随机过程作业和答案第三章

钱敏平龚光鲁随机过程答案（部分）随机过程课后习题答案第⼀章第⼆题：已知⼀列⼀维分布{();1}n F x n ≥，试构造⼀个概率空间及其上的⼀个相互独⽴的随机变量序列{(,);1}n n ξ?≥使得(,)n ξ?的分布函数为()n F x 。

解：有引理：设ξ为[0, 1]上均匀分布的随机变量，F(x)为某⼀随机变量的分布函数，且F(x)连续，那么1()F x η-=是以F(x)为分布的随机变量。

所以可以假设有相互独⽴的随机变量12,,...,n θθθ服从u[0, 1]分布，另有分布{()}n F x ，如果令1(,)()n n n F ξθ-?=，则有(,)n ξ?为服从分布()n F x 的随机变量。

⼜由假设条件可知，随机变量{(,),1}n n ξ?≥之间相互独⽴，则其中任意有限个随机变量12(,), (,),...,(,)n i i i ξξξ的联合分布为：11221122{(,),(,),...,(,)}()()()i i n in i i i i in in P i x i x i x F x F x F x ξξξ?≤?≤?≤=再令112{,,...,,...},,{|()[0,1],1,2,...}n i i i i w w w w A A x F x i -Ω=∈=∈=，令F 为Ω所有柱集的σ代数，则由Kolmogorov 定理可知，存在F 上唯⼀的概率测度P 使得：11221122{(,),(,),...,(,)}()()()i i n in i i i i in in P i w i w i w F w F w F w ξξξ?≤?≤?≤=则所构造的概率空间为(Ω,F , P)。

第⼋题：令{};1n X n ≥是⼀列相互独⽴且服从(0,1)N （正态分布）的随机变量。

⼜令1n n S X X =++22(1)n S n n ξ+=1(,,)n n F X X σ=试证明：,;1n n F n ξ≥（）是下鞅（参见23题）。

1.Players A and B take turns in answering trivia questions, starting with player A answering the first question. Each time A answers a question, she has probability p 1 of getting it right. Each time B plays, he has probability p 2 of getting it right.(a)If A answers m questions, what is the PMF of the number of questions she gets right?The r.v.is Bin(m,p 1),so the PMF is mkp k 1(1 p 1)m k for k 2{0,1,...,m }.(b)If A answers m times and B answers n times,what is the PMF of the total number of questions they get right (you can leave your answer as a sum)?Describe exactly when/whether this is a Binomial distribution.Let T be the total number of questions they get right.To get a total of k questions right,it must be that A got 0and B got k ,or A got 1and B got k 1,etc.These are disjoint events so the PMF isP (T =k )=k X j =0✓mj ◆p j 1(1 p 1)m j ✓n k j◆p k j 2(1 p 2)n (k j )for k 2{0,1,...,m +n },with the usual convention that n k is 0for k >n .This is the Bin(m +n,p )distribution if p 1=p 2=p ,as shown in class (using the story for the Binomial,or using Vandermonde’s identity).For p 1=p 2,it’s not a Binomial distribution,since the trials have di ↵erent probabilities of success;having some trials with one probability of success and other trials with another probability of success isn’t equivalent to having trials with some “e ↵ective”probability of success.(c)Suppose that the first player to answer correctly wins the game (with no prede-termined maximum number of questions that can be asked).Find the probability that A wins the game.Let r =P (A wins).Conditioning on the results of the first question for each player,we have r =p 1+(1 p 1)p 2·0+(1 p 1)(1 p 2)r,which gives r =p 11 (1 p 1)(1 p 2)=p 1p 1+p 2 p 1p 2.1SI 241 Probability & Stochastic Processes, Fall 2016Homework 3 Solutions随机过程2016作业及答案2.A message is sent over a noisy channel.The message is a sequence x1,x2,...,x n of n bits(x i2{0,1}).Since the channel is noisy,there is a chance tha t any bit might be corrupted,resulting in an error(a0becomes a1or vice versa).Assume that the error events are independent.Let p be the probability that an individual bit has an error(0<p<1/2).Let y1,y2,...,y n be the received message(so y i=x i if there is no error in that bit,but y i=1 x i if there is an error there).To help detect errors,the n th bit is reserved for a parit y check:x n is defined to be 0if x1+x2+···+x n 1is even,and1if x1+x2+···+x n 1is odd.When the message is received,the recipient checks whether y n has the same parit y as y1+y2+···+y n 1. If the parity is wrong,the recipient knows that at least one error occurred;otherwise, the recipient assumes that there were no errors.(a)For n=5,p=0.1,what is the probabilit y that the received message has errors which go undetected?Note that P n i=1x i is even.If the number of errors is even(and nonzero),the errors will go undetected;otherwise,P n i=1y i will be odd,so the errors will be detected.The number of errors is Bin(n,p),so the probability of undetected errors when n=5,p=0.1is✓52◆p2(1 p)3+✓54◆p4(1 p)⇡0.073.(b)For general n and p,write down an expression(as a sum)for the probability that the received message has errors which go undetected.By the same reasoning as in(a),the probability of undetected errors isX k even,k 2✓n k◆p k(1 p)n k.(c)Give a simplified expression,not involving a sum of a large number of terms,for the probabilit y that the received message has errors which go undetected.Hint for(c):Lettinga=X k even,k 0✓n k◆p k(1 p)n k and b=X k odd,k 1✓n k◆p k(1 p)n k,the binomial theorem makes it possible tofind simple expressions for a+b and a b, which then makes it possible to obtain a and b.2Let a,b be as in the hint.Thena +b =X k 0✓n k ◆p k (1 p )n k =1,a b =X k 0✓n k ◆( p )k (1 p )n k =(1 2p )n .Solving for a and b gives a =1+(1 2p )n 2and b =1 (1 2p )n2.Xk even,k 0✓n k ◆p k (1 p )n k =1+(1 2p )n 2.Subtrac ting o ↵the possibility of no errors,we haveX k even,k 2✓n k ◆p k (1 p )n k =1+(1 2p )n 2 (1 p )n .Miracle check :note that letting n =5,p =0.1here gives 0.073,which agrees with (a);letting p =0gives 0,as it should;and letting p =1gives 0for n odd and 1for n even,which agai n makes sense.33.Let X be a r.v. whose possible values are 0, 1, 2,...,with CDF F .In some countries, rather than using a CDF, the convention is to use the function G defined by G (x )=P (X <x ) to specify a distribution. Find a way to convert from F to G , i.e., if F is a known function show how to obtain G (x )for all real x .Write G (x )=P (X x ) P (X = x )=F (x ) P (X = x ).If x is not a nonnegative integer, then P (X = x )=0so G (x )=F (x ). For x a nonnegative integer,P (X = x )=F (x ) F (x 1/2)since the PMF corresponds to the lengths of the jumps in the CDF. (The 1/2was chosen for concreteness; we also have F (x 1/2) = F (x a )for any a 2 (0, 1].)Thus,G (x )=(F (x )if x /2{0,1,2,...}F (x 1/2)if x 2{0,1,2,...}.t More compact ly, we can also write G (x )=lim !x F (t ), where the denotes taking a limit from the left (recall that F is right continuous), and G (x )=F (d x e 1),where d x e is the “ceiling” of x (the smallest integer greater than or equal to x ).4.There are n eggs, each of which hatches a chick with probability p (independently).Eac h of these chicks survives with probability r , independently. What is the distri-bution of the number of chicks that hatch? What is the distribution of the number of chicks that survive? (Give the PMFs; also give the names of the distributions and their parame ters, if they are distributions we have seen in class.)⇤⇥ ©⇤⇥ x ⇤⇥ ⇤⇥ ⇤⇥ ©⇤⇥ ©⇤⇥ x ⇤⇥ ©⇤⇥ ⇤⇥©Let H be the number of eggs that hatch and X be the number of hatchlings that survive.Think of each egg as a Ber noulli trial,where for H we define “success”to mean hatching,while for X we define “success”to mean surviving.For example,in the picture above,where ⇤⇥ ©denotes an egg that hatches with the chick surviving,⇤⇥ x denotes an egg that hatched but whose chick died,and ⇤⇥ denotes an egg that hatch,the events H =7,X =5occurred.By the of the Binomial,H ⇠Bin(n,p ),with PMF P (H =k )= n k p k (1 p )n k for k =0,1,...,n .The eggs independently have probability pr each of hatching a chick that survives.By the story of the Binomial,we have X ⇠Bin(n,pr ),with PMF P (X =k )= n k (pr )k (1 pr )n k for k =0,1,...,n .5.A scientist wishes to study whether men or women are more likely to have a certain disease, or whether they are equally likely. A random sample of m women and n men is gathered, and each person is tested for the disease (assume for this problem that the test is completely accurate). The numbers of women and men in he sa B n(n,w p ho ha He ve re p h e di and seas p e ar are e X unkno and Y wn,re p s and ec w tiv e e r a ly,Y i 2.1 2 e w in ith tereste d ⇠Bi in n(testin g p 1) a the le mp t t X ,m nd ⇠) “null hypothesis” p 1 = p 2.(a) Consider a 2 by 2 ta ble listing with rows corresponding to disease status and columns corresponding to gender, with each entry the count of how many people have that disease status and gender (so m + n is the sum of all 4 entries). Supp ose that it is observed that X + Y = r .The Fisher exact test is based on conditioning on both the row and column sums, so m, n, r are all treated as fixed, and then seeing if the observed value of X is “extreme” compared to this conditional distribution. Assuming the null hypothesis, use Ba yes’ Rule to find the conditional PMF of X given X + Y = r .Is this a distribution we have studied in class? If so, say which (and give its paramet ers).First let us build the 2 ⇥ 2 table (conditioning on the totals m, n, and r ).4Women Men Total Disease x r No Diseasem x r r x +n x m +n r Total n m m +nNext,let us compute P (X =x |X +Y =r ).By Ba yes’rule,P (X =x |X +Y =r )=P (X +Y =r |X =x )P (X =x )P (X +Y =r )=P (Y =r x )P (X =x )P (X +Y =r ).Y Assum Bi i n n (g n,th p e )w nu i l t l h h X ypot inde h p esi e s nde an n d t l of etti Y ng ,s p o =X p +1Y =p 2Bi ,w n(e n h +ave m,X p ).⇠T Bi h n us (,m,p )and ⇠⇠r x p r x p p )r n (1 r +x n m x p x (1 p )m x (1m + n r p )m +n r P (X =x |X +Y =r )== m nx m +r n rx .So the conditional distribution is Hypergeometric with parameters m, n, r.(b) Give an intuitive explanation for the distribution of (a), explaining how this problem relates to other problems we’ve seen, and why p 1 disappears (magica lly?) in the distribution found in (a).This problem has the same structure as the elk (capture-recapture) problem. In the elk problem, we take a sample of elk from a population, where earlier some were tagged, and we want to know the distribution of the number of tagged elk in the sample. By analogy, think of the women as corresponding to tagged elk, and men as corresponding to unta gged elk. Having r people be infected with the disease corresponds to capturing a new sample of r elk the number of women among the r diseased individuals corresponds to the number of tagged elk in the new sample.Under the null hypothesis and given that X + Y = r ,the set of diseased people is equally likely to be any set of r people.It makes sense that the conditional distribution of the number of diseased women does not depend on p ,since once we know tha t X + Y = r ,we can work directly in terms of the fact that we have a population with r diseased and m + n r undiseased people, without worrying about the value of p that originally generated the population characteristics.5。

第三章 马尔科夫过程1、将一颗筛子扔多次。

记X n 为第n 次扔正面出现的点数，问{X(n) , n=1,2,3,···}是马尔科夫链吗？如果是，试写出一步转移概率矩阵。

又记Y n 为前n 次扔出正面出现点数的总和，问{Y(n) , n=1,2,3,···}是马尔科夫链吗？如果是，试写出一步转移概率矩阵。

解：1)由已知可得，每次扔筛子正面出现的点数与以前的状态无关。

故X(n)是马尔科夫链。

E={1,2,3,4,5,6} ,其一步转移概率为:P ij = P ij =P{X(n+1)=j ∣X(n)=i }=1/6 (i=1,2,…,6,j=1,2,…,6) ∴转移矩阵为2)由已知可得，每前n 次扔正面出现点数的总和是相互独立的。

即每次n 次扔正面出现点数的总和与以前状态无关，故Y(n)为马尔科夫链。

其一步转移概率为其中2、一个质点在直线上做随机游动，一步向右的概率为p , (0<p<1)，一步向左的概率为 q , q =1-p 。

在x = 0 和x = a 出放置吸收壁。

记X(n)为第n 步质点的位置，它的可能值是0,1,2,···,a 。

试写出一步转移概率矩阵。

解：由已知可得, 其一步转移概率如下：故一步转移概率为3、做一系列独立的贝努里试验，其中每一次出现“成功”的概率为p ( 0<p<1 ) ,出现“失败”的概率为q , q = 1-p 。

如果第n 次试验出现“失败”认为 X(n) 取得数值为零；如果第n 次试验出现“成功”，且接连着前面k 次试验都出现“成功”，而第 n-k 次试验出现“失败”，认为X(n)取值k ，问{X(n) , n =1,2,···}是马尔科夫链吗？试写出其一步转移概率。

解：由已知得：故为马尔科夫链，其一步转移概率为616161616161616161616161616161616161P ={6,,2,1,6/1,,8,7,,0)1,(+++=<++==+i i i j i j i i i j ij n n P 或)1(6,,2,1;6,,2,1,+++=++=n n n j n n n n i {}α,,2,1,0 =E )(0,1;)0(0,1)1,1(0,,1,,2,1101,1,ααααα≠==≠==+-≠===-=-+j P P j P P i i j P q P P P x j j ij i i i i 而时，当 10000000000000001Pp q p q p q ={}{}m m m m m m i n X l n X i n X i n X i n X l n X P ==+=====+)(0)()(,,)(,)(0)(2211 {}{}mm m m m m in X k l n X i n X i n X i n X k l n X P ==+=====+)()()(,,)(,)()(22114、在一个罐子中放入50个红球和50个蓝球。

随机过程第三章与第四章习题解答3.1 解：令()N t 表示(0,)t 时间内的体检人数，则()N t 为参数为30的poisson 过程。

以小时为单位。

则((1))30E N =。

4030(30)((1)40)!k k P N e k -=≤=∑。

3.2 解：法一：（1）乘坐1、2路汽车所到来的人数分别为参数为1λ、2λ的poisson 过程，令它们为1()N t 、2()N t 。

1N T 表示1()N t =1N 的发生时刻，2N T 表示2()N t =2N 的发生时刻。

1111111111()exp()(1)!N NN T f t t t N λλ-=-- 2221222222()exp()(1)!N NN T f t t t N λλ-=--1212121221112,12|12211122212(,)(|)()exp()exp()(1)!(1)!N N N N N NNN N T T T T T f t t f t t f t t t t t N N λλλλ--==---- 12212121112211122210012()exp()exp()(1)!(1)!NNt N N N N P T T dt t t t t dt N N λλλλ∞--<=----⎰⎰（2）当1N =2N 、1λ=2λ时，12121()()2N N N N P T T P T T <=>=法二：（1）乘车到来的人数可以看作参数为1λ+2λ的泊松过程。

令1Z 、2Z 分别表示乘坐公共汽车1、2的相邻两乘客间到来的时间间隔。

则1Z 、2Z 分别服从参数为1λ、2λ的指数分布，现在来求当一个乘客乘坐1路汽车后，下一位乘客还是乘坐1路汽车的概率。

212211122210()exp()exp()z p P Z Z dz z z dz λλλλ∞=<=--⎰⎰112λλλ=+。

故当一个乘客乘坐1路汽车后，下一位乘客乘坐2路汽车的概率为1-p 212λλλ=+上面的概率可以理解为：在乘客到来的人数为强度1λ+2λ的泊松过程时，乘客分别以112λλλ+概率乘坐公共汽车1，以212λλλ+的概率乘坐公共汽车2。