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, that is to say, E(Tsi − Ts j |D j ) = µ ji. (6) (7) Substituting (7) into (6) gives (2). (c) Now let the outcomes of successive tosses be (Sn; n ≥ 1) and define the Markov chain (Yn; n ≥ 0) by Y0 = φ = s Y1 = S1 Y2 = S1S2 Yn = Sn−2Sn−1Sn; n ≥ 3. Thus, on the third step, the chain enters the closed irreducible subset o... |
T H is 2 3. (v) If the showman wants to make on average 30 p per game, what prize money should he offer: (a) if customers choose sequences at random? (b) if customers make the best possible choice? Remark This game was named Penney–Ante by W. Penney in 1969. Worked Examples and Exercises 461 9.17 Example: Poisson Proc... |
n(t), we find using (1) that = +λ(t)(z − 1)G. Because X (0) = 0, we have G(z, 0) = 1, and so, by inspection, (2) has solution G(z, t) = exp (z − 1) 0 t λ(u) du. This of course is the p.g.f. of the Poisson distribution with parameter "(t), as required. Now we note that P(T1 > t) = P(X (t) = 0) = G(0, t) = exp(−"(t)). (b)... |
∞. If λ(t) = λe−λt for λ > 0, show that limt→∞ P(X (t) = k) = 1/ek! Let (Yn; n ≥ 1) be independent and identically dis- X (t) n=1 Yn. + Show that E(eθ Z (t)) = exp( t 0 λ(u) du(M(θ) − 1)), where M(θ ) = E(eθ Y1 ). Exercise: Doubly Stochastic Poisson Process Suppose that X (t) is a nonhomogeneous Poisson process with r... |
of Y at t is E((ze−µ(t−u) + 1 − e−µ(t−u))Y ) = GY (ze−µ(t−u) + 1 − e−µ(t−u)). Worked Examples and Exercises 463 Hence, E(z S) = E(E(z S|U )) = 1 t t 0 GY ((z − 1)e−µv + 1) dv. Finally, recalling that the total number of particles at t are the survivors of a Poisson number of such batches we obtain (1). (2) (3) Exercis... |
we could proceed by writing down forward equations. However, it is neater to use the properties of the Poisson process directly as follows. We start by assembling some facts established in earlier sections. At time t, let C(t) be the time that has elapsed since the most recent disaster. From Example 8.17, we recall th... |
, E(X (t)) → ∞ as t → ∞, whereas if λ < δ, E(X (t)) → ν δ−λ. An expression for E(s X (t)) is found by following exactly the same sequence of successive conditional expectations. Thus, given that C(t) = x, N (x) = k, and Y1 = y1, this arrival initiates a simple birth process whose size at time y1 has generating function... |
them. Exercise parameters λ and µ. Show that limt→∞ E(X (t)) < ∞ if and only if δ > λ − µ. [See 9.21]. Suppose that each immigrant gives rise to a simple birth and death process with Suppose that an ordinary immigration–death process with parameters ν and µ is δ+µ. subject to disasters. Show that the population size X... |
then the product in (5) diverges to zero as n → ∞, for θ = 0, and fT (t) = 0. Furthermore, from (3), (6) pn(t) = P(N (t) = n) = P(Tn ≤ t) − P(Tn+1 ≤ t) = t 0 fn(u)du − t 0 fn+1(u)du. Now using (6), for θ < 0, we have ∞ 0 eθt pn(t) dt = 1 = 1 λn θ (Mn(θ) − Mn+1(θ)) on integrating by parts Mn+1(θ) using (5). Because Mn(... |
+ o(h). All individuals act independently in these activities. Write down the forward equations for pn(t) = P(X (t) = n) and show that if X (0) = I, then (1) E(s X (t)) = where θ(s) = (λs − µ)/(1 − s). I s + λt(1 − s) 1 + λt(1 − s) µ exp (t(λ − µ)) + θ (s) λ exp (t(λ − µ)) + θ (s) I if λ = µ if λ = µ, Worked... |
�s ∂t = (λs − µ)(s − 1) − (λ + µ)s ∂G ∂s ∂G ∂s ∂G ∂s = λs2 + µ. (2) Because X (0) = I, we have G(s, 0) = s I, and it is straightforward but dull to verify that (1) satisfies (2) and the initial condition X (0) = I. (3) Exercise Let η be the probability that the population ever falls to zero. Show that η = 1 µ λ I ... |
, where M(t) = exp{λD(t) − 1 2 λ2t − λµt}. (b) Let Tb be the first passage time of D(t) to b > 0. Show that Ee−θ Tb = exp{b(µ − % µ2 + 2θ )}. Solution we know to be a martingale from 9.9.27. (a) By definition D(t) = µt + W (t), so M(t) = exp{λW (t) − 1 2 λ2t}, which λ2 > (b) Because Tb is a stopping time, 1 = EM(Tb ∧ t).... |
��rst passage time of W (t) to b > 0. Show Worked Examples and Exercises 469 We can deduce from (5) that P(Tb < ∞) = 1, but see also the next Exer- Remark cise (6). Use Example 9.9.33 to show that P(Tb < ∞) = 1. Exercise Let X (t) and Y (t) be independent Wiener processes, and let Tb be the first passage time Exercise o... |
n, n). The result follows, on noting that E|v(X n, n)| < ∞. Let X n be a Markov chain with state space {0, 1,..., b}, such that X n is also a Exercise martingale. Show that 0 and b are absorbing states, and that if absorption occurs with probability one in finite time, then the probability of absorption at b is X 0/b. ... |
t is a martingale. Hence, for (1) 0 = EW (T ∧ n)2 − E(T ∧ n). Because W (t)2 ≤ a2 + b2, we can let n → ∞ in (1) to obtain, by using Example 9.9.33, E(W (T ∧ n)2) E(T ∧ n) = lim n→∞ ET = lim n→∞ = EW (T )2 = a2P(W (T ) = a) + b2P(W (T ) = b) = a2b b − a = −ab. − ab2 b − a (2) (3) Exercise Exercise Let Tb be the first pa... |
. s t Solution Let Z be the event that W (t) does have at least one zero in (s, t). Recall that Tw is the first passage time of W (t) to w, with density fT given in Corollary 9.9.25. By the symmetry of the Wiener process, if W (s) = w, then P(Z ) = P(Tw ≤ t − s) = P(T−w ≤ t − s). Therefore, conditioning on W (s), we hav... |
t − u) πu exp − 1 2 (y − c)2 t − u − 1 2 c2 u!, (4) (1) (2) Exercise Let U (t) be the time at which W (t) attains its maximum in [0, t]. [It can be shown that U (t) exists and is unique with probability 1.] Use the previous exercise, and the fact that U (t) = Tx on the event M(t) = x to show that M(t) and U (t) have jo... |
K (−α) Problems 473 For the problem in question, we can write S(T ) = ae Z, where a = S(0) and Z is normal σ 2)T, σ 2T ). Inserting these values of a, γ, and τ in the above, shows that N ((r − 1 2 v = Ee−r T (S(T ) − K )+ = e−r T {S(0)e(r − 1 2 σ 2)T + 1 2 σ 2T (τ − α) − K (−α)} = S(0) = S(0) as required. τ 2 − log−r ... |
1. (a) Is Zn irreducible? (b) If X and Y are reversible and also aperiodic, show that Z is reversible. Let X be a Markov chain. Show that the sequence (Xi ; i ≥ 0) conditional on X m = r still has the Markov property. Show that Definition 9.1.1 is equivalent to each of (9.1.6), (9.1.7), and (9.1.8) as asserted. Let Yn ... |
sequence (X nd ; n ≥ 0) is a Markov chain. Let A be a subset of the states of a regular chain X. Let T1 < T2 < T3 <... be the successive times at which the chain visits A. Show that (X Tr ; r ≥ 1) is a Markov chain. Let (X n; n ≥ 0) and (Yn; n ≥ 0) be Markov chains with the same state space S, and distinct transition ... |
. Show that X n is a Markov chain; find its transition probabilities and the stationary distribution. Let ( fk; k ≥ 0) be a probability mass function. Let the irreducible Markov chain X have transition probabilities, p jk = fk− j+1 if k − j + 1 ≥ 0, j ≥ 1 ∞ and p0k = p1k. Show that X is recurrent and nonnull if Let ( fk... |
}, having stationary distribution {π0, π1}. (a) Show that as τ →∞ P(X (0) = 1|X (−τ ), X (τ )) → π1. 25 t > 0. (b) Find cov (X (s), X (s + t)); (c) What is lims→∞ cov(X (s), X (s + t))? Let N (t) and M(t) be independent Poisson processes with parameters λ and µ, respectively. (a) Is N (t) +M(t) a Poisson process? (b) I... |
nd E(X ) and var (X ). Let (X n; n ≥ 0) be an irreducible Markov chain, with state space S, stationTruncation Again ary distribution (πi ; i ∈ S), and transition probabilities pi j. Let A be some subset of S, and suppose that (Zn; n ≥ 0) is a Markov chain with state space A and transition probabilities qi j = pi j pi A... |
f (s, x; t, y) of the Wiener process satisfies the Chapman– Kolmogorov equations 32 33 34 35 f (s, x; u, z) = R f (s, x; t, y) f (t, y; u, z) dy, s < t < u. Let W (t) be the Wiener process. Show that for s < t < u, 36 E(W (t)|W (u), W (s)) = [(t − s)W (u) + (u − t)W (s)]/(u − s) and var(W (t)|W (u), W (s)) = [u(t − s) ... |
is reversible in equilibrium with a stationary distribution such that π0 = q/(m + q). (c) Deduce that the expected number w of trips between occasions when I must set off in the rain is w = (m + q)/( pq). Show that w takes its smallest possible value s when Problems 477 p = m + 1 − [m(m + 1)] 1 2 and that in this case... |
a) ;. 1.8.3 1.8.4 1.9.2 1.9.3 1.9.4 11 18 5 12 b) (a) zero; 1.10.1 1.10.2 Let C j be the event that the jth cup and saucer match. Then, i = j, P(Ci ∩ C j ∩ Ck) = 1 24 ; P(Ci ∩ C j ) = 1 12 P(. 478 479 Hence, by (1.4.5), (1.4.8), (1.6.1), P 4 Appendix 12 − 4 · 1 24 + 1 24 = 3 8. i=1 ; P(C) = 25 91 1.11.1 P(A) = 36 91 1.... |
C) = 1 4 ; P(B3) = 1 8 ; P(B336)2 (a) P(B2) = 3 8 (c) P(B2) = 1 8 1 (a) 0; (b) ; (c) 4 + n11 p2 = n7 36 36 p3 = p2 + 2 (36)3 436 990 (b) (a) ; 526 990 1.12.3 1.12.4 1.13.1 1.13.2 1.13.3 1.14.3 p1 = 361 990 ∞ 1 2, p3 = p4 = 502 990 2r 1 2 = 2 3. r =0, p5 = 601 990 1.13.4 Let p j be the chance of winning when the first di... |
b) p23 = 4 p24. 12 (a) Use induction; (b) 1 9. (Hint: The cups can be arranged in 90 distinct ways.) (b) 1 4 ; (c) 3 16 ; (d) 1 8. 0.16 9 10 11 (a) 5 ; 6 35 36 24 19 36 2 3 1 3 13 14 15 17 19 21 in each case. (a 16 (a) 4 − 3 6 4 2 6 (b) n ↓ 0; 5 6 (bac) You would get the same answers. (c) Use induction. n 12 1 n 1 n ; ... |
=1 pkπk−1sn−k; (c) 1 − (1 − p)n; n(n − 1) p2(1 − p)n−2. 1 2 2.8.7 ∞ k=1 (1 − pk) > 0, if and only if ∞ k=1 pk < ∞. 2.9.1 α(1 − α) + α2(1 − γ )/(1 − (1 − α)(1 − γ )) 2.9.2 (2 − α + αγ (1 − γ ))−1 2.9.3 Biggles 2.10.1 2.10.2. (a) P(E) = 0.108; P(Ac|D) = 45 86 (b) P(E) = 0.059; P(Ac|D) = 5 451 p(µ(1 − π) + ν(1 − µ)) p(µ(1... |
2(λ + µ) = 0. When λ = µ. 2.12.3 (a) (µ3 + λ3)/(µ2 + λ2) (b) µn + λn µn−1 + λn−1 → max{λ, µ}, as n → ∞ 2.12.4 2.12.5 (a) µ λ + µ ; µn µn + λn (b) λ λ + µ → 1, as n → ∞, if µ > λ. 2.13.3 (a) 2 p1 3 p1 + p2(1 − p1) 2.13.4 Yes, if p1 < 1 2 2.14.1 3 p1 3 p1 + p2(1 − p1). (b) ; and p2 = p1 1 − p1 (b) a = b = c.. (a) a = 1,... |
b. 14 15 Let P(A) = α and P(B) = β. If the claim is false, then (1 − α)(1 − β) < 4 9 and αβ < 4 9 plane is empty, so the claim is true. and α(1 − β) + β(1 − α) < 4 9. The intersection of these three regions in the α − β 16 P(E|A = tail) = (1 − (1 − p)s−1)P(E|A = head) P(E) = pr −1(1 − (1 − p)s) 1 − (1 − pr −1)(1 − (1 ... |
0 = 0, pk n = pk−1 n−1 = 1 − m pk m n −1 k=0 60/k + m n = 1. (b) Choose x = 6. (c) 29 (i) pk m (ii) p0 pk m− 30 (a) 31 (a) 32 (i) 6 1 6 k=1 1 − p1 3 − p1 1 16 ; 33 (a) 1 (n!)2 (b) pt = ; (b) (1 − p1)2 2 + (1 − p1)2 ; (c) 1 − p1 3 − p1 − p2 − p3. (ii) 5 metres and one step; (iii) 5 metres. for 2 ≤ t ≤ j ∧ k ... |
all k such that 6 n(n − 1) 5!(n − 5)(n − 6) 2n(n − 1)(n − 2)(n − 3) 6(n − 3) n(n − 1) n! 1 k!(r − 2k)!(n − r + k)! nr max{0, r − n} ≤ 2k ≤ r ≤ 2n. 1 nr n−r p(m, n) − p(m − 1, n), where p(m, n) is given in Example 3.11. k=0 n r (−)k Mk, where r ≥ 2n and Mk = (n − k)r −k(n − k + 1)k. n n k 3.12.3 3.13.3 Rotational symme... |
18.7 (a) 1 n!r! n−r (−)k(n − r − k)!/k!; k=0 3.18.8 (a) zero; (b) zero; (c) e−2. (b) n k=0 (−)k (n − k)! n!k! ; (c) ( p(n, 0))2. Problems ; (b) 1 a!b!c! ; (c) 6 (a + b + c)!. 49 153 (a) 1 2 3 (a) 6a!b!c! (a + b + c)! 4!48! (12!)4 52! (13!)4 ; 26 13 ; (b) − 16 52 13 72 + 52 13 39 13 52 13 52 13 ; (d) 4 2 48 9 52 13 72 3... |
1 2m. 23 (a) (n − 4)(n − 3) (n − 2)(n − 1) whether Arthur sits at random or not. (b) Number the knights at the first sitting, and then use Problem 21. n (−)k 25 (a) k=0 n k (2n − k)! (2n)! → e− 1 2. (b) Problem 21 again. The limit is e−1. 1 = 1 2 5 (ii) 7 7 3 ; 2−7 + (iii 26 27 36 2n n ( pq)n = (4 pq)n − 1 2 n 1 ≤ (4 p... |
a) exp(λ(e − 1)); (b) e−λ; (c) λ; (d) λk. 4.11.3 Choose t to minimize L(t) = a x≤t (t − x) f (x) + bP(X > t). " " (a) ˆt = (c) ˆt = b a log a − log(a + bp) log q # # + 1; (b) ˆt = ) if this lies in [−n, n]; otherwise, ˆt = n if b/a > 2n. (What if a or b can be negative?). 4.11.4 (a) Any median of X ; (b) E(X ). 4.11.5 ... |
.4 a+b k=0 mk a + b k 2−(a+b) = 4.12.5 (i) p = q; k − (a + b) q − p + p (q − p)2 q p ; (a + b)(a + b − 1) − k(k − 1). (ii) p = 1 2 M + 1 j( j + 1)M (b) 4.13.1 ; (c) 1 − exp(−λj) − 1 2 λM(M + 1) 1 − exp k. q p −. 4.13.2 E(X A) = 2. ∞ 4.13.3 (b) n=1 m + 1 m + n = ∞; (c) e− 1 2 m(m+1) ∞ n=1 e− 1 2 λ(m+n)(m+n+1) < ∞. 4... |
− dr s) pr in nonincreasing order. 4.17.4 q 2(1 + p)/(1 − pq) = (1 − p2)2/(1 + p3) (1 − q 2)2/(1 + q 3) 4.17.5 2 1 − pq 4.17.6 E(X |Bc) = 4.17.7 P(A1|B(A2|B) = qp 1 + p. 4.17.8 Every number in [2, 3) is a median. 4.17.9 P(B) = p2(1 − q 3)/(1 − (1 − p2)(1 − q 2)) E(X ) = (1 + pq)(1 − 2 pq)/(1 − pq(1 − p)(1 + q)) 4.17.1... |
m 4−m > (2m − 2)... 4.2 (2m − 1)... 3.1 = 1 2m 2m m −1. 4−m Appendix 489 4.19.6 For the left inequality, prove and use the fact that for any collection of probabilities p1,..., pr, we have − pi log pi < − pi log pi. Equality holds when g(.) is a one–one map. For the right-hand inequality, note that i i i fi = exp(−cg(... |
conditioning gives E(T ) = 1 2 (1 + E(T )) + 1 4 (2 + E(T )) + 1 8 (3 + E(T )) + 3 8. Hence, E(T ) = 14. To find E(U ), consider the event that a sequence of n tosses including no H T H is followed by H T H. Hence, either U = n + 1 or U = n + 3, and so P(U > n) 1 8 = P(U = n + 1) 1 4 + P(U = n + 3). Summing over n give... |
!)3 ; 27 (a) pkq n−k (ii) 6 5 n − 1 k − 1 + 16 15 + 5 3 = 59 15. ; (b) (r p)k(1 − r p)n−k ; n − 1 k − 1 28 (1 − pr )−n; E(X ) = n(c) (1 − r )k pk(1 − p)n−k n k (i) Choose A if 2000(1 − p) < 1000 p (i.e., if ii) Choose A if 2000 < 1000 ).. 4 + 5 p (c) (1 − p)−1. 32 (a) pn; (b) (1 − p) pn−1; 1 + p as n → ∞. 29 31 M(n)/n ... |
pk = k=0 = 0 n 2k−2n. 2n − k − 1 n 2n − k n 1 (2n − k + 1) E(2n + 1 − H ) − 2n + 1 22n+1 2k−1−2n 2n n. = 1 2 So E(H ) = 2n + 1 22n q(1 − q)k−1ak 2n n − 1. k 44 45 With an obvious notation, m = 1 + 1 2 m13 (and two similar equations), also m12 = 1 + 5 6 1 m12 + 1 3 6 equations). Solve to get m. m2 + 1 6 m1 + 1 3 m3, al... |
if Xi ≥ k if Xi < k. 492 Appendix Then and f R(k) = E n i=1 I (Xi ≥ k) n i=1 Xi = nE I (X 1 ≥ k) n i=1 Xi E(R) = E k k I (X 1 ≥ k) 1 Xi /n n. 5.14.6 Let X be uniform on {x1,..., xn} and Y uniform on {y1,..., yn}. 5.15.1 S + (n − S) p − pγ 1 − γ p 5.15.2 N γ 5.15.3 ρ(N, S) = 1 2 γ (1 − p) 1 − γ p 5.15.4 j (i) P(T = k, ... |
and r > 0, E(V ) = r < 0, E(V ) = | p − q|. Likewise, if p > q, r 1 r p q E(V ) = p q 1 | p − q| 1 | p − q| 1 | p − q| ; for p < q and r < 0 r > 0. Appendix 493 P(SX = SY = 0) = n 1 42n (2n − 1)2(2n − 3)2... 12 (2n)2(2n − 2)2... 22 ≥ n k=0 n 1 2n = ∞. (2n)! (k!)2((n − k)!)2 = n 1 42n 2 2n n 5.20.8 E(V ) = = ... |
+1 pn; cov (U, V ) = (4 pq − 1)/( pq); ρ(U, V ) = −| p − q|. (ii) P(correct) = p3 + 3 p(1 − p)2. ; 2 10 You need results like n i=1 Then cov (X, Y ) = − n + 1 12 i 2 = 1 3 n(n2 − 1) + 1 2 ; ρ(X, Y ) = − 1 11 (b) a − 4a2; (c) E(X |Y = 0) = a 5 6 2. 2 cov (U, V ) = 7 6 (a) E(|X Y |) = E(|X ||Y |) = E(|X |)E(|Y |) < ∞; √ ... |
2 ai k 2 xi. 1 ai k 1 xi 17 (c) P(X 1 = x1) = 18 (a) 1 − pc 1 − p ; (b) P(min{X, Y } > n) = pn 1 pn β mq n 1 − β mq n. 26 Let (x, y, z) take any of the 8 values (±1, ±1, ±1). Then pβ 1 − qβ ; 2 so E(Z ) = αp βq (b) (a) 20. 1 1 − p1 p2 1 − x y = |1 − x y| = |(1 − x y)||(−x z)| because |(−x z)| = 1, = |(1 − x y)(−x z)| ... |
6 Appendix 495. 34 Use the argument of Theorem 5.6.7; pr 3 − r + q = 0; r = (− p + ( p2 + 4 pq) = e−22r r! ; mean = variance = 12. r 3n−r e−88n n! 3n−r 4n ; 4n r n r ∞ n r 35 n=r r +1 1 2 )/(2 p). 38 (a) P(M ≥ r ) = p q ; P(M = r ) = (b) P(M = r |S0 = −k) = αβ k 1 − p q βp q 40 For a neat method, see Example (6.6.6). ... |
we have E(X 2) = E(X 2|X > 0)P(X > 0) ≥ (E(X |X > 0))2P(X > 0) = E(X |X > 0)E(X ). (b) Hence, E(Znρ−n|Zn > 0) ≤ E 6.13.10 Let E(s Z ∗ n ) = G∗ n(s). Then ρ−2n Z 2 n E(s Z ∗ n t Z ∗ n+m ) = G∗ m−1(t)G∗ n(sGm(t)) → sG∗ m−1(t)Gm(t)(ρ − 1)/(ρsGm(t) − 1). 496 Appendix 6.14.4 Set z = y + 1 in (3) and equate coefficients. 6.1... |
= 1 − pr sr 1 − s + q pr sr +1. Problems s−n − sn+1 1 − s ; s = 0 (a) G = 1 n P(X ≤ k)sk+1 (b) G = 1 P(X < k)sk = n − sn+1 1 − s (c) G = 1 − (1 − s−1) log(1 − s); (d) G = 1 − 1 2 (e1 − s) log(1 − s−1) − 1 2 + cs−1 1 − cs−1 1 + cs 2n + 1 1 − cs |s| ≤ 1 1 2 3 (1 − s−1) log(1 − s);! |s| = 1 ; |s| = 1. (a) A p.g.f. wherev... |
G X, if and only if G = (1 + µ)−1 1 − s(1 + µ)−1 E(s N ) + s2 13 Use conditioning. So E(s N ) = s 4 2 (2 + E(N )) + 1 2 (1 + E(N )) + 1 4 (b) P(R = r ) = E(. (a) 14, so X is geometric. E(s N ) + s2 4 ; ; E(N ) = 6. 1 − 1 ps Hence, (1 − s)G R(s) = cd(1 − p)s + cs. ER = 1 2 15 By the independence var(H − T ) = var(H ) +... |
) G X (s) = s Do not differentiate this to find the mean! (ii) GY (s) = s. (m − 1)s m − s 3 − 2s 20. 21 Use L’Hopital’s rule. 22 Gn(s) = s 2 − Gn−1(s) ; Gn(1) = 1; G n(1) = n. 24 Differentiate. λ a 1 + a − s 25 27 28 X + Y, µ(k) = (λ + µ)k. ( ps + q)n( p + qs−1)msm = ( ps + q)m+n. (a) αp 1 − [1 − p + p(1 − α)s]t ; (b) p... |
q = β1 + β2 = 1 − p. Sn is symmetric so (a) E(T ) = ∞ and (c) Let Un = X n − Yn and Vn = X n + Yn, so (b) E(s T1 )|s=1 = 1. E(sU1 t V1 ) = 1 4 = 1 2 (st + st −1 + ts−1 + s−1t −1) (s + s−1) 1 2 (t + t −1). Hence, Un and Vn are independent simple random walks and E(s X T −YT ) = E(E(s VT |T )) s + s−1 2 = E T = F1 s + s... |
�� U, so you toss a coin twice and set if you get two heads otherwise. β y γ = 1 − exp − β y γ (a) (π(1 + x 2))−1; −∞ < x < ∞ (b) 2(π(1 + x 2))−1; 0 ≤ x < ∞. 7.11.7 7.12.7 First note that the coefficient of x n in Hn is 1, so Dn Hn = n! Now integrating by parts ∞ −∞ Hn Hmφ = [(−)m−1 Hn Hm−1φ]∞ −∞ + D Hn(−)m−1 Dm−1φ. ... |
( ˆx) + c ˆx = k + c ˆy + λ∗( ˆy). = 0 yields 0 = c − h ∂µ ∂ y 7.15.6 Let g(s) = log s − (s − 2)(s + 1). At s = 1, we have g(1) = 2 > 0; at s = e4, we have g(e4) = 4 − (e4 − 2)(e4 + 1) = (3 − e4)(2 + e4) < 0. There is thus at least one root. However, log s lies below its tangent and s2 − s − 2 lies above, so there can ... |
− E(FT" )) + (E( fT" ))2. Now because FTλ is DFR, we have 1 2 )2 ≤ E(− f T" )E(1 − FT" ) by Cauchy–Schwarz. 7.17.4 (i) d dt 1 t t 0 r (v)dv = r (tv)dv = 1 t 2 0 t [r (t) − r (v)]dv > 0 if r (v) > 0 for all v. Hence, IFR ⇒ IFRA. (ii) Use (7.8.6) and Theorem 7.8.7. 7.17.5 ∞ ∞ (i) E(T − t|At ) = by Definition 7.7.8 (iii).... |
B(a, b) x a(1 − x)b−2d x. 1 2 7.19.6 Remember (or prove) that (n) = (n − 1)! when n is an integer. 7.19.7 By the reflection principle, the number of paths that visit b on the way from (0, 0) to (2n, 0) is the same as the number of paths from (0, 0) to (2n, 2b), namely,. 2n n − b 2n n − b n2n+1 2n n Hence, the probabili... |
a, b) π/ sin−1 x 1/2 exp(− exp(−x)) 6 7 E(Y ) < ∞ for λ > 2a > 0 φ(x + a x )r (x + a x ) → e−a, 1 − x + a x (1 − (x)) = φ(x)r (x) using the properties of Mills ratio r (x), and φ(x). (i) F(x) = (b − 4π 2ml0x −2)/(b − a) for 2π (ii) E(X ) = (4π(ml0) 1 2 )/( a + √ √ b). 1 2 ml0 b ≤ x ≤ 2π 1 2. ml0 a 12 Choose xl such tha... |
ercises 8.11.3 By definition, fY |X (y|x) = f (x, y)/ f X (x) = 1 (2π(1 − ρ2)) ρ Xt + 1 2 1 2 − (y − ρx)2 2(1 − ρ2). exp 1 − ρ2 t 2. This is N (ρx, 1 − ρ2); therefore, E(etY |X ) = exp 8.11.4 By conditional expectation, E(esX +tY ) = E(E(esX +tY |X )) = E(e(s+ρt)X )e 1 2 (1−ρ2)t 2 = exp s + ρt (1 − ρ2)t 2 1 2 2 + 1 2 8.... |
x − y)n, so f = n(n − 1)(1 − x − y)n−2. 8.12.8 Given neither point is on the diameter, the density of the angle they make at the midpoint of the diameter is given by (1) with a = π. Hence, the expected area in this case is π 2(π − x) π 2 1 2 sin xd x = 1 π. 0 Given one on the diameter and one not, they are jointly uni... |
.4 ∞ 1 2. = 0 λα+β uα+β−1e−λudu. Integrate (2) remembering that (1) is a density and (α + β) 504 Appendix 8.14.5 Let X and Y be independent with density x − 1 2 e−x. Then U = X + Y has density 1 1 2 −2 1 2 e−u u 0 v− 1 2 (u − v)− 1 2 dv = πe−u −2. 1 2 The result follows. 8.15.6 Set (1 + x 2)−1 = v and use Exercise 5. 8... |
N = k is binomial with parameters k and p. Hence, E(s Z |N = k) = ( ps + 1 − p)k; hence, E(s Z ) = E(( ps + 1 − p)N ) = exp(24λ( ps + 1 − p)) = exp(λE(Y )(s − 1)). 8.17.5 Argue as in (4). Given N (t) = n, then R1 and R2 have a trinomial mass function with p.g.f. E(x R1 y R2 |N = n) = ( p1x + p2 y + 1 − p1 − p2)n. Henc... |
zg(z). (a) c = 1; (b) 2e−1; f Z (z) = (c). z. Independence is impossible. 0 (a) c = (2π)−1; (b) y 0 (a + y2)− 3 2 dy = y a(a + y2) 1 2, so X has a Cauchy density. f (x, y) = 4y3x(1 − x) for 0 < x < 1, 0 < y < 1 x 1 with density 6x(1 − x) by forming U 2 + V 2 ) and accepting it as a value of X if 2 + V U 6x(1 − x). Hen... |
� 4π dθ = 1 4π. π/2 Likewise, P (any one of the three remains) = 3P (given one remains) π 2 < θ < π, (π − θ )/(2π). π = 3 θ sin θ 4π 4π 16 P(U = X ) = P(Y > X ) = π/2 dθ = 3(π − 1) λ λ + µ.. 17 Use induction 1 18 2 (a) (X + Y ) by symmetry. (b) E(X |X + Y = V ) = 19 P(T ≥ j + 1) = P j i=1 σ 2 + ρσ τ σ 2 + 2ρσ τ + τ 2 V... |
−λt λ 0 + 1 λ2 e−λ(t−u) t 1 u λ − te−λt λ − 1 λ − e−λt λ2 λe−λudu − 1 2 t 2e−λt. So cov (C(t), X 1) = 1 2 t 2e−λt. 24 Let the condition be A. Now we notice that P(X ≤ x, A) ∝ x x α−1e− α−1 α e− x α d x = x α−1e−x. x α e− α−1 ≤ x α Hence, the result is true by the rejection method Example 8.16, provided that α x ≤ 1 for... |
1. E(et Z ) = E(et X )E(e−t V ) = (e − 1)et−1 1 − et−1 1 e−tv ev e − 1 0 dv = 1 1 − t. Hence, Z is exponential with parameter 1. Alternatively, you can find FV directly. X Z has a beta density. 29 30 Consider their joint m.g.f. E(es X +t Z ) = E(e(s+tρ)X +t(1−ρ2) 35 f (x) = e−x. Calculate: E(etU (X 1+X 2)) = E E(etU (X... |
1, 1), arguments similar to those of the example show that E(s T ) = UV (s) U (s), where and U (s) = 1 2 1 1 − s2 + 1 1 − ( p − q)2s2 UV (s) = s2 2 1 1 − s2 − ( p − q)2 1 − ( p − q)2s2, which yields E(T ) after some plod. More simply, by conditional expectation we have E(T ) = 1 + p(1 + pE(T )) + q(1 + qE(T )), which y... |
X nb) E(X n(X 0) − βm α + β + βm α + β. 9.14.8 Because vn is a renewal sequence, there is a Markov chain Vn such that vn = P(Vn = 0|V0 = 0). Let Un and Vn be independent. Then ((Un,Vn); n ≥ 0) is a Markov chain and unvn = P((Un,Vn) = (0,0)|(U0,V0) = (0,0)), thus (unvn; n ≥ 0) is a renewal sequence. 9.14.9 Consider the ... |
�nite n and m. Hence, p j j (m + r + n) ≥ p ji (m) pii (r ) pi j (n). Now sum over r to get p j j = ∞ if pii (r ) = ∞. So if i is persistent so is j. Interchange the roles of i and j. If j has period t, let r = 0 to find that when p j j (m + n) ≥ 0, m + n is a multiple of t. Hence, the right-hand side is nonzero only wh... |
= E((Eeθ Y )X ). You can also get this using forward equations with a lot more work. 510 9.17.7 E(z X ) = E(E(z X |Y )) = E Appendix " (z − 1) exp # t 0 Y (u)du t G X (z)|z=1 = M (0) = X (1))2 = M (0) − (M (0))2 + M (0) E(Y (u))du. 0 G X (1) + G X (1) − (G 9.18.2 λ µ (z − 1)(1 − e−µt ) 9.18.3 The forward equations are... |
(t)eθt dt, as required. 0 p k(t) = λn pn(t), so 9.20.10 Use (5) and partial fractions. 9.21.3 η = lim P(X (t) = 0) = lim t→∞ t→∞ 9.21.4 P(T > t) = 1 − G(0, t). If λ < µ then extinction is certain, so G(0, t) E(T ) = P(T > t)dt = 0 (µ − λ) exp((λ − µ)t) µ − λ exp((λ − µ)t) dt = 1 λ log µ µ − λ. ∞ However, if λ > µ, then... |
h) = µi h pii (h) = 1 − (ν + λi)h − µi h + o(h). Hence, dpn dt = µ(n + 1) pn+1 − (ν + (λ + µ)n) pn + (ν + λ(n − 1)) pn−1. Setting dpn dt = 0, you can check that the given πk satisfies the resulting equations. Problems 1 (α) X n is, with pi j = P(X 1 = j). (β) Sn is, with pi j = P(X 1 = j − i). (γ ) Mn is, with pi j = P(... |
ρk fk. Hence, it is a ρn fn = 1. It follows that there is a Markov chain such that 14 In the obvious notation, we require Q2 = 1 4 (PX + PY )2 shows that this requires (PX − PY )2 = 0. Hence, W is Markov if PX = PY. 17 No. Pick j = s = i, then P(X n+1 = j|X n = s,X n−1 = i) = pi j (2) = P(X n+1 = j|X n = s). 18 The la... |
1 if π0 > 0 and G(1) < 1. 20 Seek a stationary distribution that satisfies πk = ∞ j=k−1 π j f j−k+1 for k ≥ 1, with π0 = ∞ ∞ π j j=0 i= j+1 fi. θ r fr = G(θ). In an optimistic spirit, we seek a solution of the form π j = (1 − θ )θ j, θ k = ∞ j=k−1 θ j f j−k+1 = ∞ r =0 θ k+r −1 fr = θ k−1G(θ ), Let giving and ∞ θ j θ 1 ... |
(−τ ) = i,X (τ ) = j) = P(X (−τ ) = i) pi1(τ ) p1 j (τ ) P(X (−τ ) = i) pi j (2τ ) (b) P(X (s + t) = X (s) = 1) − P(X (s) = 1)P(X (s + t) = 1) as s → ∞. (a) Yes, with parameter λ + µ; 25 26 "(s) 28 Let A(t) be any event defined by X (s) for s ≤ t. Then by the Markov property π1π j π j (b) No. →. (c) → π1( p11(t) − π1),... |
and Stirzaker, D.R. (2001) Probability and random processes (3rd edn.) Clarendon Press, Oxford. (2001) One thousand exercises in probability, Clarendon Press, Oxford. Ross, S.M. (2003) Introduction to probability models (8th edn.) Academic Press, Orlando. Combinatorics A classic text on combinatorics for probabilists ... |
–74, 87 disasters, 463–465 family planning, 44–45 forward eqns, 428–431 general, 465–466 bivariate generating fns, 90 bivariate normal density, 342, 348–349, 352, 359–360, 373–376, 440 bivariate rejection, 392 Black-Scholes formula, 472–473 bookmaker example, 448–449 Boole’s inequalities, 36, 39, 50 bounded convergence... |
otto, 476 Murphy’s law in, 46–47 notation, 27 runs, 103–104, 111, 275–276 simulations, 301 visits of a rw, 219 colouring, 106 combinations, 86–87 complacency example, 67–68 complements, 17–18, 26, 39 composition, 302 compound Poisson pr, 462 conditional density, 310–312, 319, 355–361, 365–366, 373, 440 conditional dist... |
373 countable additivity, 34–35 countable sets, 27 countable union, 27 counting, 83–113 coin tosses, 103–104 colouring, 106 combinations, 86–87, 95 derangements, 88, 96 dice example, 84 first principles, 83–84 generating fns, 90–93, 95 Genoese Lottery, 98–99 identity example, 102–103 inclusion-exclusion, 87–88 lottery ... |
See joint density Laplace, 320 marginal, 341, 356, 371 mixtures, 297, 318, 335 mgf and, 306–310 multinormal, 373–374 multivariate normal, 373–374, 394 normal. See normal density order statistics, 363–364 Pareto, 303 standard normal, 296, 299 sums, products, and quotients, 348–351 triangular, 228 trinormal, 373–374, 38... |
120 sample space and, 287–288 sequences of, 130–131 standard trivariate normal, 373–374 stationary, 412–418, 434–435, 436, 449–450, 454–455 triangular, 228 trinormal, 374, 387–388 Index 519 two-sided exponential, 294–295, 297 uniform, 123, 178, 288, 291, 338–339, 345–346 dog bites, 139–141, 143–144 dogfight example, 68–... |
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