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 ¤pØ8Ü¿µA S B = A S (B − A) = A S B − AB§2d VÇ\5" (4) ÏB ⊂ A§¤±A = B S A − B§dd" 1.3 A T B 6= φ§σ(A)k16"A T B = φ§σ(A)k8"ØJÑ 1.4 K8¥A3 = {(a, b) : a, b ∈ R} (1) Ï∀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§ddx /∈ An§gñ"¤± \∞ n=1 An ⊂ A"ddA = \∞ n=1 An"aq B = [∞ n=1 Bn" (2) Ï∀a ∈ R,(−∞, a] = [∞ n=1 (−n, a]§¤±A1 ⊂ σ(A2)"Ï∀a, b ∈ R,(a, b] = (−∞, b] − (−∞, a]§¤±A2 ⊂ σ(A1)"nþσ(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 ] ddσ(A1) = σ(Ai), 2 ≤ i ≤ 5 1.6 (3) ÏA2k+1 ⊂ A2k§¤± [∞ k=n Ak = ½ S∞ k=n/2 A2k = ( 1 4 , 1 2 + 1 n ] nóê S∞ k=(n+1)/2 A2k = ( 1 4 , 1 2 + 1 n+1 ] nÛê 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 ] ddP à \∞ n=1 [∞ k=n Ak ! = F( 1 2 ) − F( 1 4 )" \∞ k=n Ak = φ ⇒ [∞ n=1 \∞ k=n Ak = φ ddP à [∞ n=1 \∞ k=n Ak ! = 0" 1.10 (1) d4½Â ω ∈ {ω : limn→∞ Xn = X} ⇔ ∀m ≥ 1 ∃n ≥ 1 ∀k ≥ n |Xk(ω) − X| < 1/m ⇔ ∀m ≥ 1 ∃n ≥ 1 ω ∈ \ k≥n {ω : |Xk − X| < 1/m} ⇔ ∀m ≥ 1 ω ∈ [ n≥1 \ k≥n {ω : |Xk − X| < 1/m} ⇔ ω ∈ \ m≥1 [ n≥1 \ k≥n {ω : |Xk − X| < 1/m} 1.14 EN = X∞ n=1 nP(N = n) = X∞ n=1 Xn m=1 P(N = n) = X∞ n=1 X∞ m=n P(N = m) = X∞ n=1 P(N > 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 + N2N3Õá" (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 ùpÑ1nKy²§cüaq" (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 éÜVÇÝ¼êµ f(X,Y )(x, y) = e −(x+y) I(x>0,y>0) d X = UV /(1 + V ), Y = U/(1 + V ) JacobiÝ 1ªµ J = −u/(1 + v) 2 ddµ f(U,V )(u, v) = e −u u (1 + v) 2 I(u > 0, v > 0) ©OÈ©>©Ùµ 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) ddZ©Ù¼ê§2é©Ù¼ê¦VÇݼê" 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 ) Ï i 6= j E(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 ) ,aq" 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©ÙÓ" E(Sk|N(t) = n) = E(U(k)) = kt n + 1 , (k ≤ n) 5