ×B eff.(1.24)rotIf B eff=0,J is constant in the rotating system.The condition for this isΩ=−γB(1.25) as we have previously found in Eq.1.18.1.3.3Larmor’s theoremTreating the effects of a magneticfield on a magnetic moment by transforming to a rotat-ing co-ordinate system is closely related to Larmor’s theorem,which asserts that the effect of a magneticfield on a free charge can be eliminated by a suitable rotating co-ordinate transformation.1.4.Motion in a Rotating Magnetic Field7Consider the motion of a particle of mass m,charge q,under the influence of an applied force F0and the Lorentz force due to a staticfield B:F=F0+q v×B.(1.26) Now consider the motion in a rotating coordinate system.By applying Eq.1.20twice to r,we have¨r rot=¨r inert−2Ω×v rot−Ω×(Ω×r).(1.27)F rot=F inert−2m(Ω×v rot)−mΩ×(Ω×r),(1.28) where F rot is the apparent force in the rotating system,and F inert is the true or inertial force.Substituting Eq.1.26givesF rot=F inert+q v×B+2m v×Ω−mΩ×(Ω×r).(1.29)If we chooseΩ=−(q/2m)B,and take B=ˆz B,we haveF rot=F inert−mΩ2B2ˆz×(ˆz×r).(1.30) The last term is usually small.If we drop it we haveF rot=F inert.(1.31)The effect of the magneticfield is removed by going into a system rotating at the Larmor frequency qB/2m.Although Larmor’s theorem is suggestive of the rotating co-ordinate transformation, Eq.1.22,it is important to realize that the two transformations,though identical in form, apply to fundamentally different systems.A magnetic moment is not necessarily charged-for example a neutral atom can have a net magnetic moment,and the neutron possesses a magnetic moment in spite of being neutral-and it experiences no net force in a uniform magneticfield.Furthermore,the rotating co-ordinate transformation is exact for a mag-netic moment,whereas Larmor’s theorem for the motion of a charged particle is only valid when the B2term is neglected.1.4Motion in a Rotating Magnetic Field1.4.1Exact resonanceConsider a moment µprecessing about a staticfield B0,which we take to lie along the z axis.Its motion might be described byµz=µcosθ,µx=µsinθcosω0t,µy=−µsinθsinω0t(1.32) whereω0is the Larmor frequency,andθis the angle the moment makes with B o.8MIT8.421Notes,Spring2006 Now suppose we introduce a magneticfield B1which rotates in the x-y plane at the Larmor frequencyω0=−γB0.The magneticfield isB(t)=B1(ˆx cosω0t−ˆy sinω0t)+B0ˆz.(1.33) The problem is tofind the motion of µ.The solution is simple in a rotating coordinate system.Let systemˆx ,ˆy ,ˆz =ˆz)precess around the z-axis at rate−ω0.In this system the field B1is stationary(andˆx is chosen to lie along B1),and we haveB(t)eff=B(t)−(ω0/γ)ˆz=B1ˆx +(B0−ω0/γ)ˆz=B1ˆx .(1.34) The effectivefield is static and has the value of B1.The moment precesses about the field at rateωR=γB1,(1.35) often called the Rabi frequency.This equation contains a lot of history:the RF magnetic resonance community con-ventionally calls this frequencyω1,but the laser resonance community calls it the Rabi FrequencyωR in honor of Rabi’s invention of the resonance technique.If the moment initially lies along the z axis,then its tip traces a circle in theˆy −ˆz plane. At time t it has precessed through an angleφ=ωR t.The moment’s z-component is given byµz(t)=µcosωR t.(1.36) At time T=π/ωR,the moment points along the negative z-axis:it has“turned over”.1.4.2Off-resonance behaviorNow suppose that thefield B1rotates at frequencyω=ω0.In a coordinate frame rotating with B1the effectivefield isB eff=B1ˆx +(B0−ω/γ)ˆz.(1.37)The effectivefield lies at angleθwith the z-axis,as shown in Fig.1.3Thefield is static, and the moment precesses about it at rate(called the effective Rabi frequency)ω R=γB eff=γ (B0−ω/γ)2+B21= (ω0−ω)2+ω2R(1.38) whereω0=γB0,ωR=γB1,as before.Assume that µpoints initially along the+z-axis.Findingµz(t)is a straightforward problem in geometry.The moment precesses about B effat rateω R,sweeping a circle1.5.Adiabatic Rapid Passage:Landau-Zener Crossing9 as shown.The radius of the circle isµsinθ,where sinθ=B1/ (B0−ω/γ)2+B21=ωR/ (ω−ω0)2+ω2R.In time t the tip sweeps through angleφ=ω R t.The z-component of the moment isµz(t)=µcosαwhereαis the angle between the moment and the z-axis after it has precessed through angleφ.As the drawing shows,cosαis found from A2=2µ2(1−cosα).Since A=2µsinθsin(ω R t/2),we have4µ2sin2θsin2(ω R t/2)= 2µ2(1−cosα)andµz(t)=µcosα=µ(1−2sin2θsin2ω R t/2)=µ 1−2ω2R(ω−ω0)+ω2R sin212 (ω−ω0)2+ω R t(1.39)=µ 1−2(ωR/ω R)2sin2(ω R t/2) (1.40)The z-component of µoscillates in time,but unlessω=ω0,the moment never completely inverts.The rate of oscillation depends on the magnitude of the rotatingfield;the amplitude of oscillation depends on the frequency difference,ω−ω0,relative toωR.The quantum mechanical result will turn out to be identical.1.5Adiabatic Rapid Passage:Landau-Zener CrossingAdiabatic rapid passage is a technique for inverting a spin population by sweeping the system through resonance.Either the frequency of the oscillatingfield or the transition frequency(e.g.,by changing the applied magneticfield)is slowly varied.The principle is qualitatively simple in the rotating coordinate system.The problem can also be solved analytically.In this section we give the qualitative argument and then present the analytic quantum result.The solution is of quite general interest because this physical situation arises frequently,for example in inelastic scattering,where it is called a curvecrossing.10MIT 8.421Notes,Spring 2006B 1dB eff(t )dt =1B 11γdωdtγB 1,(1.41)or using ωR =γB 1,dωdtω2R ,(1.42)In this example we have shown that a slow change from ω γB 0to ω γB 0will flip the spin;the same argument shows that the reverse direction of slow change will also flip1.5.Adiabatic Rapid Passage:Landau-Zener Crossing11 the spin.For a two-state system the problem can be solved rigorously.Consider a spin1/2system in a magneticfield B effwith energiesW±=±12¯hγB eff.(1.43)For a uniformfield B0(with B1=0),the effectivefield in the rotating frame is B0−ω/γ, andW±=±12¯h(ω0−ω),(1.44)whereω0=γB0.Asωis swept through resonance,the two states move along their charging eigenenergies.The energies change,but the states do not.There is no coupling between the states,so a spin initially in one or the other will remain so indefinitely no matter how ωchanges relative toω0.In the presence of a rotatingfield,B1,however,the energy levels look quite different:in-stead of intersecting lines they form non-intersecting hyperbolas separated by energy±¯hωR. If the system moves along these hyperbolas,then↑→↓and↓→↑.2ωRFigure1.5.An avoided crossing.A system on one level can jump to the other if the parameter q that governs the energy levels is swept sufficiently rapidly.1.5.2Quantum treatment-Landau ZenerWhether or not the system follows an energy level adiabatically depends on how rapidly the energy is changed,compared to the minimum energy separation.To cast the problem in quantum mechanical terms,imagine two non-interacting states whose energy separation,ω,depends on some parameter x which varies linearly in time,and vanishes for some value x0.Now add a perturbation having an off-diagonal matrix element V which is independent12MIT8.421Notes,Spring2006 of x,so that the energies at x0are±V,as shown in Fig.1.11e.The probability that the system will“jump”from one adiabatic level to the other after passing through the“avoided crossing”(i.e.,the probability of non-adiabatic behavior)isP na=e−2πΓ(1.45) whereΓ=|V|2¯h2dωdt −1(1.46)This result was originally obtained by Landau and Zener.The jumping of a system as it travels across an avoided crossing is called the Landau-Zener effect.Further description, and reference to the initial papers,can be found in[5].Inserting the parameters for our magneticfield problem,we haveP na=exp −π2ω2R dω/dt (1.47) Note that the factor in the negative exponential is related to the inequality in Eq.1.42. When Eq.1.42is satisfied,the exponent is large and the probability of non-adiabatic be-havior is exponentially small.Incidentally the“rapid”in adiabatic rapid passage is something of a misnomer.The technique was originally developed in nuclear magnetic resonance in which thermal relax-ation effects destroy the spin polarization if one does not invert the population sufficiently rapidly.In the absence of such relaxation processes one can take as long as one pleases to traverse the anticrossings,and the slower the crossing the less the probability of jumping.1.6Resonance Of Quantized Spin1.6.1Expected projection of quantized spinBefore solving the quantum mechanical problem of a magnetic moment in a time varying field,it is worthwhile demonstrating that its motion is classical.By“its motion is classical”we mean the time evolution of the expectation value of the magnetic moment operation obeys the classical equation of motion.Specifically,we shall show thatddtµop =γ µop ×B.(1.48) Proof:Recall that,for any operator Od dt O =i¯h[H,O] +∂O∂t.(1.49)If the operator is not explicitly time dependent the last term vanishes.The interaction of µwith a staticfield B0z is1.6.Resonance Of Quantized Spin13H=− µop·B0=−γJ·B0=−γB0J z,(1.50) Note that J has dimensions of angular momentum.Thusd µopdt=−iγB0[J z, µop]/¯h.(1.51) Using µop=γJ,we can rewrite this asd Jdt=−iγB0[J z,J]/¯h.(1.52) The commutation rules for J are[J x,J y]=i¯h J z,etc.,or J×J=i¯h J.(This is a shorthand way of writing[J i,J j]= ij,J k.)Hence˙Jx=γB0J y(1.53)˙Jy=−γB0J x(1.54)˙Jz=0(1.55) These describe the uniform precession of J about the z axis at a rate−γB0.ThusddtJ =γ J ×B(1.56) and since µop=γJ,this directly yields Eq.1.48:ddtµop =γ µop ×B.(1.57) Thus the quantum mechanical and classical equation of motion are identical.This fact underlies the great utility of classical magnetic resonance in providing intuition about res-onance in quantum spin systems.1.6.2Resonance of quantized spin11.6.2.1The Rabi transition probabilityFor a spin1/2particle we can push the classical solution further and obtain the amplitudes and probabilities for each state.Consider µz /¯h=γ J z =γm,where m is the usual “magnetic”quantum number.For a spin1/2particle m has the value+1/2or−1/2.Let the probabilities for having these values be P+and P−respectively.ThenJ z =12P+−12P−,(1.58)or,since P++P−=1,J z =12(1−2P−),(1.59)µz =12γ¯h(1−2P−).(1.60)14MIT8.421Notes,Spring2006 If µlies along the z axis at t=0,thenµz(0)=γ¯h/2,and we haveµz(t)=µz(0)(1−2P−).(1.61) In this case,P−is the probability that a spin in state m=+1/2at t=0has made a transition to m=−1/2at time t,P↑→↓(t).Comparing Eq.1.61with1.39,we seeP↑→↓(t)=ω2Rω2R+(ω−ω0)2sin212 ω2R+(ω−ω0)2t(1.62)P↑→↓(t)=(ωR/ω R)2sin2(ω R t/2)(1.63) This result is known as the Rabi transition probability.It is important enough to memo-rize.We have derived it from a classical correspondence argument,but it can also be derived quantum mechanically.In fact,such a treatment is essential for a complete understanding of the system.1.6.2.2The Hamiltonian of a quantized spin12Now we investigate the time dependence of the wave functions for a quantized spin12systemwith moment µ=γ¯h S is placed in a uniform magneticfield B=ω0k/γand,starting at t =0,subject to afield B(t)which rotates in the x−y plane with frequencyω.Thesefields are the same as thefields discussed in the preceding section on the motion of classical spin and a time-varyingfield.The only difference is that now we are discussing their effect on a quantized system,so we must use Schr¨o dingerr’s equation rather than the laws of classical Electricity and Magnetism to discuss the dynamics of the system.The basis states are(using the standard column vector representation):|1 = 10 (1.64)|2 = 01 (1.65) with state|1 lower in energy.The unperturbed Hamiltonian is(remember,ω0=γB0,and µis an operator)H0=− µ·B0=−¯h S zω0=−12¯hω0 100−1 =−12¯hω0σz(1.66)whereσz is a Pauli spin matrix.The energies areE1=−¯hω0/2ω1=−ω0/21.7.Quantum Mechanical Solution for Resonance in a Two-State System15E2=+¯hω0/2ω2=+ω0/2(1.68) The perturbation Hamiltonian is simplified by usingωR=γB R where B R is the magnetic field which rotates in the x−y plane.(In the magnetic resonance community the subscript R is often replaced by1.)H (t)=− µ·B R(t)=− µ·(ωR/γ)[ˆx cosωt−ˆy sinωt]=−ωR[S x cosωt−S y sinωt]=−¯hωR2 0110cosωt− 0−i i0 sinωt=−¯hωR2 0e iωte−iωt0 (1.69)ˆx andˆy are unit vectors,and µ·ˆx=γS x.In the penultimate line we have replaced the operators S x and S y with(¯h/2)σx and(¯h/2)σy whereσx andσy are Pauli spin matrices. The perturbation matrix element is just the entry H 21(t)in row2and column1:2|H |1 =−¯hωR2e−iωt(1.70)We have thus derived the following Hamiltonian for the spin12problem:H=¯h2 −ω0−ωR e+iωt−ωR e−iωtω0 (1.71)This Hamiltonian is of the famous form for the“dressed atom”which we will discuss elsewhere.1.7Quantum Mechanical Solution for Resonance in a Two-State SystemAs has been emphasized,a two-state system coupled by a periodic interaction is an archetype for large areas of atomic/optical physics.The quantum mechanical solution can be achieved by a variety of approaches,the most elegant of which is the dressed atom picture in which the atom and radiationfield constitute a single quantum system and onefinds its eigenstates. That approach will be introduced later.Here we follow a rather different approach,less elegant,but capable of being generalized to a variety or problems including multi-level resonance and radiative decay in the presence of oscillatingfields.The starting point is the interaction representation.1.7.1Interaction representationWe consider a complete set of eigenstates to a Hamiltonian H0,ψ=|1>,|2>...,such thatH0|j>=E j|j>.(1.72) The problem is tofind the behavior of the system under an interaction V(t),i.e.tofind solutions to。

随机游动中首达概率的研究与分析作者：李斯儒谭静来源：《现代信息科技》2021年第08期DOI：10.19850/ki.2096-4706.2021.08.004摘要：众所周知，随机游动作为一种特殊的马尔科夫链在诸多领域有着广泛的应用，而研究系统的阈值状态的首达概率则是重中之重。

文章首先对对称一维随机游动某状态的首达概率、首达时进行计算分析;然后采用了对递推公式求母函数的方法求解对称一维随机游动的首次返回概率，最后通过蒙特卡洛方法求其首次返回概率的模拟值与理论值对照，通过大样本容量的计算证实理论解的精确性。

关键词：随机游动;首达时间;首达概率;马尔科夫链中图分类号：O211.1 文献标识码：A 文章编号：2096-4706（2021）08-0013-04Research and Analysis on the First Arrival Probability in Random WalkLI Siru，TAN Jing（Nanhang Jincheng College，Nanjing 211156，China）Abstract：As we all know，the random walk is a special Markov chain that has wide applications in many fields，and the research on the first arrival probability of the threshold state of the system is the most important thing. First，the paper calculates and analyzes the first arrival probability and first arrival time of a certain state of a symmetric one-dimensional random walk;then uses the method of finding the generating function of the recurrence formula to solve the first return probability of the symmetric one-dimensional random walk，and finally uses Monte Carlo method compares the simulated value of the probability of its first return with the theoretical value，and confirms the accuracy of the theoretical solution through the calculation of a large sample volume.Keywords：random walk;first arrival time;first arrival probability;Markov chain0 引言在马尔科夫链中有一类非常特殊的随机过程被称为随机游走[1]（Random Walk，RW）。

2015 AMC 12B竞赛真题Problem 1What is the value of ?Problem 2Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?Problem 3Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?Problem 4David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Marta finished 1 place behind Jack. David finished 2 places behind Hikmet. Jack finished 2 places behind Todd. Todd finished 1 place behind Rand. Marta finished in 6th place. Who finished in 8th place?Problem 5The Tigers beat the Sharks 2 out of the 3 times they played. They then played more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for ?Problem 6Back in 1930, Tillie had to memorize her multiplication facts fromto . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?Problem 7A regular 15-gon has lines of symmetry, and the smallest positive angle for which it has rotational symmetry is degrees. What is ?Problem 8What is the value of ?Problem 9Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a playerknocks the bottle off the ledge is , independently of what has happened before. What is the probability that Larry wins the game?Problem 10How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?Problem 11The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?Problem 12Let , , and be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation?Problem 13Quadrilateral is inscribed in a circle withand . What is ?Problem 14A circle of radius 2 is centered at . An equilateral triangle with side4 has a vertex at . What is the difference between the area of the regionthat lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?Problem 15At Rachelle's school an A counts 4 points, a B 3 points, a C 2 points, and a D 1 point. Her GPA on the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get As in both Mathematics and Science, and at least a C in each of English and History.She thinks she has a chance of getting an A in English, and a chance of getting a B. In History, she has a chance of getting an A, and a chanceof getting a B, independently of what she gets in English. What is the probability that Rachelle will get a GPA of at least 3.5?Problem 16A regular hexagon with sides of length 6 has an isosceles triangle attached to each side. Each of these triangles has two sides of length 8. The isosceles triangles are folded to make a pyramid with the hexagon as the base of the pyramid. What is the volume of the pyramid?Problem 17An unfair coin lands on heads with a probability of . When tossed times,the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of ?Problem 18For every composite positive integer , define to be the sum of the factors in the prime factorization of . For example, because the prime factorization of is , and . What is the range of the function , ?Problem 19In , and . Squares and are constructed outside of the triangle. The points , , , and lie on a circle. What is the perimeter of the triangle?Problem 20For every positive integer , let be the remainder obtained when is divided by 5. Define a functionrecursively as follows:What is ?Problem 21Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of ?Problem 22Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?Problem 23A rectangular box measures , where , , and are integers and. The volume and the surface area of the box are numerically equal. How many ordered triples are possible?Problem 24Four circles, no two of which are congruent, have centers at , , , and , and points and lie on all four circles. The radius of circleis times the radius of circle , and the radius of circle is timesthe radius of circle . Furthermore, and . Let be the midpoint of . What is ?Problem 25A bee starts flying from point . She flies inch due east to point . For , once the bee reaches point , she turns counterclockwise and then flies inches straight to point . When the bee reaches she is exactly inches away from , where , , andare positive integers and and are not divisible by the square of any prime. What is ?2015 AMC 12B竞赛真题答案1.C2.b3.a4.b5.b6.a7.d8.d9.c 10.c 11.e 12.d 13.b 14.d 15.d 16.c 17.d 18.d 19.c 20.b 21.d 22.d 23.b 24.d 25.b。

Harvard Math TournamentFebruary 27, 19991. If a@b = , for how many real values of a does a@1 = 0?2. For what single digit n does 91 divide the 9-digit number 12345n789?3. Alex is stuck on a platform floating over an abyss at 1 ft/s. An evil physicist has arranged for the platform to fall in (taking Alex with it) after traveling 100ft. One minute after the platform was launched, Edward arrives with a second platform capable of floating all the way across the abyss. He calculates for 5 seconds, then launches the second platform in such a way as to maximize the time that one end of Alex’s platform is between the two ends of the new platform, thus giving Alex as much time as possible to switch. If both platforms are 5 ft long and move with constant velocity once launched, what is the speed of the second platform (in ft/s)?4. Find all possible values of d where a2 6ad + 8d2 = 0, a = 0.5. You are trapped in a room with only one exit, a long hallway with a series of doors and land mines. To get out you must open all the doors and disarm all the mines. In the room is a panel with 3 buttons, which conveniently contains an instruction manual. The red button arms a mine, the yellow button disarms two mines and closes a door, and the green button opens two doors. Initially 3 doors are closed and 3 mines are armed. The manual warns that attempting to disarm two mines or open two doors when only one is armed/closed will reset the system to its initial state. What is the minimum number of buttons you must push to get out?6. Carl and Bob can demolish a building in 6 days, Anne and Bob can do it in 3, Anne and Carl in 5. How many days does it take all of them working together if Carl gets injured at the end of the rst day and can’t come back? Express your answer as a fraction in lowest terms.7. Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 piles but ends up with one sheet left over each time. How many piles does he need?8. If f (x) is a monic quartic polynomial such that f ( 1) = 1, f (2) = 4, f ( 3) = 9, andf (4) = 16, nd f (1).9. How many ways are there to cover a 3 × 8 rectangle with 12 identical dominoes?10. Pyramid EARLY is placed in (x y z) coordinates so that E = (10 10 0), A = (10 10 0), R = ( 10 10 0), L = ( 10 10 0), and Y = (0 0 10). Tunnels are drilled through the pyramid in such a way that one can move from (x y z) to any of the 9 points (x y z 1), (x ± 1 y z 1), (x y ± 1 z 1),(x ± 1 y ± 1 z 1). Sean starts at Y and moves randomly down to the base of thepyramid, choosing each of the possible paths with probabilityeach time. What is the probability9that he ends up at the point (8 9 0)?1Algebra SolutionsHarvard-MIT Math TournamentFebruary 27, 1999Problem A1 [3 points]If a@b = a a 3b 3 , for how many real values of a does a@1 = 0?e S x o t l r u a t neo ion:us If ss ,in 0c e t h th en a t a ma k e 1s ,he 0 d o e r n 2o f +th a o g in =a l 0 .ex pres Thus si o a n 1,o r a wh i is c h a i r s oo a t nof a 2 + a + 1. But this quadratic has no real roots, in particular its roots are − 1±2√−3 . Thereforethere are no such real values of a, so the answer is 0.Problem A2 [3 points]For what single digit n does 91 divide the 9-digit number 12345n789?Solution 1: 123450789 leaves a remainder of 7 when divided by 91, and 1000 leaves a remainder of 90, or -1, so adding 7 multiples of 1000 will give us a multiple of 91.Solution 2: For those who don’t like long division, there is a quicker way. First notice that 91 = 7 13, and 7 11 13 = 1001. Observe that 12345n789 = 123 1001000+45 n 1001 123 1001+123 45n+789 It follows that 91 will divide 12345n789 i ff 91 divides 123 45n + 789 = 462n. The number 462 is divisible by 7 and leaves a remainder of 7 when divided by 13.Problem A3 [4 points]Alex is stuck on a platform floating over an abyss at 1 ft/s. An evil physicist has arranged for the platform to fall in (taking Alex with it) after traveling 100ft. One minute after the platform was launched, Edward arrives with a second platform capable of floating all the way across the abyss. He calculates for 5 seconds, then launches the second platform in such a way as to maximize the time that one end of Alex’s platform is between the two ends of the new platform, thus giving Alex as much time as possible to switch. If both platforms are 5 ft long and move with constant velocity once launched, what is the speed of the second platform (in ft/s)?Solution: The slower the second platform is moving, the longer it will stay next to the first platform. However, it needs to be moving fast enough to reach the first platform before it’s too late. Let v be the velocity of the second platform. It starts 65 feet behind the first platform, so it reaches the s．i pdel edhhlines up with the front of the first platform at the instant that the first platform has travelled 100ft, which occurs after 100 seconds. Since the second platform is launched 65 seconds later and has to travel 105 feet, its speed is 105/35 = 3ft/s.12Find all possible values of a dwhere a 2 − 6ad + 8d 2 = 0, a = 0. Solution: Dividing a 2 − 6ad + 8d 2 = 0 by a 2 , we get 1 − 6a d + 8(ad )2 = 0. The roots of this quadratic are 21 , 41 .Problem A5 [5 points]You are trapped in a room with only one exit, a long hallway with a series of doors and land mines. To get out you must open all the doors and disarm all the mines. In the room is a panel with 3 buttons, which conveniently contains an instruction manual. The red button arms a mine, the yellow button disarms two mines and closes a door, and the green button opens two doors. Initially 3 doors are closed and 3 mines are armed. The manual warns that attempting to disarm two mines or open two doors when only one is armed/closed will reset the system to its initial state. What is the minimum number of buttons you must push to get out?Solution: Clearly we do not want to reset the system at any time. After pressing the red button r times, the yellow button y times, and the green button g times, there will be 3 + r − 2y armed mines and 3 + y − 2g closed doors, so we want the values of r , y , and g that make both of these quantities 0 while minimizing r + y + g. From the number of doors we see that y must be odd, from the number of mines we see y = (3 + r)/2 ≥ 3/2, so y ≥ 3. Then g = (3 + y)/2 ≥ 3, and r = 2y − 3 ≥ 3, so r + y + g ≥ 9. Call the red, yellow, and green buttons 1, 2, and 3 respectively for notational convenience, then a sequence of buttons that would get you out is 123123123. Another possibility is 111222333, and of course there are others. Therefore the answer is 9.Problem A6 [5 points]Carl and Bob can demolish a building in 6 days, Anne and Bob can do it in 3, Anne and Carl in5. How many days does it take all of them working together if Carl gets injured at the end of the first day and can’t come back? Express your answer as a fraction in lowest terms.Solution: Let a be the portion of the work that Anne does in one day, similarly b for Bob and c for Carl. Then what we are given is the system of equations b + c = 1/6, a + b = 1/3, and a + c = 1/5. Thus in the first day they complete a + b + c = 21(1/6 + 1/3 + 1/5) = 7/20, leaving 13/20 for Anne and Bob to complete. This takes 113//320= 39/20 days, for a total of 2059 .Matt has somewhere between 1000 and 2000 pieces of paper he’s trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries 2, 3, 4, 5, 6, 7, and 8 piles but ends up with one sheet left over each time. How many piles does he need?Solution: The number of sheets will leave a remainder of 1 when divided by the least common multiple of 2, 3, 4, 5, 6, 7, and 8, which is 8 · 3 · 5 · 7 = 840. Since the number of sheets is between 1000 and 2000, the only possibility is 1681. The number of piles must be a divisor of 1681 = 412 , hence it must be 41.23If f (x) is a monic quartic polynomial such that f (−1) = −1, f (2) = −4, f (−3) = −9, and f (4) = −16, find f (1).Solution: The given data tells us that the roots of f (x) + x 2 are -1, 2, -3, and 4. Combining with the fact that f is monic and quartic we get f (x) + x 2 = (x + 1)(x − 2)(x + 3)(x − 4). Hence f (1) = (2)(−1)(4)(−3) − 1 = 23.Problem A9 [7 points]How many ways are there to cover a 3 × 8 rectangle with 12 identical dominoes?Solution: Trivially there is 1 way to tile a 3 × 0 rectangle, and it is not hard to see there are 3 waysto tile a 3 × 2. Let T n be the number of tilings of a 3 × n rectangle, where n is even. From the diagram below we see the recursion T n = 3T n −2 + 2(T n −4 + T n −6 + . . . + T 2 + T 0). Given that, we can just calculate T 4 = 11, T 6 = 41, and T 8 is 153.Problem A10 [8 points]Pyramid EARLY is placed in (x,y , z) coordinates so that E = (10, 10, 0), A = (10, −10, 0), R = (−10, −10, 0), L = (−10, 10, 0), and Y = (0, 0, 10). Tunnels are drilled through the pyramid in such a way that one can move from (x,y , z) to any of the 9 points (x,y , z − 1), (x ± 1, y , z − 1), (x, y ± 1, z − 1),(x ± 1, y ± 1, z − 1). Sean starts at Y and moves randomly down to the base of the pyramid, choosing each of the possible paths with probability 91each time. What is the probability that he ends up at the point (8, 9, 0)?. . . . . . etc...Solution 1: Start by figuring out the probabilities of ending up at each point on the way down the pyramid. Obviously we start at the top vertex with probability 1, and each point on the next level down with probability 1/9. Since each probability after n steps will be some integer over 9n, we will look only at those numerators. The third level down has probabilities as shown below. Think of this as what you would see if you looked at the pyramid from above, and peeled offthe top two layers.1 2 3 2 1246423 6 9 6 3246421 2 3 2 134What we can observe here is not only the symmetry along vertical, horizontal, and diagonal axes, but also that each number is the product of the numbers at the ends of its row and column (e.g. 6 = 2 · 3). This comes from the notion of independence of events, i.e. that if we east and then south, we end up in the same place as if we had moved south and then east. Since we are only looking for the probability of ending up at (8, 9, 0), we need only know that this is true for the top two rows of the square of probabilities, which depend only on the top two rows of the previous layer. This will follow from the calculation of the top row of each square, which we can do via an algorithm similar to Pascal’s triangle. In the diagram below, each element is the sum of the 3 above it.11 1 11 2 3 2 11 3 6 7 6 3 11 4 10 16 19 16 10 4 11 5 15 30 45 51 45 30 15 5 1Now observe that the first 3 numbers in row n , where the top is row 0, are 1, n , n (n2+1) . This factis easily proved by induction on n , so the details are left to the reader. Now we can calculate the top two rows of each square via another induction argument, or by independence, to establish that the second row is always n times the first row. Therefore the probability of ending up at the point (8,9,0) is 910550 .Solution 2: At each move, the ① and y coordinates can each increase by 1, decrease by 1, or stay the same. The y coordinate must increase 9 times and stay the same 1 times, the ① coordinate can either increase 8 times and stay the same 1 time or decrease 1 time and increase 9 times. Now we consider every possible case. First consider the cases where the ① coordinate decreases once. If the ① coordinate decreases while the y coordinate increases, then we have 8 moves that are the sameand 2 that are di fferent, which can be done in 8!10! = 90 ways. If the ① coordinate decreases whilethe y coordinate stays the same, then we have 9 moves that are the same and 1 other, which can bedone in 9!10! = 10 ways. Now consider the cases where the ① coordinate stays the same twice. If the y coordinate stays the same while the ① coordinate increases, then we have 7 moves that are thesame, 2 that are the same, and 1 other, which can be done in 71!20!!= 360 ways. If the y coordinate stays the same while the ① coordinate stays the same, then we have 8 moves that are the same and 24 that are di fferent, which can be done in 8!10! = 90 ways. Therefore there are 360 + 90 + 90 + 10 = 550paths to (8,9,0), out of 91 0 possible paths to the bottom, so the probability of ending up at the point (8,9,0) is 910550 .。

⼤数定律（LawofLargeNumbers）的原理及Python实现本⽂以抛掷硬币(tossing coins)为例, 来理解⼤数定律(Law of Large Numbers), 并使⽤ Python 语⾔实现.原理⼤数定律, 简单来说, 就是随着抛掷硬币的次数的增多, 正⾯向上出现的⽐例(the ratio of heads)会越来越接近正⾯朝上的概率(the probability of heads).Python 实现在⽰例代码中, 假定正⾯朝上的概率(the probability of heads)为0.51, 模拟进⾏10个系列的硬币投掷(coin tosses), 每个投掷系列, 投掷硬币10000 次, 然后, 将正⾯朝上的⽐例(the ratio of heads)随着投掷次数的变化进⾏显⽰, 并保存到 images/ ⽬录下. 具体代码如下:#-*- coding: utf8 -*-from__future__import print_functionimport numpy as npimport matplotlib.pyplot as pltimport osdef law_of_large_numbers(num_series=10, num_tosses=10000, heads_prob=0.51, display=True):""" Get `num_series` series of biased coin tosses, each of which has `num_tosses` tosses, and the probability of heads in each toss is `heads_prob`."""# 1 when less than heads_prob; 0 when no less than heads_probcoin_tosses = (np.random.rand(num_tosses, num_series) < heads_prob).astype('float32')cumulative_heads_ratio = np.cumsum(coin_tosses, axis=0)/np.arange(1, num_tosses+1).reshape(-1,1)if display:plot_fig(cumulative_heads_ratio, heads_prob)def save_fig(fig_id, dirname="images/", tight_layout=True):print("Saving figure", fig_id)if tight_layout:plt.tight_layout()# First, ensure the directory existsif not os.path.isdir(dirname):os.makedirs(dirname)# Then, save the fig_id imagenameimage_path = "%s.png" % os.path.join(dirname, fig_id)plt.savefig(image_path, format='png', dpi=300)def plot_fig(cumulative_heads_ratio, heads_prob, save=True):# Get the number of tosses in a seriesnum_tosses = cumulative_heads_ratio.shape[0]# Set the width and height in inchesplt.figure(figsize=(8, 3.5))# Plot cumulative heads ratioplt.plot(cumulative_heads_ratio)# Plot the horizontal line of value `heads_prob`, with black dashed linetypeplt.plot([0, num_tosses], [heads_prob, heads_prob], "k--", linewidth=2, label="{}%".format(round(heads_prob*100, 1)))# Plot the horizontal line of value 0.5 with black solid linetypeplt.plot([0, num_tosses], [0.5, 0.5], "k-", label="50.0%")plt.xlabel("Number of coin tosses")plt.ylabel("Heads ratio")plt.legend(loc="lower right")# Set x ranges and y rangesxmin, xmax, ymin, ymax = 0, num_tosses, 0.42, 0.58plt.axis([xmin, xmax, ymin, ymax])if save:save_fig("law_of_large_numbers_plot")plt.show()if__name__ == '__main__':num_series, num_tosses = 10, 10000heads_proba = 0.51law_of_large_numbers(num_series, num_tosses, heads_proba)显⽰结果, 如下图所⽰参考资料[1] Aurélien Géron. Hands-On Machine Learning with Scikit-Learn and TensorFlow. O'Reilly Media, 2017.。

Good morning and welcome—to my colleagues here on stage, to the family members who are with us today, and most of all, to the Class of 2021! And a special shout-out to Marvin Chun, beginning his first year as the new dean of Yale College.大家早上好，欢迎各位同事、各位家长，尤其是2021届的本科新生们！也特别欢迎马文，我们今年新上任的本科生院院长。

A few years ago, I helped a friend—a member of the Yale College Class of 1982, in fact—teach a Yale College seminar called “Great Big Ideas.” Each week, students in the seminar considered a “big idea” from a different field of study. For homework, they watched video lectures delivered by various experts and read primary sources. Then they came to class ready to debate each week’s “big idea.” By the end of the course, they had become conversant in major debates and questions in art history, political philosophy, evolutionary biology, and other fields. My friend described the educational impact of the course as “a mile wide and an inch deep.”几年前，我帮一位朋友上过一门耶鲁本科生的讨论课，名字叫做“伟大的思想”——这位朋友其实是耶鲁本科1982届的校友。

Sample Selection Bias as a Specification ErrorJames J.HeckmanEconometrica,Vol.47,No.1.(Jan.,1979),pp.153-161.Stable URL:/sici?sici=0012-9682%28197901%2947%3A1%3C153%3ASSBAAS%3E2.0.CO%3B2-J Econometrica is currently published by The Econometric Society.Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use,available at/about/terms.html.JSTOR's Terms and Conditions of Use provides,in part,that unless you have obtained prior permission,you may not download an entire issue of a journal or multiple copies of articles,and you may use content in the JSTOR archive only for your personal,non-commercial use.Please contact the publisher regarding any further use of this work.Publisher contact information may be obtained at/journals/econosoc.html.Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission.JSTOR is an independent not-for-profit organization dedicated to and preserving a digital archive of scholarly journals.For more information regarding JSTOR,please contact support@.Wed Apr1811:20:522007You have printed the following article:Sample Selection Bias as a Specification ErrorJames J.HeckmanEconometrica ,Vol.47,No.1.(Jan.,1979),pp.153-161.Stable URL:/sici?sici=0012-9682%28197901%2947%3A1%3C153%3ASSBAAS%3E2.0.CO%3B2-JThis article references the following linked citations.If you are trying to access articles from anoff-campus location,you may be required to first logon via your library web site to access JSTOR.Please visit your library's website or contact a librarian to learn about options for remote access to JSTOR.[Footnotes]2Wage Comparisons--A Selectivity BiasReuben GronauThe Journal of Political Economy ,Vol.82,No.6.(Nov.-Dec.,1974),pp.1119-1143.Stable URL:/sici?sici=0022-3808%28197411%2F12%2982%3A6%3C1119%3AWCSB%3E2.0.CO%3B2-L2Comments on Selectivity Biases in Wage ComparisonsH.Gregg LewisThe Journal of Political Economy ,Vol.82,No.6.(Nov.-Dec.,1974),pp.1145-1155.Stable URL:/sici?sici=0022-3808%28197411%2F12%2982%3A6%3C1145%3ACOSBIW%3E2.0.CO%3B2-YReferences1Regression Analysis when the Dependent Variable Is Truncated NormalTakeshi AmemiyaEconometrica ,Vol.41,No.6.(Nov.,1973),pp.997-1016.Stable URL:/sici?sici=0012-9682%28197311%2941%3A6%3C997%3ARAWTDV%3E2.0.CO%3B2-FLINKED CITATIONS -Page 1of 2-2Specification Bias in Estimates of Production FunctionsZvi GrilichesJournal of Farm Economics ,Vol.39,No.1.(Feb.,1957),pp.8-20.Stable URL:/sici?sici=1071-1031%28195702%2939%3A1%3C8%3ASBIEOP%3E2.0.CO%3B2-V 4Wage Comparisons--A Selectivity BiasReuben GronauThe Journal of Political Economy ,Vol.82,No.6.(Nov.-Dec.,1974),pp.1119-1143.Stable URL:/sici?sici=0022-3808%28197411%2F12%2982%3A6%3C1119%3AWCSB%3E2.0.CO%3B2-L8Dummy Endogenous Variables in a Simultaneous Equation SystemJames J.HeckmanEconometrica ,Vol.46,No.4.(Jul.,1978),pp.931-959.Stable URL:/sici?sici=0012-9682%28197807%2946%3A4%3C931%3ADEVIAS%3E2.0.CO%3B2-F9Asymptotic Properties of Non-Linear Least Squares EstimatorsRobert I.JennrichThe Annals of Mathematical Statistics ,Vol.40,No.2.(Apr.,1969),pp.633-643.Stable URL:/sici?sici=0003-4851%28196904%2940%3A2%3C633%3AAPONLS%3E2.0.CO%3B2-311Comments on Selectivity Biases in Wage ComparisonsH.Gregg LewisThe Journal of Political Economy ,Vol.82,No.6.(Nov.-Dec.,1974),pp.1145-1155.Stable URL:/sici?sici=0022-3808%28197411%2F12%2982%3A6%3C1145%3ACOSBIW%3E2.0.CO%3B2-Y 12Specification Errors and the Estimation of Economic RelationshipsH.TheilRevue de l'Institut International de Statistique /Review of the International Statistical Institute ,Vol.25,No.1/3.(1957),pp.41-51.Stable URL:/sici?sici=0373-1138%281957%2925%3A1%2F3%3C41%3ASEATEO%3E2.0.CO%3B2-BLINKED CITATIONS -Page 2of 2-。