UA MATH574 概率論 均勻分布例題

2018May第四題。

Part a
Let $Z_i=?\ln U_i$,
$$\begin{gathered} P(Z_i\le z)=P(?\ln U_i\le z)=P(U_i\ge e^{?z})=1?F_U(e^{?z}) \\ {f}_Z(z)=e^{?z},z>0 \end{gathered}$$

This means $Z_i～EXP(1){=}_d\Gamma (1,1)$. By additivity, ${\sum }_{i=1}^n{Z}_i～\Gamma (n,1)$. Note that $V_n=e^{?{\sum }_{i=1}^n{Z}_i}$,
$$\begin{gathered} P(V_n\le v)=P(e^{?{\sum }_{i=1}^n{Z}_i}\le v)=P(\sum _{i=1}^n{Z}_i\ge ?\ln v) \\ {f}_{V_n}(v)=\frac{1}{v}\frac{(n?1)\ln (1/v)}{\Gamma (n)}{e}^{\ln v}=?\frac{n?1}{\Gamma (n)}\ln v,v\in (0,1) \end{gathered}$$

Part b
Notice $U_i/S_n$s should have the same distribution,
$$E[U_1/S_n]=\frac{1}{n}E[\sum _{i=1}^n{U}_i/S_n]=\frac{1}{n}E[S_n/S_n]=\frac{1}{n}$$

Part c
Define $Y=\ln (V_n{)}^{?1/S_n}=\frac{1}{S_n}(?\ln V_n)=\bar{Z}/\bar{U}$. By LLN, $\bar{U}{\to }_{p}1/2$, $\bar{Z}{\to }_{p}1$. By property of convergence in probability, $\bar{Z}/\bar{U}{\to }_{p}2$ and $(V_n{)}^{?1/S_n}{\to }_p{e}^2$

Part d
1

總結

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