Financial Statistics and Econometrics: Homework 1

Financial Statistics and Econometrics(金融统计与计量)第 1 次作业

第2次作业可点击: Financial Statistics and Econometrics: Homework 2

Econometrics
Financial Statistics and Econometrics: Homework 1

第一部分

  1. X_1,X_2,\cdots,X_n 为来自韦伯分布的独立同分布的随机样本,PDF 为:
f\left(x,\theta\right)=\left\{\begin{array}{l}\frac\alpha\beta X^{\alpha-1}exp^{-X^\alpha/\beta},\;\;X\;>\;0\\0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;X\;\leq\;0\end{array}\right.

其中,α>0,β>0。假设 α 已知,β 未知。

(a) 求 β 的极大似然估计量

极大似然函数为:

\mathrm L(\mathrm\theta\vert{\mathrm x}_1,{\mathrm x}_2,\cdots,{\mathrm x}_\mathrm n)=\prod_{\mathrm i=1}^\mathrm n\frac{\mathrm\alpha}{\mathrm\beta}{\mathrm X}_\mathrm i^{\mathrm\alpha-1}\mathrm e^{-\frac{\mathrm X^\mathrm\alpha}{\mathrm\beta}}

取极大似然函数的对数:

\mathrm{lnL}(\mathrm\theta\vert{\mathrm x}_1,{\mathrm x}_2,\cdots,{\mathrm x}_\mathrm n)=\mathrm{nln}\left(\mathrm\alpha\right)-\mathrm{nln}\left(\mathrm\beta\right)+\left(\mathrm\alpha-1\right)\sum_{\mathrm i=1}^\mathrm n\ln\left({\mathrm X}_\mathrm i\right)-\frac1{\mathrm\beta}\sum_{\mathrm i=1}^\mathrm n{\mathrm X}_\mathrm i^\mathrm\alpha

令:

\frac{\partial{\mathrm{lnL}(\mathrm\theta\vert{\mathrm x}_1,{\mathrm x}_2,\cdots,{\mathrm x}_\mathrm n)}}{\partial\mathrm\beta}=-\frac{\mathrm n}{\mathrm\beta}+\frac1{\mathrm\beta^2}\sum_{\mathrm i=1}^\mathrm n\mathrm X_\mathrm i^\mathrm\alpha=0

得:

\widehat{\mathrm\beta}=\frac1{\mathrm n}\sum_{\mathrm i=1}^\mathrm n\mathrm x_\mathrm i^\mathrm\alpha

(b) 上述极大似然估计量是无偏估计量吗?

\begin{array}{l}E(\widehat\beta)=E(\frac1n\sum_{i=1}^nX_i^\alpha)\\=\frac1n\sum_{i=1}^nE(X^\alpha)\\=\int_0^\infty X^\alpha\frac\alpha\beta X^{\alpha-1}e^{-\frac{X^\alpha}\beta}\operatorname dx\\=\frac\alpha\beta\int_0^\infty X^{2\alpha-1}\frac{-1}{\displaystyle\frac{\alpha X^{\alpha-1}}\beta}\operatorname de^{-\frac{X^\alpha}\beta}\\=-\int_0^\infty X^\alpha\operatorname de^{-\frac{X^\alpha}\beta}\\=-\left(\left[X^\alpha e^{-\frac{X^\alpha}\beta}\right]_0^\infty-\int_0^\infty e^{-\frac{X^\alpha}\beta}\operatorname dX^\alpha\right)\\=-\beta\int_0^\infty e^{-\frac{X^\alpha}\beta}\operatorname d\frac{-X^\alpha}\beta\\=-\beta\left[e^{-\frac{X^\alpha}\beta}\right]_0^\infty\\=\beta\\\end{array}

所以上述极大似然估计量是无偏估计量。

(c) 上述的极大似然估计量达到了 Cramer-Rao 下界了吗?

首先,计算该估计量的二阶矩:

\begin{array}{l}\begin{array}{l}E((\widehat\beta)^2)=E(\frac1{n^2}(\sum_{i=1}^nX_i^\alpha)^2)\\=\frac1{n^2}\sum_{i=1}^nE(X^{2\alpha})+{\textstyle\frac2{n^2}}(E(X^\alpha))^2\cdot(\frac{n(n+1)}2-n)\\=\frac1{n^2}\sum_{i=1}^nE(X^{2\alpha})+\frac{n-1}n\cdot(E(X^\alpha))^2\\=\frac1{n^2}\sum_{i=1}^nE(X^{2\alpha})+\frac{n-1}n\beta^2\\=\frac1n\int_0^\infty X^{2\alpha}\cdot\frac\alpha\beta\cdot X^{\alpha-1}\cdot e^{-\frac{X^\alpha}\beta}\operatorname dx+\frac{n-1}n\beta^2\\=\frac1n\int_0^\infty X^{3\alpha-1}\cdot\frac\alpha\beta\cdot\frac{-1}{\displaystyle\frac{\alpha X^{\alpha-1}}\beta}de^{-\frac{X^\alpha}\beta}+\frac{n-1}n\beta^2\\=-\frac1n\int_0^\infty X^{2\alpha}\operatorname de^{-\frac{X^\alpha}\beta}+\frac{n-1}n\beta^2\\=-\frac1n\left\{\left[X^{2\alpha}\cdot e^{-\frac{X^\alpha}\beta}\right]_0^\infty-\int_0^\infty e^{-\frac{X^\alpha}\beta}\operatorname dX^{2\alpha}\right\}+\frac{n-1}n\beta^2\\=\frac1n\int_0^\infty e^{-\frac{X^\alpha}\beta}\operatorname dX^{2\alpha}+\frac{n-1}n\beta^2\\=\frac2n\int_0^\infty X^\alpha\cdot e^{-\frac{X^\alpha}\beta}dX^\alpha+\frac{n-1}n\beta^2\end{array}\\=-\frac{2\beta^2}n\int_0^\infty\frac{X^\alpha}\beta de^{-\frac{X^\alpha}\beta}+\frac{n-1}n\beta^2\\=-\frac{2\beta^2}n\left\{\left[\frac{X^\alpha}\beta\cdot e^{-\frac{X^\alpha}\beta}\right]_0^\infty-\int_0^\infty e^{-\frac{X^\alpha}\beta}\operatorname d\frac{X^\alpha}\beta\right\}+\frac{n-1}n\beta^2\\=-\frac{2\beta^2}n+\frac{n-1}n\beta^2\\=\frac{(n+1)\beta^2}n\\\end{array}

其次,计算该估计量的方差:

\begin{array}{l}D(\widehat\beta)=E((\widehat\beta)^2)-(E(\widehat\beta))^2\\=\frac{(n+1)\beta^2}n-\beta^2\\=\frac{\beta^2}n\\\end{array}

然后,计算该估计量的方差的 Cramer-Rao 下界:

\begin{array}{l}\frac{\partial^2}{\partial\beta^2}\ln L\left(\left.\theta\right|X_1,X_2,\cdots,X_n\right)=\frac n{\beta^2}-\frac2{\beta^3}\sum_{i=1}^nX_i^\alpha\\E(\frac{\partial^2}{\partial\beta^2}\ln L\left(\left.\theta\right|X_1,X_2,\cdots,X_n\right))=\frac n{\beta^2}-\frac2{\beta^3}\cdot\sum_{i=1}^nE(X^\alpha)=-\frac n{\beta^2}\\-\frac1{nE(\frac{\partial^2}{\partial\beta^2}\ln L\left(\left.\theta\right|X_1,X_2,\cdots,X_n\right))}=-\frac1{n(-\frac n{\beta^2})}=\frac{\beta^2}{n^2}\end{array}

因为 D(\widehat\beta)>D(\widehat\beta)_{Cramer-Rao}

所以该极大似然估计量的方差未达到其 Cramer-Rao 下界。

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