UA MATH566 统计理论 一个例题 Hierarchical Model的统计性质

UA MATH566 统计理论 一个例题 Hierarchical Model的统计性质

$Y|X\sim Pois(X)$ and $X\sim \Gamma (\alpha ,\beta )$. $Y_1,\cdots ,Y_n$ are random sample from population $Y$.

Part (a) Find $EY$
Part (b) Find $Var(Y)$
Part (c) If $\alpha =1$, find MLE of $\beta$.
Part (d) If $\alpha =1$, find a minimal sufficient statistics for $\beta$.
Part (e) Is MLE consistent?

Answer.
Part (a)
Note that $E[Y|X]=X$, by law of total expectation,
$$E[E[Y|X]]=EX=\frac{\alpha }{\beta }$$

Part (b)
Note that $Var[Y|X]=X$, by law of total variance,
$$Var(Y)=Var(E[Y|X])+E(Var[Y|X])=Var(X)+EX=\frac{\alpha (1+\beta )}{{\beta }^2}$$

Part (c)
If $\alpha =1$, density of $X$ is
$$f_X(x)=\beta e^{-\beta x},x\ge 0$$

density of $Y$ is
$$\begin{gathered} f_Y(y)={\int }_0^{\infty }{f}_{Y|X}(y|x)f_X(x)dx={\int }_0^{\infty }\frac{x^y{e}^{-x}}{y!}\beta e^{-\beta x}dx \\ =\frac{\beta }{y!}{\int }_0^{\infty }{x}^y{e}^{-(1+\beta )x}dx \end{gathered}$$

Let $I(y)={\int }_0^{\infty }{x}^y{e}^{-(1+\beta )x}dx$
$$\begin{gathered} I(y)=\frac{-1}{1+\beta }{\int }_0^{\infty }{x}^{y}d{e}^{-(1+\beta )x}=\frac{-x^y{e}^{-(1+\beta )x}}{1+\beta }{|}_0^{\infty }+\frac{y}{1+\beta }{\int }_0^{\infty }{x}^{y-1}{e}^{-(1+\beta )x}dx \\ =\frac{y}{1+\beta }I(y-1) \end{gathered}$$

when $y=0$，
$$I(0)={\int }_0^{\infty }{e}^{-(1+\beta )x}dx=\frac{1}{1+\beta }$$

By iteration
$$I(y)=\frac{y}{1+\beta }I(y-1)=\frac{y}{1+\beta }\frac{y-1}{1+\beta }I(y-2)=\cdots =\frac{y!}{(1+\beta {)}^{y+1}}$$

(Alternative $I(y)={\int }_0^{\infty }{x}^y{e}^{-(1+\beta )x}dx=\frac{1}{(1+\beta {)}^{y+1}}{\int }_0^{\infty }(x(1+\beta ){)}^y{e}^{-(1+\beta )x}dx(1+\beta )=\frac{\Gamma (y+1)}{(1+\beta {)}^{y+1}}$ )

So
$$f_Y(y)=\frac{\beta }{y!}\frac{y!}{(1+\beta {)}^{y+1}}=\frac{\beta }{(1+\beta {)}^{y+1}},y=0,1,\cdots$$

Joint likelihood of sample is
$$\begin{gathered} L(\beta )=\frac{{\beta }^n}{(1+\beta {)}^{n+{\sum }_{i=1}^n{Y}_i}} \\ \ln L(\beta )=n\ln \beta -(n+\sum _{i=1}^n{Y}_i)\ln (1+\beta ) \\ \frac{\partial \ln L(\beta )}{\partial \beta }=\frac{n}{\beta }-\frac{n+n\bar{Y}}{1+\beta }=0\Rightarrow \hat{\beta }=\frac{1}{\bar{Y}} \end{gathered}$$

Part (d)
Consider another set of sample $Z_1,\cdots ,Z_n$,
$$\frac{L(\beta |\text{Z})}{L(\beta |\text{Y})}=\frac{\frac{{\beta }^n}{(1+\beta {)}^{n+{\sum }_{i=1}^n{Z}_i}}}{\frac{{\beta }^n}{(1+\beta {)}^{n+{\sum }_{i=1}^n{Y}_i}}}=(1+\beta {)}^{{\sum }_{i=1}^n{Y}_i-{\sum }_{i=1}^n{Z}_i}$$

Only when ${\sum }_{i=1}^n{Y}_i={\sum }_{i=1}^n{Z}_i$, likelihood ratio is independent of $\beta$, so ${\sum }_{i=1}^n{Y}_i$ is minimal sufficient statistics.

Part (e)
$$EY=\sum _{y=0}^{\infty }\frac{y\beta }{(1+\beta {)}^{y+1}}=\frac{1}{\beta }$$

By LLN, $\bar{Y}{\to }_{p}1/\beta$. By property of convergence in probability,
$$\frac{1}{\bar{Y}}{\to }_p\beta$$

So MLE is consistent.

总结

以上是生活随笔为你收集整理的UA MATH566 统计理论 一个例题 Hierarchical Model的统计性质的全部内容，希望文章能够帮你解决所遇到的问题。

如果觉得生活随笔网站内容还不错，欢迎将生活随笔推荐给好友。