UA MATH566 統計理論 一個例題 Hierarchical Model的統計性質

$Y|X～Pois(X)$ and $X～\Gamma (\alpha ,\beta )$. $Y_1,?,Y_n$ 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
$$I(y)=\frac{y}{1+\beta }I(y?1)=\frac{y}{1+\beta }\frac{y?1}{1+\beta }I(y?2)=?=\frac{y!}{(1+\beta {)}^{y+1}}$$

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

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

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

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.

總結

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