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习题五

1

\(E\xi=0 \times \frac{1}{2}+2 \times \frac{1}{2}=1 . \quad Var \xi=(0-1)^2 \times \frac{1}{2}+(2-1)^2 \times \frac{1}{2}=1\)

\(E Z_n=\mu^n=1 \quad Var Z_n=\sigma^2 \mu^{n-1}\left(1+\mu \cdots+\mu^{n-1}\right)=n\).

2

\(E \xi=a \times 0+b \times 1+c \times 2=b+2 c=b+2(1-a-b)=-2 a-b+2\). \(Var \xi=4 a+b-(2 a+b)^2\)

\(E Z_n=(-2 a-3+2)^n\) \(Var Z_n=\left[(4 a+b)-(2 a+b)^2\right]^2(-2 a-b+2)^{n-1} \sum\limits_{i=0}^{n-1}(-2 a-b+2)^i\).

3

\(\mu=E\xi=\frac{1}{2} . \quad \varphi(s)=E s^{\xi}=\frac{1}{2} \times s^0+\frac{1}{2} \times s^{\prime}=\frac{1}{2}(s+1)\).

由数学归纳法知道 \(\varphi_n(s)=\frac{1}{2^n}s +1-\frac{1}{2^n}\)

于是 \(\alpha_n = P\left(Z_n=0\right)=1-\frac{1}{2^n}\)

4

\(\mu=\lambda,Var\xi=\lambda\).

\(E Z_n=\lambda^n \quad Var Z_n=\lambda^{2n}+\cdots+\lambda^{n+1}\)

5

\(E W=E \sum\limits_{n=0}^{\infty} Z_n= \sum\limits_{n=0}^{\infty} E Z_n=\sum\limits_{n=0}^{\infty} \mu^n=\frac{1}{1-\mu}\)

6

  1. \(\varphi(s)=q+p s\)

    由数学归纳法知 \(\varphi_n(s)=p^n s+q \frac{1-p^n}{1-p}\)

  2. \(P\left(Z_n=0\right)=q\frac{1-p^n}{1-p}\)

  3. \(\alpha_n^{\prime}=(\alpha_n)^k=\left(q \frac{1-p^n}{1-p}\right)^k\)

7

\[ \begin{aligned} & \varphi(0)=p_0=P(\xi=0) \\ & \varphi_n(s)=E s^n=\sum_{k=1}^{\infty} p(\eta=k) s^k=\sum_{k=1}^{\infty} p(\xi=k\mid \xi>0) s^k \\ & =\sum_{k=1}^{\infty} \frac{p(\xi=k, \xi>0)}{p(\xi>0)} s^k=\frac{1}{1-p(\xi=0)}\left(\sum_{k=0}^{\infty} p(\xi=k) s^k-p(\xi=0)\right) \\ & =\frac{1}{1-\varphi(0)} \varphi(s)-\frac{\varphi(0)}{1-\varphi(0)} \\ & \implies \varphi_\eta(s)=\frac{\varphi(s)-\varphi(0)}{1-\varphi(0)} \\ \end{aligned} \]

8

\[x=\varphi(x) \implies a x^2+(b-1) x+c=a x^2-(a+c) x+c=0\]
\[ \Rightarrow x=\frac{c}{a} \text { 或 } 1 \]

又 \(\mu =2a+b>1 \implies a>c\)

于是 \(\tau=\frac{c}{a}\)

9

只需验证 \(\varphi(s)|_{s=1}=1\),显然符合

用数学归纳法可得

\[ \varphi_n(s)=1-p^n(1-s)^{\beta n} \]

10

\(Z_m\) 的每一个后代视为一个祖先, 则有

\[ \begin{aligned} EZ_n Z_m & =E \sum_{k=0}^{\infty}\left(k Z_{n-m}\right) Z_m P\left(Z_m=k\right) \\ & =E \sum_{n=0}^{\infty} k P\left(Z_m=k\right) Z_m Z_{n-m} \\ & =E Z_m^2 E Z_{n-m}=\mu^{n-m} E Z_m^2 \end{aligned} \]

11

\(E\xi=\varphi'(1)=1\implies EZ_n=1\implies EW_n=n+1\)

\(\implies EW_1=2,E(2W_2-W_3)=2\)

12

先求 \(\alpha_n=P(Z_n=0)\)

\(p\neq \frac12\) 时容易得到

\[ \varphi(s)=\frac{p}{1-s(1-p)} \]

于是

\[ \alpha_{n+1}= \varphi(\alpha_n)=\frac{p}{1-\alpha_n(1-p)},\alpha_0=p \]

解得

\[ \alpha_n=\frac{p^{n+1} - 2p(1-p)^{n+1}}{p^{n+1}-(1-p)^{n+1}} \]

于是

\[ P(T=n)=\alpha_n-\alpha_{n-1}=\frac{ (2p-1)^2p^n(1-p)^n }{ (p^n-(1-p)^n)(p^{n+1}-(1-p)^{n+1}) } \]

\(p=\frac12\) 时,

\[ P(T=n)=\frac{1}{n(n+1)} \]

2.

当 \(p=\frac12: ET=\sum\limits_{n=0}^\infty \frac{1}{n+1}=\infty\)

当 \(p\neq \frac12: P(T=n)\sim \frac{1}{\beta^n},\beta>1,n\to\infty\)

于是 \(ET<\infty\)

13

\(E\xi = \frac{7}{6}\)

  1. \(EZ_{30} = 3EZ'_{30}=3\times (\frac{7}{6})^{30}\)

  2. \(\alpha=\varphi(\alpha)\implies \alpha = \frac{2}{3}\) 其中 \(\varphi(s) = \frac{1}{3}+\frac16 s+\frac12 s^2\)

\(\implies \lim\limits_{n\to\infty} P(Z_n=0)=\alpha^3 = \frac{8}{27}\)

  1. 视为一个种族即可, 容易得到 \(P=\frac{13}{36}\)

14

  1. \(P=\prod\limits_{k=0}^{n-1}(\frac{1}{4})^{2^k} = (\frac{1}{4})^{2^n-1}=2^{2-2^{n+1}}\)

  2. \(\alpha=\frac{1}{3}\) (把白细胞视为没有即可)