當前位置：首頁 > 编程资源 > 编程问答 >内容正文

编程问答

UA MATH566 统计理论一个例题 Hierarchical Model的统计性质

發布時間：2025/4/14 编程问答 31 豆豆

生活随笔收集整理的這篇文章主要介紹了 UA MATH566 统计理论一个例题 Hierarchical Model的统计性质小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

UA MATH566 統計理論一個例題 Hierarchical Model的統計性質

\sim Pois(X)

and

\sim \Gamma(\alpha,\beta)

Y1,?,YnY_1,\cdots,Y_n

are random sample from population

Y

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

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

Part (b)
Note that $V a r [Y ∣ X] = X$ , by law of total variance,
$\frac{\alpha (1+\beta)}{\beta^2}$

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

density of $Y$ is
$fY(y)=∫0∞fY∣X(y∣x)fX(x)dx=∫0∞xye?xy!βe?βxdx=βy!∫0∞xye?(1+β)xdxf_Y(y) =\int_{ 0}^{\infty} f_{Y|X}(y|x)f_X(x) dx= \int_{ 0}^{\infty} \frac{x^y e^{-x}}{y!} \beta e^{-\beta x} dx \\ = \frac{\beta}{y!}\int_{ 0}^{\infty} x^y e^{-(1+\beta )x} dx$

Let $\int_{ 0}^{\infty} x^y e^{-(1+\beta )x} dx$
$\frac{-1}{1+\beta}\int_{ 0}^{\infty} x^y de^{-(1+\beta )x} = \frac{-x^ye^{-(1+\beta )x}}{1+\beta}|_0^{\infty} + \frac{y}{1+\beta}\int_{ 0}^{\infty} x^{y-1} e^{-(1+\beta )x} dx \\ = \frac{y}{1+\beta}I(y-1)$

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

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

(Alternative $\int_{ 0}^{\infty} x^y e^{-(1+\beta )x} dx =\frac{1}{(1+\beta)^{y+1}} \int_{ 0}^{\infty} (x(1+\beta))^y e^{-(1+\beta )x} dx(1+\beta) = \frac{\Gamma(y+1)}{(1+\beta)^{y+1}}$ )

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

Joint likelihood of sample is
$L(β)=βn(1+β)n+∑i=1nYiln?L(β)=nln?β?(n+∑i=1nYi)ln?(1+β)?ln?L(β)?β=nβ?n+nYˉ1+β=0?β^=1YˉL(\beta) = \frac{\beta^n}{(1+\beta)^{n+\sum_{i=1}^n Y_i}} \\ \ln L(\beta) = n\ln \beta - (n + \sum_{i=1}^n Y_i) \ln (1 + \beta) \\ \frac{\partial \ln L(\beta)}{\partial \beta} = \frac{n}{\beta} - \frac{n + n\bar{Y}}{1 + \beta} = 0 \Rightarrow \hat\beta = \frac{1}{\bar{Y}}$

Part (d)
Consider another set of sample $Z1,?,ZnZ_1,\cdots,Z_n$ ,
$L(β∣Z)L(β∣Y)=βn(1+β)n+∑i=1nZiβn(1+β)n+∑i=1nYi=(1+β)∑i=1nYi?∑i=1nZi\frac{L(\beta|\textbf{Z})}{L(\beta|\textbf{Y})} = \frac{\frac{\beta^n}{(1+\beta)^{n+\sum_{i=1}^n Z_i}}}{\frac{\beta^n}{(1+\beta)^{n+\sum_{i=1}^n Y_i}}} = (1+\beta)^{\sum_{i=1}^n Y_i - \sum_{i=1}^n Z_i}$

Only when $∑i=1nYi=∑i=1nZi\sum_{i=1}^n Y_i= \sum_{i=1}^n Z_i$ , likelihood ratio is independent of $β\beta$ , so $∑i=1nYi\sum_{i=1}^n Y_i$ is minimal sufficient statistics.

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

By LLN, $Yˉ→p1/β\bar{Y} \to_p 1/\beta$ . By property of convergence in probability,
$1Yˉ→pβ\frac{1}{\bar{Y}} \to_p \beta$

So MLE is consistent.

總結

以上是生活随笔為你收集整理的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。

上一篇： UA MATH564 概率论样本均值的
下一篇： UA MATH566 例题 Poisso

编程问答

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

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

總結

UA MATH566 统计理论一个例题 Hierarchical Model的统计性质

UA MATH566 統計理論一個例題 Hierarchical Model的統計性質