概率论大学习

题设

买饮料有若干种方案，每次购买有$p_i$的概率选择第$i$种方案，花$a_i$元买到$b_i$瓶饮料，$a_i \in \mathbb{Z}, b_i \in \mathbb{Z}, \operatorname{E}\left(a\right) < \infty, \operatorname{E}\left(b\right) < \infty, \sum p_i=1$。

1

问题

$\operatorname{E}\left(b\right) > 0, b_i \text{有上界}$

现在重复购买直到拥有至少$k$瓶饮料时停下，记最后总共花了$A$元，得到了$B$瓶饮料，求$\lim_{k \to \infty} \operatorname{E}\left(\frac{A}{B}\right)$。

解答

买饮料的过程是一个Renewal Process，记最后在第$N_k$轮结束时停止，第$i$轮花$A_i$元买到$B_i$瓶饮料。

$$\begin{aligned} &\lim_{k \to \infty} \operatorname{E}\left(\frac{A}{B}\right) \\ =&\lim_{k \to \infty} \sum_{x,y} \operatorname{P}(A=x \wedge B=y) \frac{x}{y} \\ =&\lim_{k \to \infty} \sum_{y=0}^C \sum_{x} \operatorname{P}(A=x \wedge B=k+y) \frac{x}{k+y} \quad &\text{($b_i$有上界，存在常数$C$使$B-k \leq C$)} \\ =&\sum_{y=0}^C \lim_{k \to \infty} \frac{1}{k+y} \sum_{x} x \operatorname{P}(A=x \wedge B=k+y) \\ =&\lim_{k \to \infty} \frac{1}{k} \sum_{x} x \operatorname{P}(A=x) \sum_{y=0}^C \operatorname{P}(B=k+y|A=x) \\ =&\lim_{k \to \infty} \frac{1}{k} \operatorname{E}(A) \\ =&\lim_{k \to \infty} \frac{1}{k} \operatorname{E}\left(\sum_{i=1}^{N_k} A_i\right) \\ =&\lim_{k \to \infty} \frac{1}{k} \operatorname{E}\left(\sum_{i=1}^{\infty} A_i \operatorname{\mathbb{I}}_{\{N_k \geq i\}} \right) \\ \end{aligned}$$

考虑应用Fubini–Tonelli定理，

$$\begin{aligned} &\sum_{i=1}^{\infty}\operatorname{E}\left(\left|A_i \operatorname{\mathbb{I}}_{\{N_k \geq i\}}\right|\right) \\ =&\sum_{i=1}^{\infty}\operatorname{E}\left(\left|A_i\right|\right) \operatorname{E}\left(\operatorname{\mathbb{I}}_{\{N_k \geq i\}}\right) \quad &\text{(Indicator只依赖于前$i-1$次，与$A_i$独立)} \\ =&\operatorname{E}\left(\left|A_1\right|\right) \operatorname{E}\left(N_k\right) < \infty \\ \end{aligned}$$

于是原式，

$$\begin{aligned} =&\lim_{k \to \infty} \frac{1}{k} \sum_{i=1}^{\infty} \operatorname{E}\left(A_i \operatorname{\mathbb{I}}_{\{N_k \geq i\}} \right) \\ =&\lim_{k \to \infty} \frac{\operatorname{E}\left(A_1\right)}{k} \sum_{i=1}^{\infty} \operatorname{E}\left(\operatorname{\mathbb{I}}_{\{N_k \geq i\}} \right) \\ =&\lim_{k \to \infty} \frac{\operatorname{E}\left(A_1\right)}{k} \operatorname{E}\left(N_k\right) \\ =&\operatorname{E}\left(A_1\right) \lim_{k \to \infty} \operatorname{E}\left(\frac{N_k}{k}\right) \\ =&\frac{\operatorname{E}\left(A_1\right)}{\operatorname{E}\left(B_1\right)} \quad &\text{(Renewal Process结论)} \\ =&\frac{\sum p_i a_i}{\sum p_i b_i} \\ \end{aligned}$$

2

问题

$b_i > 0$

固定买$k$次停下，仍然求$\lim_{k \to \infty} \operatorname{E}\left(\frac{A}{B}\right)$。

解答

记$d=\frac{\operatorname{E}\left(a\right)}{\operatorname{E}\left(b\right)}, p(\epsilon_1, \epsilon_2)=\operatorname{P}\left(\overline{A}_k=\operatorname{E}\left(a\right)+\epsilon_1 \wedge \overline{B}_k=\operatorname{E}\left(b\right)+\epsilon_2\right)$。

$$\begin{aligned} &\lim_{k \to \infty} \operatorname{E}\left(\frac{A}{B}\right) \\ =&\lim_{k \to \infty} \operatorname{E}\left(\frac{\overline{A}_k}{\overline{B}_k}\right) \\ =&\lim_{k \to \infty} \sum_{\epsilon_1, \epsilon_2} p(\epsilon_1, \epsilon_2) \frac{\operatorname{E}\left(a\right)+\epsilon_1}{\operatorname{E}\left(b\right)+\epsilon_2} \\ =&\lim_{k \to \infty} \sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2)\frac{\operatorname{E}\left(a\right)+\epsilon_1}{\operatorname{E}\left(b\right)+\epsilon_2} \\ =&\lim_{k \to \infty} \sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2) \left(d+\frac{\epsilon_1}{\operatorname{E}\left(b\right)}\right) \left(1-\frac{\epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}\right) \\ =&\lim_{k \to \infty} \sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2) \left(d-\frac{d \epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}+\frac{\epsilon_1}{\operatorname{E}\left(b\right)}\left(1-\frac{\epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}\right)\right) \\ =&d+\lim_{k \to \infty} \sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2) \left(-\frac{d \epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}+\frac{\epsilon_1}{\operatorname{E}\left(b\right)}\left(1-\frac{\epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}\right)\right) \\ \end{aligned}$$

考虑余项的绝对值，

$$\begin{aligned} &\lim_{k \to \infty} \left|\sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2) \left(-\frac{d \epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}+\frac{\epsilon_1}{\operatorname{E}\left(b\right)}\left(1-\frac{\epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}\right)\right)\right| \\ \leq &\lim_{k \to \infty} \sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2) \left(\left|\frac{d \epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}\right|+\left|\frac{\epsilon_1}{\operatorname{E}\left(b\right)}\left(1-\frac{\epsilon_2}{\operatorname{E}\left(b\right)+\epsilon_2}\right)\right|\right) \\ \leq &\lim_{k \to \infty} \sum_{\epsilon_1} \sum_{\epsilon_2} p(\epsilon_1, \epsilon_2) \left(d\left|\epsilon_2\right|+\left|\epsilon_1\right|\right) \quad &\text{(由$b_i>0$有$\overline{B}_k \geq 1$)} \\ =&0 \quad &\text{(弱大数定律夹逼)} \\ \end{aligned}$$

于是，$\lim_{k \to \infty} \operatorname{E}\left(\frac{A}{B}\right)=d=\frac{\sum p_i a_i}{\sum p_i b_i}$。

3

问题

几乎同1，但不要求$b_i$有上界

解答

我也不知道

鸣谢

timetraveler314