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�ne,..., Mn = X 1 1, M2 = Sn−1 n − 1 M1 = Sn n, Let Sn = n 1 Xr, where the Xr are independent 196 5 Random Vectors: Independence and Dependence which is to say that Mm = Sn−m+1/(n − m + 1). Then obviously, E|Mn| < ∞. Also by symmetry, for 1 ≤ r ≤ m, E(Xr |Sm) = Sm/m. Hence, also by symmetry, E(Mm+1|M1,..., Mm−m n − m |... |
1) P $ $ $ $ 1 n $ $ $ $ > Nn − p ≤ 2 exp (−n2/4) → 0 as n → ∞. Roughly speaking, this says that the proportion of experiments in which A occurs approaches the probability of A as n increases. (It is pleasing that this agrees with our intuitive notions about events and their probabilities.) 5.8 The Law of Averages 197 ... |
$ $ $ > Sn − µ ≤ 1 n22 E = n−2−2 n (Xi − µ) i=1 n var (Xi ) = σ 2/(n2) and (6) is proved. i=1 → 0 as n → ∞, 198 5 Random Vectors: Independence and Dependence Actually, when Sn is a binomial random variable, we have already shown in Exercise 4.18.3 that more can be said. (8) Theorem If (Xi ; i ≥ 1) are independent indi... |
uniformly in R. Then (10) P(P1 ∈ B) = 1 |R| b f (x) d x = p say. a Now we pick a series (Pj ; j ≥ 1) of such points independently in R, and let I j be the indicator of the event that Pj lies in B. Then by the weak law above 1 n n j=1 P→ p, I j as n → ∞. Hence, for large n, we may expect that (|R|/n) approximation to I... |
run behaviour of random variables and their distributions. This seems an appropriate moment to point out the connections between these concepts. First, we recall our earlier definitions. Here (X n; n ≥ 1) is a sequence of random variables with corresponding distributions (Fn(x); n ≥ 1). Also, F(x) is the distribution o... |
X n − X | > ). Likewise, Hence, F(x − ) = P(X ≤ x −, X n ≤ x) + P(X ≤ x −, X n > x) ≤ Fn(x) + P(|X n − X | > ). F(x − ) − P(|X n − X | > ) ≤ Fn(x) ≤ F(x + ) + P(|X n − X > ). Now allowing n → ∞ and → 0 yields the result. Furthermore, it is not always necessary to postulate the existence of a limit random variable X. A ... |
it follows that |X | ≤ Z. Hence, |Zn| ≤ 2Z. Introduce the indicator function I (A), which takes the value 1 if the event A occurs, and is otherwise 0. Then, for ε > 0, E|Zn| = E[Zn I (Zn ≤ ε)] + E[Zn I (Zn > ε)] ≤ ε + 2E[Z I (Zn > ε)]. Because E|Z | < ∞ and Zn P→ 0, the last term decreases to zero as n → ∞. To see thi... |
part (a) of Remark the theorem at T ∧ n, and then let n → ∞, using Dominated or Monotone convergence to obtain the required result. Proof We proved this when (a) holds in (5.7.14), where we also showed that E(X T ∧n − X 0) = 0. Allowing n → ∞, we have X T ∧. Now suppose (b) holds. Because the sequence is bounded, usin... |
E(YkYi ) − E(Y j Yi ) = 0. Thus, after some algebra, E[(Yk − Y j )2|Y0,..., Yi )] = E Y 2 k |Y0,..., Yi − E Y 2 j |Y0,..., Yi. n ; n ≥ 1) is nondecreasing and Therefore, 0 ≤ E(Yk − Y j )2 ≤ EY 2 k bounded, and therefore converges. Finally, we deduce that E(Yk − Y j )2 → 0 as k, j → ∞, and this shows Yk j. Now (EY 2 ms... |
∈C In particular, we have the Joint distribution: F(x, y) = P(X ≤ x, Y ≤ y) = u≤x v≤y f (u, v). Marginals: Functions: f X (x) = y f (x, y), fY (y) = x f (x, y). P(g(X, Y ) = z) = x,y:g=z f (x, y). 204 Sums: 5 Random Vectors: Independence and Dependence P(X + Y = z) = x f (x, z − x) = y f (z − y, y). Independence: X and... |
Y = y) = x f X |Y (x|y) = x f (x, y)/ fY (y). x 5.10 Review and Checklist for Chapter 5 205 For any pair of random variables where both sides exist, E[E(X |Y )] = EX. Key properties: the conditional expectation E(X |Y ) satisfies E(X |Y ) = EX, if X and Y are independent. E(Xg(Y )|Y ) = g(Y )E(X |Y ), the pull-through p... |
, Y, and Z. (b) What is the marginal distribution of X? (c) Show that the correlation coefficient of X and Y is ρ(X, Y ) = − pq (1 − p)(1 − q) 1 2. Solution Let Ak be the event that Arnold wins the kth hole, Bk the event that he loses the kth hole, and Hk the event that the kth hole is halved. We illustrate the possibil... |
X = x) = (2) = = y,z 18−x y=0 18 x P(X = x, Y = y, Z = z) 18! px x!(18 − x)! (18 − x)! y!(18 − x − y)! q yr 18−x−y px (q + r )18−x, which is binomial with parameters 18 and p. II Either Arnold succeeds with probability p in winning each hole, or he fails with probability 1 − p = q + r. Hence, by Example 5.4.1, the mass... |
number of surviving couples in which both the partners are alive. Show that E(S|A) = A(A − 1) 2(2m − 1). Remark This problem was discussed by Daniel Bernoulli in 1768. Solution methods of solution; you can choose your favourite, or of course find a better one. Let Sa be the number of surviving couples given that A = a.... |
Exercises 209 or E(Sa) = a(a − 1) 2m(2m − 1) E(S2m) = a(a − 1) 2(2m − 1). III Number the couples 1,..., m. Let Yi be the indicator of the event that the male of the ith couple survives and Xi the indicator of the event that the female of the ith couple survives. Then Sa = m 1 Xi Yi. Now P(Y1 = 1|X 1 = 1) = (a − 1)/(2m... |
nd E(Rn), and show that as n → ∞ E(Rn) → 3. Solution The losing finalist comes from the half of the draw not containing the topranked player. These 2n−1 players have ranks N1 < N2 < · · · < N2n−1, which are drawn at random from the 2n − 1 integers {2, 3, 4,..., 2n}, and Rn = N1. Let X 1, X 2,... be the numbers of player... |
same colour. Show that the expected number remaining is r/(b + 1) + b/(r + 1). What is the probability pr that they are all red? Exercise Let X 1, X 2,... be independent and identically distributed. What is m when m ≤ n? E 1 n Xi Xi 1 5.14 Example: Congregations Suppose that n initially separate co... |
with mean µ. If you pick an Exercise individual at random find the probability that she is the kth born of her family, and the expectation of her order of birth in her family. Compare this with µ. Exercise Use the Cauchy–Schwarz inequality Lemma 5.3.13 to prove Cauchy’s inequality (3). 5.15 Example: Propagation A plant... |
) j (1 − γ p)n− j. Thus, S is binomial with parameters n and γ p. (d) Finally, P(N = i|S = j) = P(N = i ∩ S = j) P(S = j) = n−− j ; j ≤ i ≤ n. Thus, the variable N − S given that S = j germinate is binomial with parameters n − j and ( p − pγ )/(1 − γ p). (1) (2) (3) (4) (5) (6) Exercise Exercise Exercise Exercise succu... |
y = 1, Hence, (1) log y ≤ y − 1 with equality if and only if y = 1. Therefore, − i f (i) log f (i) + i f (i) log ai = ≤ i i = 0, f (i) log f (i) ai f (i) ai f (i) − 1 by (1) with equality if and only if f (i) = ai for all i. (b) The positive numbers f X (x) fY (y) satisfy f X (x) fY (y) = 1. Therefore, by part (a), f ... |
X by Y. Exercise Exercise H (Z |X ) = 0. Show that H (X |X ) = 0. Show that if H (X |Y ) = 0 = H (Y |X ) and H (Y |Z ) = H (Z |Y ) = 0, then H (X |Z ) = 5.17 Example: Cooperation Achilles and his two friends, Briseis and Chryseis, play a cooperative game. They possess a die with n faces, and each of them rolls it once... |
? Find the mean and variance of Z = I (AB) + I (BC) + I (AC). If n women play a similar game, and I (Ai, A j ) is the indicator of the event that the i= j I (Ai A j ). 5.18 Example: Strange But True Let (Sn; n ≥ 0) be a simple symmetric random walk with S0 = 0. Let f0(n) = P(T0 = n) be the probability that the walk firs... |
Now, (4) Vn = V n V n V n + 1 if T0 > n − 1 if S1,..., Sn revisits zero exactly once otherwise (5) (6) (7) (8) 216 Hence, 5 Random Vectors: Independence and Dependence E(Vn) − E(Vn−1) = E(Vn) − E(V n) = P(T0 > n) − P(S1,..., Sn revisits 0 exactly once) by (4), = P(T0 > n) − = P(T0 > n) − [ n 2 ] k=1 [ n 2 ] k=1 ... |
Solution (a) I For the event {Ym = n} to occur, it is necessary that: (i) The nth animal is marked, which can occur in a ways. (ii) The preceding n − 1 animals include exactly m − 1 marked and n − m unmarked n − m) ways. animals, which may occur in ( a − 1 m − 1)( b − a The total number of ways of first selecting a dis... |
the relation between the negative binomial distribution and the binomial distribution. The hypergeometric p.m.f. is (3.16.1). (3) (4) (5) If you capture and keep a fixed number n of animals, find the variance of the number Exercise that are marked. Exercise the number of captures required to secure m marked animals, find... |
) = 1, using Theorem 5.6.7. (b) Let Jn be the indicator of a visit to the origin at the nth step, and let R be the total number of returns to the origin. Then E(R) = E ∞ Jn ∞ = E(Jn) = ∞ n=1 P(Sn = 0) where n = 2k, n=1 2k k n=1 1 22k (2k − 1)(2k − 3)... 3.1 2kk(k − 1)... 2.1 = = ≥ = ∞ k=1 ∞ k=1 ∞ k=1 ∞ k=1 1 2k = ∞. (2... |
x) ≤ FY (x). Show that E(X ) ≥ E(Y ), and deduce that FX (x) ≤ FY (x) if and only if, for all increasing functions h(.), E(h(X )) ≥ E(h(Y )). Solution From Example 4.3.3, we have ∞ E(X ) = P(X > k) − −∞ P(X < k) = ∞ (1 − FX (k)) − 0 ∞ ≥ (1 − FY (k)) − k=0 −∞ FY (k) 0 0 = E(Y ). Now if h(.) is an increasing function 0 −... |
independent random variables with respective means (µn; n ≥ 1) and finite variances (σ 2 n ; n ≥ 1). Show that n 2 n Mn = (Xr − µr ) − σ 2 r r =1 r =1 defines a martingale with respect to X n. Now assume that the X n are identically distributed, with mean µ and variance σ 2, and T. is a stopping time for (X n; n ≥ 1) wi... |
T ∧n − µn ∧ T converges in mean square as n → ∞. But from (2) above, we know it converges to YT − µT with probability 1. Hence, E(YT ∧n − µT∧n)2 → E(YT − µT )2. Now taking the limit as n → ∞ in (1) gives the required result. Worked Examples and Exercises 221 (3) Exercise Un and Vn are martingales, where Let (X n; n ≥ 1... |
0 are integers. Solution the conditions for ( q p )Sn to be a martingale follow easily. (a) We noted in (5.6.10) that E( q p )Xr = 1. Because the X n are independent, Let T be the first time at which Sn takes either of the values a or b. The probability that any consecutive sequence of X n, of length a + b, yields a + b... |
6nS2 − n; (a) Sn; n Hence find P(ST = −a), ET, and E(T ∩ {ST = −a}). Show finally that varT = ab(a2 + b2 − 2)/3. + 3n2 + 2n. − 3nSn; (b) S2 n (d) S4 n (c) S3 n 5.24 Example: You Can’t Beat the Odds Let Yn be the total net fortune of a gambler after betting a unit stake on each of n consecutive fair plays in a casino. Th... |
ale: Exercise (a) Optional skipping. At each play, the gambler skips the round or wagers a unit stake. (b) Optional starting. The gambler does not join in until the (T + 1)th play, where T is a stopping time for Yn. (c) Optional stopping. The gambler uses the system until a stopping time T, and then quits. Exercise: Op... |
3 that var(X n+1|M0,..., Mn) = 1 0 if Mn > 1 if Mn = 1. (1) (2) (3) 224 5 Random Vectors: Independence and Dependence Hence, var X n+1 ≤ 1, and we may write E((Mn+1 + n + 1)2 + Mn+1|M0,..., Mn) = (Mn + n)2 − 2(Mn + n)E(X n+1 − 1) + Mn +E((X n+1 − 1)2 − X n+1|M0,..., Mn) ≤ (Mn + n)2 + Mn. Thus, (Mn + n)2 + Mn is a nonne... |
coats outside the subgroup, simply redistributes their coats correctly, and leaves. Show that the expected number of rounds required is C/k. Exercise Suppose now that the purpose of the coats exercise is not simply to leave, but to leave in pairs. Thus, only pairs holding each others coat swap and leave; the rest, inc... |
Zn) = 1 2 [(X n + 1)(Yn − 1) + (X n − 1)(Yn + 1)] = X nYn − 1. The other two possibilities being treated similarly, we obtain the same martingale condition (1). Clearly, T is a finite-mean stopping time for this bounded martingale, and UT = 0, so ET = E(UT + T ) = E(U0 + 0) = ab + bc + ca. (2) Exercise Let S be the num... |
a + b + c + d) are martingales, and deduce that and ES = 2(abc + bcd + acd + abd) a + b + c + d, ET = ab + bc + cd + da + ac + bd. P R O B L E M S You roll two fair dice. Let X be the number of 2s shown, and Y the number of 4s. Write down the joint probability mass function of X and Y, and find cov (X, Y ) and ρ(X, Y ).... |
and Z = cos( 1 2 π X ). Find the probability of correct transmission of a symbol when the message passes through three similar independent channels. Let (X n; n ≥ 1) be a sequence of independent random variables such that P(X n = 1(X n = −1). Let U be the number of terms in the sequence before the first change of sign,... |
and Y, for E(X Y ) to exist in general. 14 Which of the following functions f (i, j) can be a joint probability mass function of two random 12 13 j |i| + | j, j < ∞ 1 ≤ i ≤ c, j ≥ 1, c an integer 1 ≤ i ≤ j < ∞. variables X and Y? (a) θ |i|+| j|; (b) θ i+ j ; (c) θ i+ j+2; (d) θ i+ j+1; (e) (i j − (i − 1) j )α ; (f) α(... |
integer and Z = min{c, X }, find E(Z ). (b) Find the distribution and expectation of min {X, Y }. Let X and Y have joint p.m.f. f (x, y) = C (x + y − 1)(x + y)(x + y + 1) ; m ≥ 1, n ≥ 1. Find the p.m.f. of X, the p.m.f. of Y, and the value of C. Let X and Y be independent random variables each with a geometric distribu... |
− λ)λm and P(Y = m) = (1 − µ)µm. P(X + Y = n) = (1 − λ)(1 − µ) λ − µ (λn+1 − µn+1), λ = µ. Find P(X = k|X + Y = n). Problems 229 (b) Find the distribution of Z = X + Y when λ = µ, and show that in this case P(X = k|Z = n) = Let X, Y, and Z be jointly distributed random variables such that each can 1/(n + 1). Bell’s In... |
; i ≥ 1) be a random walk with S0 = 0 and Sr = max1≤k≤n{Sk}. Show that P(Mn ≥ x) ≤ ( p 1 Xi. Define the maximum Mn = 3 U for large U. r q )x, x ≥ 0, and deduce that for p < q. E(Mn) ≤ q/(q − p), lim n→∞ An urn contains n balls such that each of the n consecutive integers 1, 2,..., n is carried by one ball. If k balls a... |
with probability 1 4. What is the probability that, on a day when n casualties arrive, exactly r require surgery? The number X of casualties arriving on weekdays follows a Poisson distribution with mean 8; 1 Xi be a random walk with S0 = a > 0, such that P(Xi = −1) = q, P(Xi = +2) = 3 then that is, for each day, P{X =... |
and deduce that E(Dk|Vk) = 1 3 Let (Sn; n ≥ 1) be a simple random walk, and let M be its maximum, M = maxn≥1{Sn}. (a) If S0 = 0, and p < q, show that m has a geometric distribution and find its mean. (b) If S0 is a random variable with distribution P(S0 = −k) = αβ k; k = 0, 1, 2,... find the distri- 1 ≤ k ≤ K. (K 2 − k2... |
g(E(X |Y )) ≤ E(g(X )|Y ). Jensen’s Inequality A bag initially contains r red and b blue balls, r b > 0. Polya’s urn (Example 2.7) revisited. A ball is drawn at random, its colour noted, and it is returned to the bag together with a new ball of the same colour. Let R(n) be the number of red balls in the bag after n su... |
−a + 1,..., b − 1, b}, where a, b > 0. Then provided s = 1, G(s) = b k=−a 1 a + b + 1 sk = s−a − sb+1 (a + b + 1)(1 − s). Notice that, by Theorem 4.3.4, we have from Definition 1 of G(s) that (3) G(s) = E(s X ); 232 6.1 Introduction 233 this is a particularly useful representation of G(s), and we use it a great deal in... |
b + c = 1. Now (i) If 0 ≤ c < 1, then we can write, for any n, G(s) = (a + bs)(1 + cs + · · · + (cs)n) + a + bs 1 − cs (cs)n+1 = a + (b + ac)s + (b + ac)cs2 + · · · + (b + ac)cn−1sn + bcnsn+1 + a + bs 1 − cs (cs)n+1. For |s| < c−1, we can let n → ∞ to obtain a series expansion of G(s). This has the required properties... |
X > n)sn = b−1 snP(X > n) − b−1 n = a sn+1P(X > n) n = a b−1 = sn(P(X > n) − P(X > n − 1)) + saP(X > a) n = a+1 − sbP(X > b − 1) = sa − b a P(X = n)sn = sa − G(s), as required. (8) Example 2 Revisited Here X is uniform on {−a,..., b}, and so TX (s) = s−a 1 − s − (s−a − sb−1) (a + b + 1)(1 − s)2 = (1 − s)(a + b)s−a + sb... |
notation. If (Ai ; i ≥ 1) is a collection of disjoint events with k i Ai =, then it is easy to show that (11) E(s X ) = i E(s X |Ai )P(Ai ). This result is often useful in finding E(s X ). If the random variables X and Y are jointly distributed, then in like manner we have (12) E(s X |Y = y) = k skP(X = k|Y = y) = k sk... |
random variable with parameter c. Such random variables arise quite naturally in applications. (15) Example A biased coin is tossed repeatedly until the first occasion when r consecutive heads have resulted. Let X be the number of tosses required. Find E(s X ). We suppose that the chance of a head is p, and note that i... |
k n! (n − k)! f (n)sn−k. (2) (3) (4) Second, it follows that G(s) determines the collection ( f (k); k ≥ 0). (5) Theorem (Uniqueness) for some G(s), we have Let X and Y have generating functions G X (s) and GY (s). If G X (s) = GY (s) = G(s) for |s| < 1, then X and Y have the same mass function. Proof This follows fro... |
(n − k)! 0 pk 1 ≤ k ≤ n, k > n. Hence, by (9), var (X ) = n(n − 1) p2 + np − (np)2) = npq. (12) Example: Poisson p.g.f. Let X have a Poisson distribution with parameter λ. Then G(s) = ∞ 0 e−λ λk k! sk = e+λ(s−1). Hence, we find that µ(k) = λk, for k ≥ 1. Moments can also be obtained from the tail generating function T ... |
(k) = kq k−1 (1 − q)k. We conclude this section with a note about defective probability mass functions. k = 0 f (k) < 1, then it still makes sense If X is a nonnegative random variable such that ∞ k k f (k) < ∞, to define the generating function G(s) = then G(1) = k k f (k). However, this is not now the expectation E(X ... |
= 1 Proof (a) Because X 1 and X 2 are independent, s X 1 and s X 2 are also independent. Hence, G Z (s) = E(s X 1+X 2) = E(s X 1)E(s X 2) by Theorem 5.3.8 = G1(s)G2(s). Part (b) is proved similarly. Let X and Y be independent and binomially distributed Example: Binomial Sum with parameters (m, p) and (n, p), respectiv... |
1 n j j z − zn+1 1 − z by Example 6.1.2. Hence, by Example 6.1.9, ∞ k= j zkP(S j ≤ k) = z n j (1 − zn) j (1 − z) j+1. Equating coefficients of zn on each side of (8) gives P(S j ≤ n) = 1 n j n j = P(Tn ≥ j + 1) by (7). Hence, P(Tn = j) = 1 n j− From (9), Tn has tail generating function n n z j P(Tn > j. Hence, from The... |
each die has p.g.f. G X (s) = 1 6 s(1 − s6) 1 − s, and so the p.g.f. of the total is given by Theorem 13 as G(s) = 2 − 1 6 s(1 − s6) 1 − s −1. (16) Example Let Z = N i=1 Xi, where f X (k) = −k−1 pk log(1 − p) ; k ≥ 1, 0 < p < 1 and f N (k) = λke−λ/k!; k ≥ 1, 0 < λ. Show that Z has a negative binomial mass function. So... |
es existing at time n be (X j ; 0 ≤ j ≤ Zn). Then we obtain the attractive and useful representation and, by Theorem 13, Zn+1 = Zn j=0 X j Gn+1(s) = Gn(G(s)). (18) (19) (20) 244 6 Generating Functions and Their Applications (This basic argument is used repeatedly in the theory of branching processes.) Hence, Gn+1(s) is... |
�0 = G(0) ≤ G(η) = η, and so ηn < η for all n. Hence, λ ≤ η and so λ = η. Once again, we conclude with a note about defective random variables. If X and Y are defective, then they are independent if P(X = i, Y = j) = P(X = i)P(Y = j) for all finite X and Y. Hence, we can still write and we can denote this by G X +Y (s) ... |
) (5) (6) Example ( n+k−1 k Let X have a negative binomial distribution with mass function f (k) = )q k pn, k ≥ 0. By the negative binomial expansion, G(s) = ∞ k=0 pn n + k − 1 k q ksk = n p 1 − qs, |s| < q −1. Then X has moment generating function M(t) = p 1 − qet n, t < − log q. Let us consider an example in which X ... |
MZ (t) = E(et(X +Y )) = MX (t)MY (t) by independence. G Z (1 + t) = E((1 + t)Z ) = G X (1 + t)GY (1 + t). Finally, we record the existence of yet another function that generates the moments of X, albeit indirectly. (11) Definition If the function κ(t) = log(E(e X t )) = log(MX (t)) 6.5 Joint Generating Functions 247 ca... |
used to get this result. We use this idea in the next example. (3) Example: de Moivre trials Each of a sequence of n independent trials results in a win, loss, or draw, with probabilities α, β, and γ respectively. Find the joint p.g.f. of the wins, losses, and draws, the so-called trinomial p.g.f. Solution or draw. Th... |
s=t=1. E(X ) = $ $ $ $ ∂G ∂s s=t=1, and E(Y ) = $ $ $ $ ∂G ∂t. s=t=1 Quite often, we write Gst (s, t) for ∂ 2G/∂s∂t, and so on; in this form, the covariance of X and Y is given by cov (X, Y ) = Gst (1, 1) − Gs(1, 1)Gt (1, 1). (8) (9) (10) Example 5.11 Revisited: Golf Recall that you play n holes of golf, each of which... |
+ qs−1 + r )n. Hence, dG W /ds = n( p − qs−2)( ps + qs−1 + r )n−1, and d 2G W ds2 ( ps + qs−1 + r )n−2 + 2nqs−3( ps + qs−1 + r )n−1. Therefore, = n(n − 1)( p − qs−2)2 var (W ) = n(n − 1)( p − q)2 + 2nq + n( p − q) − n2( p − q)2 = n( p + q − ( p − q)2). Finally, we record that joint generating functions provide a usefu... |
parameter λp. To find ρ(X, Y ), we first find the joint p.g.f. of X and Y, again using conditional expectation. Thus, E(s X yY ) = E(s X E(yY |X )) = E(s X ( py + 1 − p)X ) = exp (λ(s( py + 1 − p)− 1)). Hence, using (7), E(X Y ) = λ2 p + λp, and so, using the first part, ρ(X, Y ) = should compare this with the method of E... |
, we defined the convergence of a sequence of mass functions. This can be usefully connected to the convergence of corresponding sequences of generating functions. For sequences of probability generating functions, we have the following result, which we give without proof. (1) Theorem Let f (k) be a probability mass fun... |
→ F(x), as n → ∞, wherever F(x) is continuous. Then as n → ∞, for each j, µ j (n) → µ j < ∞, and (µ j ; j ≥ 1) are the moments of F(x). (ii) Conversely, for each j ≥ 1, as n → ∞, suppose that µ j (n) → µ j < ∞, where {µ j ; 1 ≤ j} are the moments of a unique distribution F(x). Then, as n → ∞, Fn(x) → F(x) wherever F(x... |
(S) = n k E(I j1... I jk ) 253 6.6 Sequences = = n k n k P (a given set of k all match!. µ(k) = 1; 0;! k ≤ n k > n, → 1 for all k Hence, as n → ∞. But these are the factorial moments of the Poisson distribution with parameter 1, and so as n → ∞ (9) P(X = k) → 1 ek!. We conclude with an example that leads into the mater... |
random variables, each uniform on [0, 1], required to produce a sum greater than 1. 6.7 Regeneration Many interesting and important sequences of random variables arise as some process evolves in time. Often, a complete analysis of the process may be too difficult, and we seek simplifying ideas. One such concept, which ... |
Examples 4.4.9, 4.19, and 5.4.13. Here is another elementary illustration. Three players A, B, and C take turns rolling a fair die in the order Example ABC AB... until one of them rolls a 5 or a 6. Let X 0 be the duration of the game (i.e., the number of rolls). Let A0 be the event that A wins, and let Ar be the event... |
interval between the (n − 1)th and nth occurrences of H. Thus, X 1 = min {n > 0: Hn occurs} X 1 + X 2 = min {n > X 1: Hn occurs} 256 6 Generating Functions and Their Applications Figure 6.2 A delayed renewal process. Here, X 1 = T1 = 3; X 2 = 2, X 3 = 5, X 4 = 1,... and so on. We suppose that (X n; n ≥ 2) are independ... |
. (The answer is yes, as we will see.) It is customary to make a further distinction between two different types of persistent renewal process. Definition process is said to be nonnull. If E(X 2) = ∞, then the process is said to be null; if E(X 2) < ∞, then the Note that E(X 2) is sometimes known as the mean recurrence ... |
s)D(s) = D(s) 1 − G(s) by (5). Thus, given G(s) [and D(s) in the delayed case] we can in principle find P(Hn), the probability that H occurs at time n, by expanding U (s) in powers of s. Conversely, given V (s) [and U (s) in the delayed case] we can find G(s) and D(s), and also decide whether the process is transient or ... |
X 1) = H (s). Show that for all n (10) P(Hn) = 1 G(1). Solution cients. Furthermore, by L’Hˆopital’s rule, From (6.1.6), we have that H (s) is a power series with nonnegative coeffi- H (1) = lim s↑1 −G(s) −E(X ) = 1 Hence, H (s) is a p.g.f. Finally, if D(s) = H (s) in (6), then 6.8 Random Walks V (s) = 1 E(X 2)(1 − s),... |
s) = E(s T j ). Find U (s) and G0(s), and show that the simple symmetric random walk is persistent null. Solution We give two methods of finding U (s) and G j (s). For the first, define T ∗ j = min {n: Sn = j − 1|S0 = −1} and let ˆT1 be a random variable having the same distribution as T1, but independent of T1. Because t... |
2 Now recall that by the negative binomial theorem ∞ 0 2n 1 2 2n n x n = (1 − x) −1 2, and (9) and (8) follow. Setting s = 1 shows that G0(1) = 1 (and U (1) = ∞) so H is persistent. However, d ds G0(s) = s (1 − s2) 1 2, and setting s = 1 shows that H is null; the expected number of steps to return to the origin is infi... |
2k f2k + 2mP(T > 2m) = s2mE(T ∧ 2m) = m−1 s2m m k=0 u2k + s + sU (s) = s2U (s) 1 − s2 2s2 (1 − s2) =. 3 2 m k=1 m u2k−2 + 2mu2m. 2ms2m−1u2m using (6.1.10) Finally, using the hitting time theorem (5.6.17), m s2mE(|S2m|) = 2 s2m 2kP(S2m = 2k) m m = 2 m d ds = 2s k=1 m k=1 s2m s2m 2m f2k(2m) m f2k(2m) m k=1 (17) (18) (19... |
1 2. Now using conditional expectation E(s L 2n ) = (E(E(s L 2n |T ) n = E(s L 2n |T = 2r ) f (2r ) + E(s L 2n |T > 2n) r = 1 Now, depending on the first step, ∞ r = n+1 f (2r ). P(L T = T ) = 1 2 and visits to zero constitute regeneration points for the process L 2n. Hence, we may rewrite (18) as = P(L T = 0), G2n(s) ... |
g.f. will produce the probability distribution (in most cases). We used them to study sums of independent random variables, branching processes, renewal theory, random walks, and limits. They have these properties: Connections: Tails: Uniqueness: Moments: MX (t) = G X (et ) and G X (s) = MX (log s). T (s) = ∞ n=0 snP(X... |
Negative binomial distribution ( ps Logarithmic distribution log(1−sp) log(1− p) 1−qs )n 1−sn+1 (n+1)(1−s) Checklist of Terms for Chapter 6 6.1 probability generating function tail generating function 6.2 uniqueness theorem factorial moments 6.3 sums and random sums branching process extinction probability 6.4 moment ... |
f (x) at a. This may or may not be equal to f (a). Accordingly, we define: Continuity The function f (x) is continuous in (α, β) if, for all a ∈ (α, β), f (x) = f (a). lim x→a Now, given a continuous function f (x), we are often interested in two principal questions about f (x). (i) What is the slope (or gradient) of f... |
��ned as a limit, but any general statements would take us too far afield. For well-behaved positive functions, you can determine the integral as follows. Plot f (x) on squared graph paper with interval length 1/n. Let Sn be the number of squares lying entirely between f (x) and the x-axis between a and b. Set In = Sn/n... |
a)g(a). x 1 (1/y)dy. + (iii) log x = Functions of More Than One Variable We note briefly that the above ideas can be extended quite routinely to functions of more than one variable. For example, let f (x, y) be a function of x and y. Then (i) f (x, y) is continuous in x at (a, y) if limx→a f (x, y) = f (a, y). (ii) f (x... |
nitely rich opponent.) Find E(s T10) = F1,0(s) (say). Solution This makes no difference to the fact that, by conditional expectation, for 0 < a < K, (a) Of course, Ta0 is defective in general because the walk may stop at K. Fa(s) = E(E(s Ta0|X 1)) = pE(s Ta0|X 1 = 1) + qE(s Ta0|X 1 = −1) = psE(s Ta+1,0) + qsE(s Ta−1,0)... |
K }. Show that E(s Ta K ) = λa 1 λK 1 − λa 2 − λK 2. (7) (8) (9) Exercise What is the p.g.f. of the duration of the game in the gambler’s ruin problem? Exercise What is E(s Ta0 ) for the unrestricted random walk? Exercise ever visits 0 and E(Ta0|Ta0 < ∞). For the unrestricted random walk started at a > 0, find the prob... |
Now the sum of Luke’s dice L 1 + L 2 has p.g.f. E(s L 1+L 2) = 1 6 (s + 2s2 + 2s3 + s4) 1 6 (s + s3 + s4 + s5 + s6 + s8) = G(s) on multiplying out the brackets. So Luke’s claim is correct. However, G(s) can be factorized as 36G(s) = s2(1 + s)2(1 − s + s2)2(1 + s + s2)2, where 1 + s + s2, 1 − s + s2 are irreducible, ha... |
pentagonal faces. to 12 inclusive. (b) Show that two such dodecahedral dice can be biased in such a way that their sum has the same distribution as the sum of the fair dice. Hint: Let f (x) = x + x 12 + (2 − + (7 − 4 √ √ 3)(x 2 + x 11) + (5 − 2 √ √ 3)(x 3 + x 10) 3)(x 4 + x 9) + (10 − 5 3)(x 5 + x 8) + (11 − 6 √ 3)(x ... |
, recall the basic identity of branching processes: namely, given Zn−1, Zn−1 Zn = X (r, n). r =1 Now let the cumulant generating function of Zn be κn(t). Then by conditional expectation, κn(t) = log (E(et Zn )) = log (E(E(et Zn |Zn−1))) ) * Zn−1 = log E E exp t X (r, n) |Zn−1 by (4) r =1 = log (E[(E(et X (1.1)))Zn−1]) ... |
) (10) Exercise (11) Exercise Find var (Zn) when µ = 1. Show that for n > m, E(Zn Zm) = µn−mE(Z 2 m). Deduce that when µ = 1, m n Find an expression for ρ(Zm, Zn) when µ = 1, and deduce that for µ > 1 as n, ρ(Zm, Zn) = 2. 1 (12) Exercise m → ∞, with n − m held fixed, ρ(Zm, Zn) → 1. (13) Exercise If r is such that r = E(... |
�n+1 − 1 − ρs(ρn − 1) ρn+2 − 1 − ρs(ρn+1 − 1) =. (4) by the induction hypothesis Because (2) is true for n = 1, by (1), the result does follow by induction. (b) Let ( ˆZn; n ≥ 1) be a collection of independent random variables such that ˆZn has the same distribution as Zn. Now Z ∗ n is the sum of the descendants of the... |
. (6) (7) 274 6 Generating Functions and Their Applications (8) Exercise (7) Continued Show that in this case ( p = 1 2 ), we have E Zn n |Zn > 0 → 1, as n → ∞. (a) Let X be any nonnegative random variable such that E(X ) = 1. Show that (9) Exercise E(X |X > 0) ≤ E(X 2). (b) Deduce that if p > 1 2 (10) Exercise When p ... |
= E X (z − 1)m m=0 X m = E((1 + z − 1)X ) = E(z X ) = Now equating coefficients of zm proves (b). n m=0 pm zm. Worked Examples and Exercises 275 (c) By Theorem 3.6.7 or Theorem 6.1.5, we have Hence, by (2), (3) Gq (z) = 1 − zG X (z) 1 − z. Gq (z) − 1 z = Gs(z − 1) − 1 z − 1. Equating coefficients of zm yields (c). (4) E... |
two tosses of the n were not H T, in which case X = n + 3. The probabilities of these two events are P(X = n + 1) pq and P(X = n + 3), respectively. Hence, P(X > n) p2q = P(X = n + 1) pq + P(X = n + 3); n ≥ 0. Multiplying (3) by sn+3, summing over n, and recalling (6.1.7), yields 1 − E(s X ) 1 − s The required result ... |
+ f 1(n + 3). Hence, Now we also have 1 − E(sY ) 1 − s pq 2s3 = pqs2G2(s) + qsG1(s) + G2(s). so solving (6), (8), and (9) for E(sY ) yields (2) as required. E(sY ) = G1(s) + G2(s) (5) (6) (7) (8) (9) (10) Exercise When p = 1 2, find E(X ) for all possible triples of the form HHH, HHT, etc. Comment on your results. (11)... |
bulb is called “unusual” if its life is shorter than a or longer than b, where a ≤ b. Let T be the time at which a bulb is first identified as being an unusual bulb. Show that (for integers a and b), E(s T ) = E(Ias X 1) + sbE(Ib) a I c b ) 1 − E(s X 1 I c. where Ia and Ib are the indicators of the events {X 1 < a} and ... |
. Evaluate this when f X (k) = qpk−1; k ≥ 1. Exercise Let L be the time until a light bulb has lasted longer than r. Show that (1) (2) (3) (4) (5) (6) (7) (8) E(s L ) = sr P(X > r ) r 1 − skP(X = k). 1 278 6 Generating Functions and Their Applications (9) Exercise A biased coin is tossed repeatedly; on each toss, it sh... |
n+1(s))]Zn = [Hn(s)]Zn by (2). Trivially, EMn = 1; it follows that Mn is a martingale. Show that η Zn is a martingale where η is the extinction probability defined in (6.3.20). Exercise If EZ1 = µ, show that Znµ−n is a martingale. Exercise Exercise Let Zn be the size of the nth generation of the branching process in whi... |
ET ≤ K < ∞, we can use the final part of the optional stopping theorem 5.7.14 to obtain EYT = 1, which is the required result. Exercise Show that ET ≤ K < ∞. Exercise (1). Show that, approximately, Let var X 1 > 0, and let T be the smallest n such that either Sn ≤ −a < 0 or Sn ≥ b > 0. Assume there is some t = 0 such t... |
= and show that (1) Qn(s) = EsYn, 0 ≤ s ≤ 1, Qn+1(s) = sG(Qn(s)). Solution copy of the branching process. Conditional on X 1, we may therefore write Note that each member of the first generation X 1 gives rise to an independent Y (1) n−1 + ˜Y (2) n−1 + · · · + ˜Y (X 1) n−1, where Y (i) finally, n−1 has the same distribu... |
λn − (µ − 2qs)µn 1 − 4spq. √, Problems P R O B L E M S 281 ∞ 0 f X (k)sk, where f X (k) = P(X = k); k ≥ 0. Show that: Find the probability generating function of each of the following distributions and indicate where it exists. (a) 1 ≤ k ≤ n. Let G(s) = ∞ (a) P(X < k)sk = sG(s)/(1 − s). 0 ∞ (b) P(X ≥ k)sk = (1 − sG(s))... |
it is not possible to load two traditional cubic dice in such a way that the sum of their scores is uniformly distributed on {2, 3,..., 12}. The three pairs of opposite faces of a fair die show 1, 2, and 3, respectively. The two faces of a fair coin show 1 and 2, respectively. (a) Find the distribution of the sum of t... |
+ (1 − λ)G2 is a p.g.f., and interpret this result. In a multiple-choice examination, a student chooses between one true and one false answer to each question. Assume the student answers at random, and let N be the number of such answers until she first answers two successive questions correctly. Show that E(s N ) = s2... |
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