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354 CHAPTER 9. CENTRAL LIMIT THEOREM Exercises 1 A die is rolled 24 times. Use the Central Limit Theorem to estimate the probability that (a) the sum is greater than 84. (b) the sum is equal to 84. 2 A random walker starts at 0 on the x-axis and at each time unit moves 1 step to the right or 1 step to the left with pro... | prob_Page_362_Chunk5201 |
9.2. DISCRETE INDEPENDENT TRIALS 355 (d) Estimate the probability that the bank will lose any money in 100 rolls. (e) How many rolls does the bank need to collect to have a 99 percent chance of a net loss? 7 A surveying instrument makes an error of −2, −1, 0, 1, or 2 feet with equal probabilities when measuring the hei... | prob_Page_363_Chunk5202 |
356 CHAPTER 9. CENTRAL LIMIT THEOREM 14 It has been suggested that Example 9.6 is unrealistic, in the sense that the probabilities of errors are too low. Make up your own (reasonable) estimate for the distribution m(x), and determine the probability that a student’s grade point average is accurate to within .05. Also d... | prob_Page_364_Chunk5203 |
9.3. CONTINUOUS INDEPENDENT TRIALS 357 -3 -2 -1 1 2 3 0.1 0.2 0.3 0.4 n = 2 n = 3 n = 4 n = 10 Figure 9.13: Density function for S∗ n (uniform case, n = 2, 3, 4, 10). µ = E(Xi) = 1 λ , σ2 = V (Xj) = 1 λ2 . Here we know the density function for Sn explicitly (see Section 7.2). We can use Corollary 5.1 to calculate the d... | prob_Page_365_Chunk5204 |
358 CHAPTER 9. CENTRAL LIMIT THEOREM -4 -2 2 0.1 0.2 0.3 0.4 0.5 n = 2 n = 3 n = 10 n = 30 Figure 9.14: Density function for S∗ n (exponential case, n = 2, 3, 10, 30, λ = 1). for all a < b, lim n→∞P(a < S∗ n < b) = 1 √ 2π Z b a e−x2/2 dx . 2 We shall give a proof of this theorem in Section 10.3. We will now look at som... | prob_Page_366_Chunk5205 |
9.3. CONTINUOUS INDEPENDENT TRIALS 359 We have already noticed that the estimate in the Chebyshev inequality is not always a good one, and here is a case in point. If we assume that n is large enough so that the density for Sn is approximately normal, then we have P Sn n −µ < .0001 = P | prob_Page_367_Chunk5206 |
360 CHAPTER 9. CENTRAL LIMIT THEOREM Sample Mean If he knows the variance σ2 of the error distribution is .0002, then he can estimate the mean µ by taking the average, or sample mean of, say, 36 measurements: ¯µ = x1 + x2 + · · · + xn n , where n = 36. Then, as before, E(¯µ) = µ. Moreover, the preceding argument shows ... | prob_Page_368_Chunk5207 |
9.3. CONTINUOUS INDEPENDENT TRIALS 361 -6 -4 -2 2 4 6 0.1 0.2 0.3 0.4 Figure 9.15: Graph of t−density for n = 1, 3, 8 and the normal density with µ = 0, σ = 1. Exercises Notes on computer problems: (a) Simulation: Recall (see Corollary 5.2) that X = F −1(rnd) will simulate a random variable with density f(x) and distri... | prob_Page_369_Chunk5208 |
362 CHAPTER 9. CENTRAL LIMIT THEOREM 2 Let {Xk}, 1 ≤k ≤n, be a sequence of independent random variables, all with mean 0 and variance 1, and let Sn, S∗ n, and An be their sum, standardized sum, and average, respectively. Verify directly that S∗ n = Sn/√n = √nAn. 3 Let {Xk}, 1 ≤k ≤n, be a sequence of random variables, a... | prob_Page_370_Chunk5209 |
9.3. CONTINUOUS INDEPENDENT TRIALS 363 9 How large must n be before Sn = X1+X2+· · ·+Xn is approximately normal? This number is often surprisingly small. Let us explore this question with a computer simulation. Choose n numbers from [0, 1] with probability density f(x), where n = 3, 6, 12, 20, and f(x) is each of the d... | prob_Page_371_Chunk5210 |
364 CHAPTER 9. CENTRAL LIMIT THEOREM distributed with mean d, the true distance, and variance 16. She performs a series of n measurements. Let An = X1 + X2 + · · · + Xn n be the average of these measurements. (a) Show that P An − 8 √n ≤d ≤An + 8 √n ≈.95. (b) When nine measurements were taken, the average of the dis... | prob_Page_372_Chunk5211 |
Chapter 10 Generating Functions 10.1 Generating Functions for Discrete Distribu- tions So far we have considered in detail only the two most important attributes of a random variable, namely, the mean and the variance. We have seen how these attributes enter into the fundamental limit theorems of probability, as well a... | prob_Page_373_Chunk5212 |
366 CHAPTER 10. GENERATING FUNCTIONS µk = kth moment of X = E(Xk) = ∞ X j=1 (xj)kp(xj) , provided the sum converges. Here p(xj) = P(X = xj). In terms of these moments, the mean µ and variance σ2 of X are given simply by µ = µ1, σ2 = µ2 −µ2 1 , so that a knowledge of the first two moments of X gives us its mean and varia... | prob_Page_374_Chunk5213 |
10.1. DISCRETE DISTRIBUTIONS 367 Examples Example 10.1 Suppose X has range {1, 2, 3, . . ., n} and pX(j) = 1/n for 1 ≤j ≤n (uniform distribution). Then g(t) = n X j=1 1 netj = 1 n(et + e2t + · · · + ent) = et(ent −1) n(et −1) . If we use the expression on the right-hand side of the second line above, then it is easy to... | prob_Page_375_Chunk5214 |
368 CHAPTER 10. GENERATING FUNCTIONS Here µ1 = g′(0) = pet (1 −qet)2 t=0 = 1 p , µ2 = g′′(0) = pet + pqe2t (1 −qet)3 t=0 = 1 + q p2 , µ = µ1 = 1/p, and σ2 = µ2 −µ2 1 = q/p2, as computed in Example 6.26. 2 Example 10.4 Let X have range {0, 1, 2, 3, . . .} and let pX(j) = e−λλj/j! for all j (Poisson distribution with mea... | prob_Page_376_Chunk5215 |
10.1. DISCRETE DISTRIBUTIONS 369 We note that g(t) is differentiable for all t, since it is a finite linear combination of exponential functions. If we compute g′(t)/g(t), we obtain x1p(x1)etx1 + . . . + xnp(xn)etxn p(x1)etx1 + . . . + p(xn)etxn . Dividing both top and bottom by etxn, we obtain the expression x1p(x1)et(x... | prob_Page_377_Chunk5216 |
370 CHAPTER 10. GENERATING FUNCTIONS For example, suppose we know that the moments of a certain discrete random variable X are given by µ0 = 1 , µk = 1 2 + 2k 4 , for k ≥1 . Then the moment generating function g of X is g(t) = ∞ X k=0 µktk k! = 1 + 1 2 ∞ X k=1 tk k! + 1 4 ∞ X k=1 (2t)k k! = 1 4 + 1 2et + 1 4e2t . This ... | prob_Page_378_Chunk5217 |
10.1. DISCRETE DISTRIBUTIONS 371 If X and Y are independent random variables and Z = X + Y is their sum, with pX, pY , and pZ the associated distribution functions, then we have seen in Chapter 7 that pZ is the convolution of pX and pY , and we know that convolution involves a rather complicated calculation. But for th... | prob_Page_379_Chunk5218 |
372 CHAPTER 10. GENERATING FUNCTIONS Hence, we have gZ(t) = gX(t)gY (t) = (pet + q)2n , or, what is the same, hZ(z) = hX(z)hY (z) = (pz + q)2n = 2n X j=0 2n j (pz)jq2n−j , from which we can see that the coefficient of zj is just pZ(j) = | prob_Page_380_Chunk5219 |
10.1. DISCRETE DISTRIBUTIONS 373 Then the Xk are independent random variables describing a Bernoulli process. Let S0 = 0, and, for n ≥1, let Sn = X1 + X2 + · · · + Xn . Then Sn describes Peter’s fortune after n trials, and Peter is first in the lead after n trials if Sk ≤0 for 1 ≤k < n and Sn = 1. Now this can happen wh... | prob_Page_381_Chunk5220 |
374 CHAPTER 10. GENERATING FUNCTIONS so that Peter is sure to be in the lead eventually if p ≥q. How long will it take? That is, what is the expected value of T? This value is given by E(T) = h′ T (1) = 1/(p −q), if p > q, ∞, if p = q. This says that if p > q, then Peter can expect to be in the lead by about 1/(p −q)... | prob_Page_382_Chunk5221 |
10.1. DISCRETE DISTRIBUTIONS 375 (a) Show that p and p′ have the same first and second moments, but not the same third and fourth moments. (b) Find the ordinary and moment generating functions for p and p′. 6 Let p be the probability distribution p = 0 1 2 0 1/3 2/3 , and let pn = p ∗p ∗· · · ∗p be the n-fold convol... | prob_Page_383_Chunk5222 |
376 CHAPTER 10. GENERATING FUNCTIONS (c) Show that hZ(z) cannot ever have the form hZ(z) = z2 + z3 + · · · + z12 11 . Hint: hX and hY must have at least one nonzero root, but hZ(z) in the form given has no nonzero real roots. It follows from this observation that there is no way to load two dice so that the probability... | prob_Page_384_Chunk5223 |
10.2. BRANCHING PROCESSES 377 The first attempt at a solution was given by Reverend H. W. Watson. Because of a mistake in algebra, he incorrectly concluded that a family name would always die out with probability 1. However, the methods that he employed to solve the problems were, and still are, the basis for obtaining ... | prob_Page_385_Chunk5224 |
378 CHAPTER 10. GENERATING FUNCTIONS 2 1 0 1/4 1/4 1/4 1/4 1/4 1/4 1/2 1/16 1/8 5/16 1/2 4 3 2 1 0 0 1 2 1/64 1/32 5/64 1/8 1/16 1/16 1/16 1/16 1/2 Figure 10.1: Tree diagram for Example 10.8. Branching processes have served not only as crude models for population growth but also as models for certain physical processes... | prob_Page_386_Chunk5225 |
10.2. BRANCHING PROCESSES 379 Let dm be the probability that the process dies out by the mth generation. Of course, d0 = 0. In our example, d1 = 1/2 and d2 = 1/2 + 1/8 + 1/16 = 11/16 (see Figure 10.1). Note that we must add the probabilities for all paths that lead to 0 by the mth generation. It is clear from the defini... | prob_Page_387_Chunk5226 |
380 CHAPTER 10. GENERATING FUNCTIONS 1 1 1 1 1 1 0 0 0 0 0 y z d > 1 d < 1 d = 1 0 y = z y y z z y = h (z) 1 1 (a) (c) (b) Figure 10.2: Graphs of y = z and y = h(z). y = h(z) can intersect the line y = z in at most two points. Since we know it must intersect the line y = z at (1, 1), we know that there are just three p... | prob_Page_388_Chunk5227 |
10.2. BRANCHING PROCESSES 381 y = z y = h(z) y z 1 p0 0 d = 0 1 d d d d 1 2 3 Figure 10.3: Geometric determination of d. the probability of dying out by the nth generation. Then we know that d1 = p0. We know further that dn = h(dn−1) where h(z) is the generating function for the number of offspring produced by a single ... | prob_Page_389_Chunk5228 |
382 CHAPTER 10. GENERATING FUNCTIONS Generation Probability of dying out 1 .2 2 .312 3 .385203 4 .437116 5 .475879 6 .505878 7 .529713 8 .549035 9 .564949 10 .578225 11 .589416 12 .598931 Table 10.1: Probability of dying out. p0 = .2092 p1 = .2584 p2 = .2360 p3 = .1593 p4 = .0828 p5 = .0357 p6 = .0133 p7 = .0042 p8 = .... | prob_Page_390_Chunk5229 |
10.2. BRANCHING PROCESSES 383 however, that we can describe the limiting behavior of Zn as n →∞. We first show that the generating function hn(z) of the distribution of Zn can be obtained from h(z) for any branching process. We recall that the value of the generating function at the value z for any random variable X can... | prob_Page_391_Chunk5230 |
384 CHAPTER 10. GENERATING FUNCTIONS Examples Example 10.10 For the branching process of Example 10.8 we have h(z) = 1/2 + (1/4)z + (1/4)z2 , h2(z) = h(h(z)) = 1/2 + (1/4)[1/2 + (1/4)z + (1/4)z2] = +(1/4)[1/2 + (1/4)z + (1/4)z2]2 = 11/16 + (1/8)z + (9/64)z2 + (1/32)z3 + (1/64)z4 . The probabilities for the number of off... | prob_Page_392_Chunk5231 |
10.2. BRANCHING PROCESSES 385 This leads us to a quadratic equation. We know that z = 1 is one solution. The other is found to be d = 1 −b −c c(1 −c) . It is easy to verify that d < 1 just when m > 1. It is possible in this case to find the distribution of Zn. This is done by first finding the generating function hn(z).7 ... | prob_Page_393_Chunk5232 |
386 CHAPTER 10. GENERATING FUNCTIONS Geometric pj Data Model 0 .2092 .1816 1 .2584 .3666 2 .2360 .2028 3 .1593 .1122 4 .0828 .0621 5 .0357 .0344 6 .0133 .0190 7 .0042 .0105 8 .0011 .0058 9 .0002 .0032 10 .0000 .0018 Table 10.3: Comparison of observed and expected frequencies. The geometric model tends to favor the larg... | prob_Page_394_Chunk5233 |
10.2. BRANCHING PROCESSES 387 5 10 15 20 25 0.5 1 1.5 2 2.5 3 Figure 10.4: Simulation of Zn/mn for the Keyfitz example. the simulations the process died out, which is consistent with the value d = .3 that we found for this example. For the other seven simulations the value of Wn tends to a limiting value which is differe... | prob_Page_395_Chunk5234 |
388 CHAPTER 10. GENERATING FUNCTIONS suitable hypotheses, as n →∞, mnP(Zn = [tmn]) →k(t) , for t > 0, and P(Zn = 0) →d . However, unlike the Central Limit Theorem for sums of independent random vari- ables, the function k(t) will depend upon the basic distribution that determines the process. Its form is known for only... | prob_Page_396_Chunk5235 |
10.2. BRANCHING PROCESSES 389 where m = p1+2p2 is the expected number of letters you sold. Thus to be favorable we just have 50m + 50m12 > 100 , or m + m12 > 2 . But this will be true if and only if m > 1. We have seen that this will occur in the quadratic case if and only if p2 > p0. Let us assume for example that p0 ... | prob_Page_397_Chunk5236 |
390 CHAPTER 10. GENERATING FUNCTIONS Z1 Z2 Z3 Z4 Z5 Z6 Z7 Z8 Z9 Z10 Z11 Z12 Profit 1 0 0 0 0 0 0 0 0 0 0 0 -50 1 1 2 3 2 3 2 1 2 3 3 6 250 0 0 0 0 0 0 0 0 0 0 0 0 -100 2 4 4 2 3 4 4 3 2 2 1 1 50 1 2 3 5 4 3 3 3 5 8 6 6 250 0 0 0 0 0 0 0 0 0 0 0 0 -100 2 3 2 2 2 1 2 3 3 3 4 6 300 1 2 1 1 1 1 2 1 0 0 0 0 -50 0 0 0 0 0 0 0... | prob_Page_398_Chunk5237 |
10.2. BRANCHING PROCESSES 391 Z1 Z2 Z3 Z4 Z5 Z6 Z7 Z8 Z9 Z10 Z11 Z12 Profit 1 2 6 7 7 8 11 9 7 6 6 5 200 1 0 0 0 0 0 0 0 0 0 0 0 -50 1 0 0 0 0 0 0 0 0 0 0 0 -50 1 1 1 0 0 0 0 0 0 0 0 0 -50 0 0 0 0 0 0 0 0 0 0 0 0 -100 1 1 1 1 1 1 2 4 9 7 9 7 300 2 3 3 4 2 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 -50 2 1 0 0 0 0 0 0 0 0 0... | prob_Page_399_Chunk5238 |
392 CHAPTER 10. GENERATING FUNCTIONS Exercises 1 Let Z1, Z2, . . . , ZN describe a branching process in which each parent has j offspring with probability pj. Find the probability d that the process even- tually dies out if (a) p0 = 1/2, p1 = 1/4, and p2 = 1/4. (b) p0 = 1/3, p1 = 1/3, and p2 = 1/3. (c) p0 = 1/3, p1 = 0,... | prob_Page_400_Chunk5239 |
10.3. CONTINUOUS DENSITIES 393 (a) Find the generating function f(z) for the number of customers who arrive in one minute and the generating function g(z) for the length of time that a person spends in service once she begins service. (b) Consider a customer branching process by considering the offspring of a customer t... | prob_Page_401_Chunk5240 |
394 CHAPTER 10. GENERATING FUNCTIONS provided the integral µn = E(Xn) = Z +∞ −∞ |x|nfX(x) dx , is finite. Then, just as in the discrete case, we see that µ0 = 1, µ1 = µ, and µ2 −µ2 1 = σ2. Moment Generating Functions Now we define the moment generating function g(t) for X by the formula g(t) = ∞ X k=0 µktk k! = ∞ X k=0 E... | prob_Page_402_Chunk5241 |
10.3. CONTINUOUS DENSITIES 395 In particular, we verify that µ = g′(0) = 1/2 and σ2 = g′′(0) −(g′(0))2 = 1 3 −1 4 = 1 12 as before (see Example 6.25). 2 Example 10.16 Let X have range [ 0, ∞) and density function fX(x) = λe−λx (exponential density with parameter λ). In this case µn = Z ∞ 0 xnλe−λx dx = λ(−1)n dn dλn Z ... | prob_Page_403_Chunk5242 |
396 CHAPTER 10. GENERATING FUNCTIONS (These moments are calculated by integrating once by parts to show that µn = (n −1)µn−2, and observing that µ0 = 1 and µ1 = 0.) Hence, g(t) = ∞ X n=0 µntn n! = ∞ X m=0 t2m 2mm! = et2/2 . This series converges for all values of t. Again we can verify that g(n)(0) = µn. Let X be a nor... | prob_Page_404_Chunk5243 |
10.3. CONTINUOUS DENSITIES 397 Hence, for all N we have N X k=0 µktk k! ≤ N X k=0 (M|t|)k k! ≤eM|t| , which shows that the power series converges for all t. We know that the sum of a convergent power series is always differentiable. 2 Moment Problem Theorem 10.4 If X is a bounded random variable, then the moment generat... | prob_Page_405_Chunk5244 |
398 CHAPTER 10. GENERATING FUNCTIONS Let X1, X2, . . . , Xn be an independent trials process with each Xi having density fX, and let Sn = X1 + X2 + · · · + Xn, and S∗ n = (Sn −nµ)/ √ nσ2 = Sn/√n. Then each Xi has moment generating function g(t), and since the Xi are independent, the sum Sn, just as in the discrete case... | prob_Page_406_Chunk5245 |
10.3. CONTINUOUS DENSITIES 399 as n →∞. Since the densities, and hence the distributions, of the S∗ n are uniquely deter- mined by their moment generating functions under our hypotheses, these conclu- sions are certainly plausible, but their proofs involve a detailed examination of characteristic functions and Fourier ... | prob_Page_407_Chunk5246 |
400 CHAPTER 10. GENERATING FUNCTIONS = +N X j=−N p(j)etj et/2 −e−t/2 2t/2 = g(t)sinh(t/2) t/2 , where we have put sinh(t/2) = et/2 −e−t/2 2 . In the same way, we find that ¯gn(t) = gn(t)sinh(t/2) t/2 , ¯g∗ n(t) = g∗ n(t)sinh(t/2√n) t/2√n . Now, as n →∞, we know that g∗ n(t) →et2/2, and, by L’Hˆopital’s rule, lim n→∞ sin... | prob_Page_408_Chunk5247 |
10.3. CONTINUOUS DENSITIES 401 This is hard to check directly, but easy to check by using characteristic functions. Note first that µ2 = E(X2) = Z +∞ −∞ x2 π(1 + x2) dx = ∞ so that µ2 is infinite. Nevertheless, we can define the characteristic function kX(τ) of x by the formula kX(τ) = Z +∞ −∞ eiτx 1 π(1 + x2) dx . This i... | prob_Page_409_Chunk5248 |
402 CHAPTER 10. GENERATING FUNCTIONS 3 Let X be a continuous random variable with values in [ 0, ∞) and density fX. Find the moment generating functions for X if (a) fX(x) = 2e−2x. (b) fX(x) = e−2x + (1/2)e−x. (c) fX(x) = 4xe−2x. (d) fX(x) = λ(λx)n−1e−λx/(n −1)!. 4 For each of the densities in Exercise 3, calculate the... | prob_Page_410_Chunk5249 |
10.3. CONTINUOUS DENSITIES 403 (c) Sn = X1 + X2 + · · · + Xn. (d) An = Sn/n. (e) S∗ n = (Sn −nµ)/ √ nσ2. 10 Let X1, X2, . . . , Xn be an independent trials process with density f(x) = 1 2e−|x|, −∞< x < +∞. (a) Find the mean and variance of f(x). (b) Find the moment generating function for X1, Sn, An, and S∗ n. (c) What... | prob_Page_411_Chunk5250 |
404 CHAPTER 10. GENERATING FUNCTIONS | prob_Page_412_Chunk5251 |
Chapter 11 Markov Chains 11.1 Introduction Most of our study of probability has dealt with independent trials processes. These processes are the basis of classical probability theory and much of statistics. We have discussed two of the principal theorems for these processes: the Law of Large Numbers and the Central Lim... | prob_Page_413_Chunk5252 |
406 CHAPTER 11. MARKOV CHAINS state. The probabilities pij are called transition probabilities. The process can remain in the state it is in, and this occurs with probability pii. An initial probability distribution, defined on S, specifies the starting state. Usually this is done by specifying a particular state as the ... | prob_Page_414_Chunk5253 |
11.1. INTRODUCTION 407 events also have probabilities that can be written as products of entries of P. Thus, we have p(2) 13 = p11p13 + p12p23 + p13p33 . This equation should remind the reader of a dot product of two vectors; we are dotting the first row of P with the third column of P. This is just what is done in obta... | prob_Page_415_Chunk5254 |
408 CHAPTER 11. MARKOV CHAINS P1 = Rain Nice Snow Rain .500 .250 .250 Nice .500 .000 .500 Snow .250 .250 .500 P2 = Rain Nice Snow Rain .438 .188 .375 Nice .375 .250 .375 Snow .375 .188 .438 P3 = Rain Nice Snow Rain .406 .203 .391 Nice .406 .188 .406 Snow .391 .203 .406 P4 = Rain Nice Snow Ra... | prob_Page_416_Chunk5255 |
11.1. INTRODUCTION 409 Theorem 11.2 Let P be the transition matrix of a Markov chain, and let u be the probability vector which represents the starting distribution. Then the probability that the chain is in state si after n steps is the ith entry in the vector u(n) = uPn . Proof. The proof of this theorem is left as a... | prob_Page_417_Chunk5256 |
410 CHAPTER 11. MARKOV CHAINS but it can also be considered from the point of view of Markov chain theory. The transition matrix is P = W P S W .5 .25 .25 P .5 .25 .25 S .5 .25 .25 . 2 Example 11.6 In the Dark Ages, Harvard, Dartmouth, and Yale admitted only male students. Assume that, at that time, 80 percent o... | prob_Page_418_Chunk5257 |
11.1. INTRODUCTION 411 Example 11.9 (Gene Model) The simplest type of inheritance of traits in animals occurs when a trait is governed by a pair of genes, each of which may be of two types, say G and g. An individual may have a GG combination or Gg (which is genetically the same as gG) or gg. Very often the GG and Gg t... | prob_Page_419_Chunk5258 |
412 CHAPTER 11. MARKOV CHAINS Example 11.11 We start with two animals of opposite sex, mate them, select two of their offspring of opposite sex, and mate those, and so forth. To simplify the example, we will assume that the trait under consideration is independent of sex. Here a state is determined by a pair of animals.... | prob_Page_420_Chunk5259 |
11.1. INTRODUCTION 413 Figure 11.1: Initial state of the stepping stone model. Figure 11.2: State of the stepping stone model after 10,000 steps. is enormous. This is an example of a Markov chain that is easy to simulate but difficult to analyze in terms of its transition matrix. The program SteppingStone simulates this ... | prob_Page_421_Chunk5260 |
414 CHAPTER 11. MARKOV CHAINS 2 In Example 11.4, let a = 0 and b = 1/2. Find P, P2, and P3. What would Pn be? What happens to Pn as n tends to infinity? Interpret this result. 3 In Example 11.5, find P, P2, and P3. What is Pn? 4 For Example 11.6, find the probability that the grandson of a man from Har- vard went to Harva... | prob_Page_422_Chunk5261 |
11.1. INTRODUCTION 415 the matrix of transition probabilities. Find the probability that a randomly chosen grandson of an unskilled laborer is a professional man. 12 In Exercise 11, we assumed that every man has a son. Assume instead that the probability that a man has at least one son is .8. Form a Markov chain with f... | prob_Page_423_Chunk5262 |
416 CHAPTER 11. MARKOV CHAINS 11.2 Absorbing Markov Chains The subject of Markov chains is best studied by considering special types of Markov chains. The first type that we shall study is called an absorbing Markov chain. Definition 11.1 A state si of a Markov chain is called absorbing if it is impossible to leave it (i... | prob_Page_424_Chunk5263 |
11.2. ABSORBING MARKOV CHAINS 417 1 2 3 0 4 1 1 1/2 1/2 1/2 1/2 1/2 1/2 Figure 11.3: Drunkard’s walk. P = TR. ABS. TR. Q R ABS. 0 I Here I is an r-by-r indentity matrix, 0 is an r-by-t zero matrix, R is a nonzero t-by-r matrix, and Q is an t-by-t matrix. The first t states are transient and the last r states... | prob_Page_425_Chunk5264 |
418 CHAPTER 11. MARKOV CHAINS mj and let p be the largest of pj. The probability of not being absorbed in m steps is less than or equal to p, in 2m steps less than or equal to p2, etc. Since p < 1 these probabilities tend to 0. Since the probability of not being absorbed in n steps is monotone decreasing, these probabi... | prob_Page_426_Chunk5265 |
11.2. ABSORBING MARKOV CHAINS 419 Definition 11.3 For an absorbing Markov chain P, the matrix N = (I −Q)−1 is called the fundamental matrix for P. The entry nij of N gives the expected number of times that the process is in the transient state sj if it is started in the transient state si. 2 Example 11.14 (Example 11.13... | prob_Page_427_Chunk5266 |
420 CHAPTER 11. MARKOV CHAINS Proof. If we add all the entries in the ith row of N, we will have the expected number of times in any of the transient states for a given starting state si, that is, the expected time required before being absorbed. Thus, ti is the sum of the entries in the ith row of N. If we write this ... | prob_Page_428_Chunk5267 |
11.2. ABSORBING MARKOV CHAINS 421 Thus, starting in states 1, 2, and 3, the expected times to absorption are 3, 4, and 3, respectively. From the canonical form, R = 0 4 1 1/2 0 2 0 0 3 0 1/2 . Hence, B = NR = 3/2 1 1/2 1 2 1 1/2 1 3/2 · 1/2 0 0 0 0 1/2 = 0 4 1 3/4 1/4 2 1/2 1/2 3 1/4 3/4 . ... | prob_Page_429_Chunk5268 |
422 CHAPTER 11. MARKOV CHAINS N = 1 2 3 4 1 1.60 1.20 .80 .40 2 1.20 2.40 1.60 .80 3 .80 1.60 2.40 1.20 4 .40 .80 1.20 1.60 ; t = 1 4.00 2 6.00 3 6.00 4 4.00 ; B = 0 5 1 .80 .20 2 .60 .40 3 .40 .60 4 .20 .80 . Note that the probability of reaching the bar before reaching hom... | prob_Page_430_Chunk5269 |
11.2. ABSORBING MARKOV CHAINS 423 Find the fundamental matrix N, and also Nc and NR. Interpret the results. 7 In Example 11.8, make states 0 and 4 into absorbing states. Find the fun- damental matrix N, and also Nc and NR, for the resulting absorbing chain. Interpret the results. 8 In Example 11.13 (Drunkard’s Walk) of... | prob_Page_431_Chunk5270 |
424 CHAPTER 11. MARKOV CHAINS it fires. The tanks fire together and each tank fires at the strongest opponent not yet destroyed. Form a Markov chain by taking as states the subsets of the set of tanks. Find N, Nc, and NR, and interpret your results. Hint: Take as states ABC, AC, BC, A, B, C, and none, indicating the tanks... | prob_Page_432_Chunk5271 |
11.2. ABSORBING MARKOV CHAINS 425 17 Show that in both Example 11.11 and the example just given, the probability of absorption in a state having genes of a particular type is equal to the proportion of genes of that type in the starting state. Show that this can be explained by the fact that a game in which your fortun... | prob_Page_433_Chunk5272 |
426 CHAPTER 11. MARKOV CHAINS 21 (Roberts7) A city is divided into 3 areas 1, 2, and 3. It is estimated that amounts u1, u2, and u3 of pollution are emitted each day from these three areas. A fraction qij of the pollution from region i ends up the next day at region j. A fraction qi = 1−P j qij > 0 goes into the atmosp... | prob_Page_434_Chunk5273 |
11.2. ABSORBING MARKOV CHAINS 427 problem is the problem of finding the probability wx of winning an amount T before losing everything, starting with state x. Show that this problem may be considered to be an absorbing Markov chain with states 0, 1, 2, . . . , T with 0 and T absorbing states. Suppose that a gambler has ... | prob_Page_435_Chunk5274 |
428 CHAPTER 11. MARKOV CHAINS 27 Consider an absorbing Markov chain with state space S. Let f be a function defined on S with the property that f(i) = X j∈S pijf(j) , or in vector form f = Pf . Then f is called a harmonic function for P. If you imagine a game in which your fortune is f(i) when you are in state i, then t... | prob_Page_436_Chunk5275 |
11.2. ABSORBING MARKOV CHAINS 429 three tosses. If H occurs, he wins 2 dollars and bets this amount that the next outcome will be T. If he wins, he wins 4 dollars and bets this amount that H will come up next time. If he wins, he wins 8 dollars and the pattern has occurred. If at any time he loses, he leaves with no wi... | prob_Page_437_Chunk5276 |
430 CHAPTER 11. MARKOV CHAINS made an amount AB. The total amount the gamblers will have made when the pattern B occurs is BB. Thus, the amount that the gamblers made after the pattern A has occurred is BB - AB. Again by the fair game argument, EA(T B) = BB-AB. For example, suppose that we start with pattern A = HT and... | prob_Page_438_Chunk5277 |
11.2. ABSORBING MARKOV CHAINS 431 31 In Example 11.11, define f(i) to be the proportion of G genes in state i. Show that f is a harmonic function (see Exercise 27). Why does this show that the probability of being absorbed in state (GG, GG) is equal to the proportion of G genes in the starting state? (See Exercise 17.) ... | prob_Page_439_Chunk5278 |
432 CHAPTER 11. MARKOV CHAINS 35 In Monte Carlo roulette (see Example 6.6), under option (c), there are six states (S, W, L, E, P1, and P2). The reader is referred to Figure 6.2, which contains a tree for this option. Form a Markov chain for this option, and use the program AbsorbingChain to find the probabilities that ... | prob_Page_440_Chunk5279 |
11.3. ERGODIC MARKOV CHAINS 433 11.3 Ergodic Markov Chains A second important kind of Markov chain we shall study in detail is an ergodic Markov chain, defined as follows. Definition 11.4 A Markov chain is called an ergodic chain if it is possible to go from every state to every state (not necessarily in one move). 2 In ... | prob_Page_441_Chunk5280 |
434 CHAPTER 11. MARKOV CHAINS Regular Markov Chains Any transition matrix that has no zeros determines a regular Markov chain. How- ever, it is possible for a regular Markov chain to have a transition matrix that has zeros. The transition matrix of the Land of Oz example of Section 11.1 has pNN = 0 but the second power... | prob_Page_442_Chunk5281 |
11.3. ERGODIC MARKOV CHAINS 435 Example 11.18 Recall that for the Land of Oz example of Section 11.1, the sixth power of the transition matrix P is, to three decimal places, P6 = R N S R .4 .2 .4 N .4 .2 .4 S .4 .2 .4 . Thus, to this degree of accuracy, the probability of rain six days after a rainy day is the s... | prob_Page_443_Chunk5282 |
436 CHAPTER 11. MARKOV CHAINS Thus, the common row of W is the unique vector w which is both a fixed row vector for P and a probability vector. Theorem 11.8 shows that any fixed row vector for P is a multiple of w and any fixed column vector for P is a constant vector. One can also state Definition 11.6 in terms of eigenva... | prob_Page_444_Chunk5283 |
11.3. ERGODIC MARKOV CHAINS 437 Now we divide this vector by the sum of the components, to obtain the final answer: w = ( .4 .2 .4 ) . This method can be easily programmed to run on a computer. 2 As mentioned above, we can also think of the fixed row vector w as a left eigenvector of the transition matrix P. Thus, if we ... | prob_Page_445_Chunk5284 |
438 CHAPTER 11. MARKOV CHAINS Equilibrium We also obtain a new interpretation for w. Suppose that our starting vector picks state si as a starting state with probability wi, for all i. Then the probability of being in the various states after n steps is given by wPn = w, and is the same on all steps. This method of sta... | prob_Page_446_Chunk5285 |
11.3. ERGODIC MARKOV CHAINS 439 is, the expected proportion of times in state sj in the first n steps if the chain starts in state si. If we call being in state sj success and any other state failure, we could ask if a theorem analogous to the law of large numbers for independent trials holds. The answer is yes and is g... | prob_Page_447_Chunk5286 |
440 CHAPTER 11. MARKOV CHAINS We note that the simulation gives a proportion of times in each of the states not too different from the long run predictions of .4, .2, and .4 assured by Theorem 11.7. To get better results we have to simulate our chain for a longer time. We do this for 10,000 days without printing out eac... | prob_Page_448_Chunk5287 |
11.3. ERGODIC MARKOV CHAINS 441 1 2 3 4 5 6 7 8 9 Figure 11.4: The maze problem. To find the fixed probability vector for this matrix, we would have to solve ten equations in nine unknowns. However, it would seem reasonable that the times spent in each compartment should, in the long run, be proportional to the number of... | prob_Page_449_Chunk5288 |
442 CHAPTER 11. MARKOV CHAINS times in state 0 is .0625 and the proportion of times in state 1 is .375. The astute reader will note that these numbers are the binomial distribution 1/16, 4/16, 6/16, 4/16, 1/16. We could have guessed this answer as follows: If we consider a particular ball, it simply moves randomly back... | prob_Page_450_Chunk5289 |
11.3. ERGODIC MARKOV CHAINS 443 4 Find the fixed probability vector w for the matrices in Exercise 3 that are ergodic. 5 Find the fixed probability vector w for each of the following regular matrices. (a) P = .75 .25 .5 .5 . (b) P = .9 .1 .1 .9 . (c) P = 3/4 1/4 0 0 2/3 1/3 1/4 1/4 1/2 . 6 Consider the Mar... | prob_Page_451_Chunk5290 |
444 CHAPTER 11. MARKOV CHAINS 14 Consider an independent trials process to be a Markov chain whose states are the possible outcomes of the individual trials. What is its fixed probability vector? Is the chain always regular? Illustrate this for Example 11.5. 15 Show that Example 11.8 is an ergodic chain, but not a regul... | prob_Page_452_Chunk5291 |
11.3. ERGODIC MARKOV CHAINS 445 21 Write a computer program to simulate the queue in Exercise 20. Have your program keep track of the proportion of the time that the queue length is j for j = 0, 1, . . . , n and the average queue length. Show that the behavior of the queue length is very different depending upon whether... | prob_Page_453_Chunk5292 |
446 CHAPTER 11. MARKOV CHAINS (4 2 4) (5 2 3) (8 2 4) (7 3 4) (8 4 4) (8 3 5) (8 4 8) (10 4 6) (12 4 8) (12 5 7) (12 6 8) (13 5 8) (16 6 8) (15 6 9) (16 6 12) (17 7 10) (20 8 12) (20 8 12) . Table 11.4: Distribution of chips. each state i, let ai be the least common multiple of the denominators of the non-zero entries ... | prob_Page_454_Chunk5293 |
11.4. FUNDAMENTAL LIMIT THEOREM 447 the starting vector a is less than or equal to the corresponding component of b. Show that, in the iteration, the components of the vectors are always increasing, and always less than or equal to the corresponding component of b. 30 (Coffman, Kaduta, and Shepp16) A computing center ke... | prob_Page_455_Chunk5294 |
448 CHAPTER 11. MARKOV CHAINS Lemma 11.1 Let P be an r-by-r transition matrix with no zero entries. Let d be the smallest entry of the matrix. Let y be a column vector with r components, the largest of which is M0 and the smallest m0. Let M1 and m1 be the largest and smallest component, respectively, of the vector Py. ... | prob_Page_456_Chunk5295 |
11.4. FUNDAMENTAL LIMIT THEOREM 449 and m0 ≤m1 ≤m2 ≤· · · . Each sequence is monotone and bounded: m0 ≤mn ≤Mn ≤M0 . Hence, each of these sequences will have a limit as n tends to infinity. Let M be the limit of Mn and m the limit of mn. We know that m ≤M. We shall prove that M −m = 0. This will be the case if Mn −mn ten... | prob_Page_457_Chunk5296 |
450 CHAPTER 11. MARKOV CHAINS Theorem 11.14 Let P be the transition matrix for a regular Markov chain with fixed vector w. Then for any initial probability vector u, uPn →w as n →∞. Proof. Let X0, X1, . . . be a Markov chain with transition matrix P started in state si. Let Y0, Y1, . . . be a Markov chain with transitio... | prob_Page_458_Chunk5297 |
11.4. FUNDAMENTAL LIMIT THEOREM 451 as n goes to ∞. But by similar reasoning to that used above, the difference between this last expression and P(Xn = j) goes to 0 as n goes to ∞. Therefore, P(Xn = j) →wj , as n goes to ∞. This completes the proof. 2 In the above proof, we have said nothing about the rate at which the ... | prob_Page_459_Chunk5298 |
452 CHAPTER 11. MARKOV CHAINS 7 Here is a trick to try on your friends. Shuffle a deck of cards and deal them out one at a time. Count the face cards each as ten. Ask your friend to look at one of the first ten cards; if this card is a six, she is to look at the card that turns up six cards later; if this card is a three,... | prob_Page_460_Chunk5299 |
11.5. MEAN FIRST PASSAGE TIME 453 1 2 3 4 5 6 7 8 9 Figure 11.5: The maze problem. chain, these quantities give the expected number of times in each of the states before reaching state sj for the first time. The ith component of the vector Nc gives the expected number of steps before absorption in the new chain, startin... | prob_Page_461_Chunk5300 |
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