概率论与随机过程 科学出版社习题解答

1、 n = n n ( i=1 xi 1 e(ba ¤± ln f = n ln + n ln + ( 1 n ln xi (b a i=1 - ln f = 0, ln f =0 n n (b a = 0 n + i=1 ln xi (b a ln(b a = 0 n = n , (ba = n ln i=1 b a xi 8.3 µ Ï P N (t = n = P Sn t P Sn+1 t, S0 = 0, n Sn = k=1 xk , n 1, P xn = 1 + P xn = 2 = 1 ¤±: (1 P N (1 =

2、k = P Sk 1 P Sk+1 1 k = 0 § Sk = S0 = 0§ ¦ Sk+1 = S0+1 = S1 1 ¤á, k 1 =2 S1 = X1 = 1 â¤á, ¤±, k = 0 , P N (1 = 0 = 1 3 3 1 k = 1 , P N (1 = 1 = 1 0 = 3 3 k 2 , éØ ù ÷v Sk 1 Ú Sk+1 1, d P N (1 = k = 0. ¤± 2 3, k = 0

3、1 P N (1 = k = 3, k = 1 0, k 2 (2 P N (2 = k = P Sk 2 P Sk+1 2 k = 0 §Sk = S0 = 0 < 2, Sk+1 = S1 = X1 = 1 < 2 ½ Sk+1 = S1 = 1 2 X1 = 2 = 2. d P N (2 = 0 = 1 ( 3 +3 =0 k = 1 , S2 = X1 + X2 = 1 + 1 = 2, 1 1 1 8 d P N (2 = 1 = ( 3 +2 3 (3 × 3 = 9 k = 2 , S2 = X1 + X2 = 1 + 1 = 2; &

4、#233;Ø ù ÷v S3 2, 1 1 1 d P N (2 = 2 = ( 3 × 3 0 = 9 k 3 , éØ ù ÷v Sk 2 Ú Sk+1 2, d P N (2 = k = 0. ¤± 8 9, k = 1 1 P N (2 = k = 9, k = 2 0, k = 0 ½ k 3 (3 P N (3 = k = P Sk 3 P Sk+1 3 16 dd µ P N (t = n = P Sn t P Sn+1 t t = j =0 (n&#

5、181;j exp(nµ j! t k=0 (n + 1µk exp(n + 1µ k! 9.1 µ Yn , n 1 'u Xn § Kk E |Yn | = E |cXn + Xn1 | |c|E |Xn | + E |Xn1 | < ¤± |c| < = = E cXn+1 + Xn |X0 , X1 , · · · , Xn cE Xn+1 |X0 , X1 , · · · , Xn + E Xn |X0 , X1 , · &#

6、183; · , Xn E Yn+1 |X0 , X1 , · · · , Xn = c(Xn + Xn1 + Xn = cXn + cXn1 + Xn = Yn = cXn + Xn1 ¤± Ïd c = 1 c= 1 > 0, > 0, + = 1 + = =1+ 9.6 y²µ q E |( Sn | p q q = E |( X0 | × · · · × E |( Xn | p p q Xn q X0 = E ( × 

7、3; · · × E ( p p q 1 q 1 q q = ( · p + ( · q × · · · × ( 1 · p + ( 1 · q p p p p = 1 × ··· × 1 = 1 < q q E ( Xn+1 · ( Sn |X0 , X1 , · · · , Xn p p q q q q ( Sn E ( Xn+1 |X0 , X1 , · · · , Xn = ( Sn E ( Xn+1 p p p p q q q ( Sn ( 1 · p + ( 1 · q