清华大学：《应用随机过程》课程教学资源（讲义）第六讲习题参考答案

1(3)将AUB拆成互不相交的集合的并:AUB=AU(B-A)=AUB-AB,再由 概率的可加性可得 (4)因为BCA,所以A=B∪A-B,由此可得 13当A∩B≠时,a(4)共有16个元素。当A∩B=时,a(A)有8个元素。不难写出 14题目中A3={(a,b):a,b∈R} (1)因为切n,ACAn所以有Ac∩An 下面证明∩AnCA。反证法:假设∈∩4n,xA,那么x-a>0,取n=「xa1+1 则x>a+,由此可得xAn,矛盾。所以∩AnCA。由此可得A=∩An。类似可 n=1 得B=UBn (2)因为a∈R(-x,=∪(=n,a,所以A1c叫(A2)。因为va,b∈R(a=(-∞ (-∞,a],所以A2(A1)。综上可得σ(A1)=σ(A2) ∩(-∞,a+) (一n,a+ m=ln ∩UL-n,a+ va,b∈R,(a,b]=(-∞,b-(-∞,a] (a,b) (-x,b-1-(- a,b=(-∞,b 由此可得a(A1)=0(41),2≤i≤5 1.6(3)因为A2k+1CA2k,所以 A ∫U=n/2Axk=(l,l+ 为偶数 Uk=(x+1)2A2k=(,+n+ln为奇数

1.1 (3) òA S B ¤p؃8Ü¿µA S B = A S (B − A) = A S B − AB§2d VÇŒ\5Œ" (4) ϏB ⊂ A§¤±A = B S A − B§ddŒ" 1.3 A T B 6= φž§σ(A)
k16‡ƒ"A T B = φž§σ(A)k8‡ƒ"ØJÑ 1.4 K8¥A3 = {(a, b) : a, b ∈ R} (1) Ϗ∀n, A ⊂ An¤±kA ⊂ \∞ n=1 An" e¡y² \∞ n=1 An ⊂ A"‡y{µb∃x ∈ \∞ n=1 An, x /∈ A§@ox − a > 0§n = d 1 x−a e + 1§ Kx > a + 1 n§ddŒx /∈ An§gñ"¤± \∞ n=1 An ⊂ A"ddŒA = \∞ n=1 An"aqŒ B = [∞ n=1 Bn" (2) Ϗ∀a ∈ R,(−∞, a] = [∞ n=1 (−n, a]§¤±A1 ⊂ σ(A2)"Ϗ∀a, b ∈ R,(a, b] = (−∞, b] − (−∞, a]§¤±A2 ⊂ σ(A1)"nþŒσ(A1) = σ(A2) (3) ∀a ∈ R,(−∞, a] = [∞ n=1 (−n, a] = \∞ m=1 (−∞, a + 1 m ) = \∞ m=1 [∞ n=1 (−n, a + 1 m ) = \∞ m=1 [∞ n=1 [−n, a + 1 m ) = [∞ n=1 [−n, a] ∀a, b ∈ R,(a, b] = (−∞, b] − (−∞, a] (a, b) = [∞ n=1 (−∞, b − 1 n ] − (−∞, a] [a, b) = [∞ n=1 (−∞, b − 1 n ] − [∞ m=1 (−∞, a − 1 m ] [a, b] = (−∞, b] − [∞ m=1 (−∞, a − 1 m ] ddŒσ(A1) = σ(Ai), 2 ≤ i ≤ 5 1.6 (3) ϏA2k+1 ⊂ A2k§¤± [∞ k=n Ak = ½ S∞ k=n/2 A2k = ( 1 4 , 1 2 + 1 n ] nóê S∞ k=(n+1)/2 A2k = ( 1 4 , 1 2 + 1 n+1 ] nÛê 1

∩∪4=∩ n=l ken 由此可得P A F F Ak =o n=1k=7 由此可得P(∪U∩ 1.10(1)由极限的定义 Ww: lim Xn=XI 兮Ⅶm≥1m≥1k≥n|Xk(u)-Ⅺ<1/m 台Ⅶm21n21u∈∩{:xk-x1<1/m 台m21u∈U∩{:|Xk-x1<1/m 台u∈∩∪∩{:X-x<1/m} 1.14 P(M S PIN EX F(r) ddf(z) dF(a)dy=/(1-F(a)dx II"F(a)= my -l dy)dF(a) dF(a)ny"dy n-(1-F(a)dx P(N=n)P(S=HN=n P(N=n)P(∑X1=N=n) n=k P(N=n)P∑X2=k)

⇒ \∞ n=1 [∞ k=n Ak = \∞ n=1 ( 1 4 , 1 2 + 1 n ] = (1 4 , 1 2 ] ddŒP à \∞ n=1 [∞ k=n Ak ! = F( 1 2 ) − F( 1 4 )" \∞ k=n Ak = φ ⇒ [∞ n=1 \∞ k=n Ak = φ ddŒP à [∞ n=1 \∞ k=n Ak ! = 0" 1.10 (1) d4½Â ω ∈ {ω : limn→∞ Xn = X} ⇔ ∀m ≥ 1 ∃n ≥ 1 ∀k ≥ n |Xk(ω) − X| n) EX = Z ∞ 0 xdF(x) = Z ∞ 0 ( Z x 0 dy)dF(x) = Z ∞ 0 Z ∞ y dF(x)dy = Z ∞ 0 (1 − F(x))dx E(Xn ) = Z ∞ 0 x n dF(x) = Z ∞ 0 ( Z x 0 nyn−1 dy)dF(x) = Z ∞ 0 Z ∞ y dF(x)nyn−1 dy = Z ∞ 0 nxn−1 (1 − F(x))dx 1.18 P(ξ = k) = X∞ n=k P(N = n)P(ξ = k|N = n) = X∞ n=k P(N = n)P( Xn i=1 Xi = k|N = n) = X∞ n=k P(N = n)P( Xn i=1 Xi = k) 2

p2(1-p) 入e n=k k!(n-k) a pk p5(1-p) k!(n-k) A(1-p) 所以E() 1.19(1) P(N1+N2=n)=>P(N+N2=n,N,=k) k=0 ∑P(N2=n-k)P(N1=k) (n一k) P(N1=kNi+ N2=n P(NI= h)P(N2=n-k) P(N+N2=n 入2 P(N1+N2|N3)=P(N1N3)+P(N2|N3)=P(N1)+P(N2)=P(N1+N2) (4) E(N1|N1+N2=n) kCk

= X∞ n=k λ ne −λ n! C k np k (1 − p) n−k = X∞ n=k λ ne −λ k!(n − k)!p k (1 − p) n−k = e −λλ kp k k! X∞ n=k λ n−k (n − k)!(1 − p) n−k = X∞ n=k λ ne −λ k!(n − k)!p k (1 − p) n−k = e −λλ kp k k! e λ(1−p) = (λp) k k! e −λp ∼ P o(λp) ¤±E(ξ) = λp, D(ξ) = λp" 1.19 (1) P(N1 + N2 = n) = Xn k=0 P(N1 + N2 = n, N1 = k) = Xn k=0 P(N2 = n − k)P(N1 = k) = Xn k=0 λ k 1 e −λ1 k! λ n−k 2 e −λ2 (n − k)! = (λ1 + λ2) n n! e −(λ1+λ2) (2) P(N1 = k|N1 + N2 = n) = P(N1 = k)P(N2 = n − k) P(N1 + N2 = n = C k n µ λ1 λ1 + λ2 ¶k µ λ2 λ1 + λ2 ¶n−k (3) P(N1 + N2|N3) = P(N1|N3) + P(N2|N3) = P(N1) + P(N2) = P(N1 + N2) ¤±N1 + N2†N3Õá" (4) E(N1|N1 + N2 = n) = Xn k=0 kP(N1 = k|N1 + N2 = n) = Xn k=0 kCk n µ λ1 λ1 + λ2 ¶k µ λ2 λ1 + λ2 ¶n−k = nλ1 λ1 + λ2 Xn k=1 C k−1 n−1 µ λ1 λ1 + λ2 ¶k−1 µ λ2 λ1 + λ2 ¶n−k = nλ1 λ1 + λ2 3

所以E(N1N1+N2)=、入(M1+N2) +入2 E(M1+N2N1)=E(N1N1)+E(N2|N1)=M1+E(N2)=N1+A2 120这里只给出第三小题的证明,前两个类似可得 E(E(ALIB)IIB, Ic)=E(P(AB)IB+P(AB)IBIIB, Ic E(P(AB)IBIIB, Ic)+E(P(AB)IBIIB, Ic) P(AB)IB+ P(AB)IE E(IAIB) 124X,Y的联合概率密度函数为 由X=UV/(1+V),Y=U/(1+V)得 Jacobi矩阵的行列式为: /(1+v) 由此可得 fuu,v)(u, U) (1+)2>0.> 分别积分可得边缘分布为 fu(u)=ue"(u>0) fv(u) P(Z≤ P(X+Y≤z) ∑P(X+y≤2X=k) 由此可得Z的分布函数,再对分布函数求导可得概率密度函数 E(E(XY, ZY ∑E∑E(XY=,Z=2)(y=y)1(z=a)Y=) ∑E(XY 2k)E((2z=4)Y=v)(y=y2) 因为当计≠j时E((y=m)1(z=4)Y=y)=0 ∑∑zP(X=zY= Y=yj)I0

¤±E(N1|N1 + N2) = λ1 λ1 + λ2 (N1 + N2)" E(N1 + N2|N1) = E(N1|N1) + E(N2|N1) = N1 + E(N2) = N1 + λ2 1.20 ùp‰Ñ1nKy²§cü‡aqŒ" (3) E(E(IA|IB)|IB, IC ) = E(P(A|B)IB + P(A|B)IB |IB, IC ) = E(P(A|B)IB|IB, IC ) + E(P(A|B)IB |IB, IC ) = P(A|B)IB + P(A|B)IB = E(IA|IB) 1.24 X, Y éÜVÇݼꏵ f(X,Y )(x, y) = e −(x+y) I(x>0,y>0) d X = UV /(1 + V ), Y = U/(1 + V ) JacobiÝ 1ªµ J = −u/(1 + v) 2 ddŒµ f(U,V )(u, v) = e −u u (1 + v) 2 I(u > 0, v > 0) ©OÈ©Œ>©Ùµ fU (u) = ue−u I(u > 0) fV (v) = 1 (1 + v) 2 I(v > 0) 1.25 P(Z ≤ z) = P(X + Y ≤ z) = Xn k=0 P(X + Y ≤ z, X = k) = Xn k=0 P(Y ≤ z − k)P(X = k) ddŒZ©Ù¼ê§2é©Ù¼ê¦ŒVÇݼê" 1.28 E(E(X|Y, Z)|Y ) = X i E( X j, k E(X|Y = yj , Z = zk)I(Y =yj )I(Z=zk) |Y = yi)I(Y =yi) = X j, k E(X|Y = yj , Z = zk)E(I(Z=zk) |Y = yj )I(Y =yj ) Ϗ i 6= j žE(I(Y =yj )I(Z=zk) |Y = yj ) = 0 = X j, k X i xiP(X = xi |Y = yj , Z = zk)P(Z = zk|Y = yj )I(Y =yj ) 4

∑∑zP(X yj)I(Y=y) ∑∑P(X=|Y=)1y=m) ∑E(XY=)1y E(XY 另一个类似可得。 22证明: P(N(s=kN(t)=n P(N(s)=k, N(t)=n) P(N(t)=n) P(N(s)=kP(N(t-s)=n-k) P(N(t=n) (A8)e-As((=9)m-e-A(t-) 23解:(1) E(N(N(s+t)) E(N(t)2+N(t(N(s+t)-N())) E(N()2)+E(N(D)E(N(s) (A)2+M+A2 E(N(s+t)IN(s)) E(N(S))+E(N(s +t)-N(SIN(s) N(s)+E(N(IN(s)) (s)+At P(E(Ns +DIN(S)=n+At)=P(N(s)=n)=Cs)e-As 25解:设U1~U0,,i=1,2,……,n,则其顺序统计量与S1,S2,…,Sn在 N(t)=n的条件下的分布相同 E(SkIN(t)=n)=E(U())=-,(k n

= X j, k X i xiP(X = xi , Z = zk|Y = yj )I(Y =yj ) = X j X i xiP(X = xi |Y = yj )I(Y =yj ) = X j E(X|Y = yj )I(Y =yj ) = E(X|Y ) ,‡aqŒ" 2.2 y²µ P(N(s) = k|N(t) = n) = P(N(s) = k, N(t) = n) P(N(t) = n) = P(N(s) = k, N(t) − N(s) = n − k) P(N(t) = n) = P(N(s) = k)P(N(t − s) = n − k) P(N(t) = n) = (λs) k k! e −λs (λ(t−s))n−k (n−k)! e −λ(t−s) (λt) k k! e−λt = C k n( s t ) k (1 − s t ) n−k (0 ≤ k ≤ n) 2.3 )µ(1) E(N(t)N(s + t)) = E(N(t) 2 + N(t)(N(s + t) − N(t))) = E(N(t) 2 ) + E(N(t))E(N(s)) = (λt) 2 + λt + λ 2 ts (2) E(N(s + t)|N(s)) = E(N(s)) + E(N(s + t) − N(s)|N(s)) = N(s) + E(N(t)|N(s)) = N(s) + λt P(E(N(s + t)|N(s)) = n + λt) = P(N(s) = n) = (λs) n n! e −λs 2.5 )µ Ui ∼ U[0, t], i = 1, 2, · · · , n§KÙ^SÚOþ† S1, S2, · · · , Sn 3 N(t) = n ^‡e©ÙƒÓ" E(Sk|N(t) = n) = E(U(k)) = kt n + 1 , (k ≤ n) 5

2.6 P(W(t)≤x) P(N(t+x)-N(t)>0) 1-P(N(t+x)-N(t)=0) 当x0) P(N(t)-N(t-x)=0) 当x≥t时,P(V(t)≤x)=1 所以 1 00,00,N(t)-N(t-y)>0) P(N(t+x)-N(t)>0)P(N(t)-N(t-y)>0) P(W(t)≤x,V(t)≤y) P(N(t+x)-N()>0) P(N(t+x)-N(t)>0) 210解:由题意知A=1/2,那么 P 2.15解:由基本更新定理 E(N()) 2 E(X 216解:令Y1=X-6,则Y~Ex(),所以由Y2生成的是泊松流 P(N(t)≥n)

2.6 )µx ≥ 0ž P(W(t) ≤ x) = P(N(t + x) − N(t) > 0) = 1 − P(N(t + x) − N(t) = 0) = 1 − e −λx x 0) = 1 − P(N(t) − N(t − x) = 0) = 1 − e −λx x ≥ tž§P(V (t) ≤ x) = 1" x ≤ 0ž§P(V (t) ≤ x) = 0" ¤± P(V (t) ≤ x)    1 − e λx , 0 0, 0 0, N(t) − N(t − y) > 0) = P(N(t + x) − N(t) > 0)P(N(t) − N(t − y) > 0) = (1 − e −λx)(1 − e −λy) y ≥ tž P(W(t) ≤ x, V (t) ≤ y) = P(N(t + x) − N(t) > 0) = P(N(t + x) − N(t) > 0) = 1 − e −λx 2.10 )µdK¿ λ = 1/2§@o P(Sn > x) = Z ∞ x λ(λt) n−1 (n − 1)! e −λtdt = 1 2 n(n − 1)! Z ∞ x t n−1 e −t/2 dt 2.15 )µdč#½n limx→∞ E(N(x)) x = 1 E(X + Y ) = 2 3 2.16 )µ- Yi = Xi − δ §K Yi ∼ Ex(ρ)§¤±d Yi )¤´Ñt6" P(N(t) ≥ n) 6

P(Sn≤t) PCX≤t P∑Y≤t-n6) [(t-n6) (s+) 入入 )( 218生:当0<x<t时 P(B4≤x)=P(W(t)+V(t)≤x) P(W()≤x-y|V(t)=y)dP(v(t)≤y) P(W()≤x-y)dP(V(t)≤y) De ydy P(4≤x)=P(W(t)+V(t)≤x) P(W(t)≤x-yV(t)=y)dP(V(t)≤y) P(W(t)≤x-ydP((t)≤y) (W(t)≤x-yV(t) P(V(t)≤ Ate-A +(1

= P(Sn ≤ t) = P( Xn i=1 Xi ≤ t) = P( Xn i=1 Yi ≤ t − nδ) = 1 − nX−1 k=0 [λ(t − nδ)]k k! e −λ(t−nδ) 2.17 )µ F˜(s) = Z ∞ 0 e −stλ 2 te−λtdt = λ 2 (s + λ) 2 ⇒ m˜ (s) = F˜(s) 1 − F˜(s) = λ 2 s(s + 2λ) ⇒ m0 (t) = λ 2 − λ 2 e −2λt ⇒ m(t) = (λ 2 t + 1 4 e −2λt − 1 4 )I(t≥0) 2.18 )µ 0 < x < t ž P(βt ≤ x) = P(W(t) + V (t) ≤ x) = Z x 0 P(W(t) ≤ x − y|V (t) = y)dP(V (t) ≤ y) = Z x 0 P(W(t) ≤ x − y)dP(V (t) ≤ y) = Z x 0 (1 − e −λ(x−y) )e −λydy = 1 − e −λx − λxe−λx  x ≥ t ž P(βt ≤ x) = P(W(t) + V (t) ≤ x) = Z x 0 P(W(t) ≤ x − y|V (t) = y)dP(V (t) ≤ y) = Z x 0 P(W(t) ≤ x − y)dP(V (t) ≤ y) = Z t − 0 (1 − e −λ(x−y) )e −λydy + Z ∞ t− P(W(t) ≤ x − y|V (t) = y)dP(V (t) ≤ y) = 1 − e −λt − λte−λt + (1 − e −λ(x−t) )e −λt = 1 − e −λx − λte−λx 2.25 )µ 7

(1)S1,S2-S1,…,Sn-Sn-1同分布但不是条件独立 (2)当n=0时 E(S1|(t)=0)=E(W(t)+tN(t)=0) t+E(W(t)) E(S1N(t)=n)=E(U(1) (3)当n≤k时 E(SkIN(t)=n)=e( w(t)+tN(t)=n 当n≥k时 (SkIN(t)=n (4)与U(),Uk)的联合分布相同,可用微元法或者积分得到 (1) P(x<S2≤x+h,y<S5≤y+h imP(N(x)=.N(x+b)-N(x)=1,N(o)-Nx+b)=2.N+b)-Nw=1) Are-Ar Ahe-Ah 2(y-I-h=e-X(y-x-h)Ahe-Ah E(S1)=E(S1NY(t)≥1)P(N(t)≥1)+E(S1|N(t)=0)P(N(t)=0) E(S1|N()21)=25)=ESLN(O=PAN()=0) P(N(t)≥1) E(S1|N(t)≥1)) E(S1)-E(SIN(=OP(N(t) P(N(t)≥1)

(1) S1, S2 − S1, · · · , Sn − Sn−1 Ó©ÙØ´^‡Õá" (2)  n = 0 ž E(S1|N(t) = 0) = E(W(t) + t|N(t) = 0) = t + E(W(t)) = t + 1 λ  n ≥ 1 ž E(S1|N(t) = n) = E(U(1)) = t n + 1 (3)  n ≤ k ž E(Sk|N(t) = n) = E(xk + xk−1 + · · · + xn+2 + W(t) + t|N(t) = n) = t + k − n λ  n ≥ k ž E(Sk|N(t) = n) = E(U(k)) = kt n + 1 (4) † U(i) , U(k) éܩكӧŒ^‡{½öÈ©" 2.26 )µ (1) f(x, y) = lim h→0 P(x < S2 ≤ x + h, y < S5 ≤ y + h h 2 = lim h→0 P(N(x) = 1, N(x + h) − N(x) = 1, N(y) − N(x + h) = 2, N(y + h) − N(y = 1)) h 2 = lim h→0 λxe−λxλhe−λhλ 2 (y−x−h) 2 2 e −λ(y−x−h)λhe−λh h 2 = λ 5x(y − x) 2 e −λy 2 (2) E(S1) = E(S1|N(t) ≥ 1)P(N(t) ≥ 1) + E(S1|N(t) = 0)P(N(t) = 0) ⇒ E(S1|N(t) ≥ 1)) = E(S1) − E(S1|N(t) = 0)P(N(t) = 0) P(N(t) ≥ 1) E(S1|N(t) ≥ 1)) = E(S1) − E(S1|N(t) = 0)P(N(t) = 0) P(N(t) ≥ 1) = 1 λ − (t + 1 λ )e −λt 1 − e−λt 8

P(xt P(S2≤x,N(t)=1) P(S2≤|N(t)=1)P(N(t)=1) P(W(t)<I-tIN(t)=IP(N(t)=1) P(W(t)<I-t)P(N()=1)

(3) f(x, y) = lim h→0 P(x t P(S1 ≤ x, N(t) = 0) = P(S1 ≤ x|N(t) = 0)P(N(t) = 0) = P(W(t) ≤ x − t|N(t) = 0)P(N(t) = 0) = P(W(t) ≤ x − t)P(N(t) = 0) = (1 − e −λ(x−t) )e −λt = e −λt − e −λx  n ≥ 1 ž e x t) = [1 − µ t − x t ¶n ] (λt) ne −λt n! I(x≤t) + (λt) ne −λt n! I(x>t) (2)  n = 0 ž§ e x ≤ t, P(S2 ≤ x, N(t) = 0) = 0 " e x > t P(S2 ≤ x, N(t) = 0) = P(N(t) = 0, N(x) − N(t) ≥ 2) = P(N(t) = 0)P(N(x) − N(t) ≥ 2) = e −λt − e −λx − λ(x − t)e −λx  n = 1 ž e x ≤ t, P(S2 ≤ x, N(t) = 1) = 0 " e x > t P(S2 ≤ x, N(t) = 1) = P(S2 ≤ x|N(t) = 1)P(N(t) = 1) = P(W(t) ≤ x − t|N(t) = 1)P(N(t) = 1) = P(W(t) ≤ x − t)P(N(t) = 1) = (1 − e −λ(x−t) )λte−λt 9

当n≥2时 若x≤0,P(S2 若x>0 P(S2≤x,N(t)=n) P(S2 (X2) 39 E(X2|X1=i)=∑jP(x2=jX1=)=∑jP E(X2|X1=1)=1 E(X2X1=2) E(X2|X1=3) →E(X2X1)=(x1=1)+2x1=2)+1(x1=3 同理E(X3|X2)=I(x2=1)+3(x2=2)+31(x2=3) (X1=11X=3)=f31=0 P(r=2X=3)=P(x2=1,X11x=3=∑P1=5 P(T=3X0=3)=P(X3=1,X1≠1,X2≠1X0 ∑∑P3P P(T>4X X E(TA4X0=3) i P(T=iXo=3)+4P(T24X0=3) P(T1=1|X0=1)=P(X1=1X0=1)=0 P(Ti1=2k|X0=1)=P2(P23P32)-P21+B13(P32P2)-1P31 10

 n ≥ 2 ž e x ≤ 0, P(S2 ≤ x, N(t) = n) = 0 " e x > 0 P(S2 ≤ x, N(t) = n) = P(S2 ≤ x|N(t) = n)P(N(t) = n) = P(U(2) ≤ x)P(N(t) = n)I(x≤t) + P(N(t) = n)I(x>t) = " 1 − n µ t − x t ¶n−1 x t − µ t − x t ¶n # (λt) ne −λt n! I(x≤t) + (λt) ne −λt n! I(x>t) = £ t n − n(t − x) n−1x − (t − x) n ¤ (λt) ne −λt n! I(x≤t) + (λt) ne −λt n! I(x>t) 3.1 (1) E(X2) = 1 × 1 9 + 2 × 2 9 + 3 × 2 3 = 23 9 E(X2|X1 = i) = X 3 j=1 j P(X2 = j|X1 = i) = X 3 j=1 j Pij ⇒    E(X2|X1 = 1) = 1 E(X2|X1 = 2) = 7 3 E(X2|X1 = 3) = 8 3 ⇒ E(X2|X1) = I(X1=1) + 7 3 I(X1=2) + 8 3 I(X1=3) Ón E(X3|X2) = I(X2=1) + 7 3 I(X2=2) + 8 3 I(X2=3) (2) P(T = 1|X0 = 3) = P(X1 = 1|X0 = 3) = P31 = 0 P(T = 2|X0 = 3) = P(X2 = 1, X1 6= 1|X0 = 3) = X 3 i=2 P3iPi1 = 1 9 P(T = 3|X0 = 3) = P(X3 = 1, X1 6= 1, X2 6= 1|X0 = 3) = X 3 i=2 X 3 j=2 P(X3 = 1, X2 = i, X1 = j|X0 = 3) = X 3 i=2 X 3 j=2 P3jPjiPi1 = 2 27 P(T ≥ 4|X0 = 3) = 1 − X 3 k=1 P(T = k|X0 = 3) = 22 27 E(T ∧ 4|X0 = 3) = X 3 i=1 i P(T = i|X0 = 3) + 4P(T ≥ 4|X0 = 3) = 100 27 (3) P(T11 = 1|X0 = 1) = P(X1 = 1|X0 = 1) = 0 P(T11 = 2k|X0 = 1) = P12(P23P32) k−1P21 + P13(P32P23) k−1P31 10