UA MATH564 概率論 QE練習題2

• 第一題
• 第二題
• 第三題

這是2014年5月理論的1-3題。

第一題

For $X_2$,
$$\begin{gathered} P(X_2=0|X_1=0)=\frac{b+1}{b+r+1} \\ P(X_2=1|X_1=0)=\frac{r}{b+r+1} \\ P(X_2=0|X_1=1)=\frac{b}{b+r+1} \\ P(X_2=1|X_1=1)=\frac{r+1}{b+r+1} \end{gathered}$$

Notice $Y\in \{0,1,2\}$.
$$\begin{gathered} P(Y=0)=P(X_1=0,X_2=0)=P(X_2=0|X_1=0)P(X_1=0)=\frac{b(b+1)}{(b+r+1)(b+r)} \\ P(Y=1)=P(X_2=1|X_1=0)P(X_1=0)+P(X_2=0|X_1=1)P(X_1=1) \\ =\frac{rb}{(b+r+1)(b+r)}+\frac{br}{(b+r+1)(b+r)}=\frac{2br}{(b+r+1)(b+r)} \\ P(Y=2)=P(X_2=1|X_1=1)P(X_1=1)=\frac{r(r+1)}{(b+r+1)(b+r)} \end{gathered}$$

So
$$\begin{gathered} EY=\frac{2br}{(b+r+1)(b+r)}+\frac{2r(r+1)}{(b+r+1)(b+r)}=\frac{2r}{b+r} \\ E{Y}^2=\frac{2br}{(b+r+1)(b+r)}+\frac{4r(r+1)}{(b+r+1)(b+r)}=\frac{4r^2+2br+4r}{(b+r+1)(b+r)} \\ VarY=E{Y}^2?(EY{)}^2=\frac{4r^2+2br+4r}{(b+r+1)(b+r)}?\frac{4r^2}{(b+r{)}^2} \\ =\frac{4r^{2}b+2b^{2}r+4br+4r^3+2b{r}^2+4r^2?4r^{2}b?4r^3?4r^2}{(b+r+1)(b+r{)}^2}=\frac{2br(b+r+2)}{(b+r+1)(b+r{)}^2} \end{gathered}$$

第二題

這道題的題干是不對的，${\tau }_Y$前漏了一個$1/n$的系數。

Now that ${\mu }_Y=0$, kurtosis is
$${\tau }_Y=\frac{E{Y}^4}{{\sigma }_Y^4}?E{Y}^4={\sigma }_Y^4{\tau }_Y$$

For $T={\sum }_{i=1}^n{Y}_i$,
$$\begin{gathered} E{T}^2=E(\sum _{i=1}^n{Y}_i{)}^2=E\sum _{i=1}^n{Y}_i^2+\sum _{i=j}E{Y}_{i}E{Y}_j=E\sum _{i=1}^n{Y}_i^2+E\sum _{i=j}{Y}_i{Y}_j=n{\sigma }_Y^2 \\ E{T}^4=E(\sum _{i=1}^n{Y}_i{)}^4=E(\sum _{i=1}^n{Y}_i^2+\sum _{i=j}{Y}_i{Y}_j{)}^2 \\ =E(\sum _{i=1}^n{Y}_i^2{)}^2+2E(\sum _{i=1}^n{Y}_i^2)(\sum _{i=j}{Y}_i{Y}_j)+E(\sum _{i=j}{Y}_i{Y}_j{)}^2 \\ =E(\sum _{i=1}^n{Y}_i^4)+3E(\sum _{i=j}{Y}_k^2{Y}_i{Y}_j)+3E(\sum _{i=j}{Y}_i^2{Y}_j^2)+E(\sum _{i=j=k=l}{Y}_i{Y}_j{Y}_k{Y}_l) \\ =n{\sigma }_Y^4{\tau }_Y+3n(n?1){\sigma }_Y^4 \end{gathered}$$

So
$${\tau }_T=\frac{E{T}^4}{(ET{)}^2}=\frac{1}{n}{\tau }_Y+\frac{3(n?1)}{n}$$

第三題

Part a
Expectation of $X$ is
$$EX={\int }_0^{\infty }\frac{x}{\theta }{e}^{?\frac{x}{\theta }}dx=?x{e}^{?\frac{x}{\theta }}{|}_0^{\infty }+{\int }_0^{\infty }{e}^{?\frac{x}{\theta }}dx=\theta$$

$$E{\hat{\theta }}_a=aE{\bar{X}}_n=\frac{a}{n}\sum _{i=1}^{n}E{X}_i=a\theta$$

Second-order moment of $X$ is
$$E{X}^2={\int }_0^{\infty }\frac{x^2}{\theta }{e}^{?\frac{x}{\theta }}dx=?x^2{e}^{?\frac{x}{\theta }}{|}_0^{\infty }+2{\int }_0^{\infty }x{e}^{?\frac{x}{\theta }}dx=2{\theta }^2$$

$$\begin{gathered} E{\hat{\theta }}_a^2=\frac{a^2}{n^2}E(\sum _{i=1}^n{X}_i{)}^2=\frac{a^2}{n^2}\sum _{i=1}^{n}E(X_i{)}^2+\frac{a^2}{n^2}\sum _{i=j}E{X}_i{X}_j \\ =\frac{2n{a}^2{\theta }^2+n(n?1)a^2{\theta }^2}{n^2}=\frac{n+1}{n}{a}^2{\theta }^2 \end{gathered}$$

Now consider
$$E({\hat{\theta }}_a?\theta {)}^2=E{\hat{\theta }}_a^2?2\theta E{\hat{\theta }}_a+{\theta }^2=(\frac{(n+1)a^2}{n}?2a+1){\theta }^2$$

To make $E({\hat{\theta }}_a?\theta {)}^2$ as small as possible,
$$a=\frac{n}{n+1}$$

Part b
By LLN, ${\bar{X}}_n{\to }_{p}EX=\theta$, so ${\bar{X}}_n$ is consistent. By CLT,
$$\frac{{\bar{X}}_n?\theta }{\sqrt{\frac{1}{n}}\theta }{\to }_{d}N(0,1)$$

So aymptotic variance is ${\theta }^2/n$.

Part c
By LLN, ${\bar{Y}}_n{\to }_{p}EY=1/\theta$. By property of convrgence in probability

$1/\bar{Y}{\to }_p\theta$. So $1/{\bar{Y}}_n$ is consistent. Now apply generalized CLT:

（參考UA MATH564 概率論V 中心極限定理的Delta方法）
We can get
$$\frac{1/{\bar{Y}}_n?\theta }{\sqrt{1/n}\theta }{\to }_{d}N(0,1)$$

So aymptotic variance is ${\theta }^2/n$.

總結

以上是生活随笔為你收集整理的UA MATH564 概率论 QE练习题2的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。