题目
5.设总体X的概率密度为-|||-(x;theta )= ^2)(e)^-dfrac ({x^2)(2{theta )^2}},xgt 1 0, . ,-|||-有样本X1,X2,···,Xn,其相应的样本值为x1,x2,···,xn ,求未知参数θ的极大似然估计值.

题目解答
答案

解析
步骤 1:写出似然函数
似然函数是概率密度函数关于参数θ的函数,对于给定的样本值x1,x2,···,xn,似然函数为:
$L(\theta )=\prod _{i=1}^{n}f(x_{i};\theta )=\prod _{i=1}^{n}\dfrac {x_{i}}{{\theta }^{2}}{e}^{-\dfrac {{x_{i}}^{2}}{2{\theta }^{2}}}$
步骤 2:对似然函数取对数
为了简化计算,我们对似然函数取对数,得到对数似然函数:
$l(\theta )=\ln L(\theta )=\sum _{i=1}^{n}\ln \left ( \dfrac {x_{i}}{{\theta }^{2}}{e}^{-\dfrac {{x_{i}}^{2}}{2{\theta }^{2}}} \right )$
$=\sum _{i=1}^{n}\left ( \ln x_{i}-2\ln \theta -\dfrac {{x_{i}}^{2}}{2{\theta }^{2}} \right )$
步骤 3:求对数似然函数关于θ的导数
为了找到极大似然估计值,我们需要求对数似然函数关于θ的导数,并令其等于0:
$\dfrac {\partial l(\theta )}{\partial \theta }=\sum _{i=1}^{n}\left ( -\dfrac {2}{\theta }+\dfrac {{x_{i}}^{2}}{{\theta }^{3}} \right )$
令导数等于0,得到:
$\sum _{i=1}^{n}\left ( -\dfrac {2}{\theta }+\dfrac {{x_{i}}^{2}}{{\theta }^{3}} \right )=0$
步骤 4:求解θ的极大似然估计值
将导数等于0的方程化简,得到:
$\sum _{i=1}^{n}\dfrac {{x_{i}}^{2}}{{\theta }^{3}}=\sum _{i=1}^{n}\dfrac {2}{\theta }$
$\sum _{i=1}^{n}{x_{i}}^{2}=2n{\theta }^{2}$
$\theta ^{2}=\dfrac {\sum _{i=1}^{n}{x_{i}}^{2}}{2n}$
$\theta =\sqrt {\dfrac {\sum _{i=1}^{n}{x_{i}}^{2}}{2n}}$
似然函数是概率密度函数关于参数θ的函数,对于给定的样本值x1,x2,···,xn,似然函数为:
$L(\theta )=\prod _{i=1}^{n}f(x_{i};\theta )=\prod _{i=1}^{n}\dfrac {x_{i}}{{\theta }^{2}}{e}^{-\dfrac {{x_{i}}^{2}}{2{\theta }^{2}}}$
步骤 2:对似然函数取对数
为了简化计算,我们对似然函数取对数,得到对数似然函数:
$l(\theta )=\ln L(\theta )=\sum _{i=1}^{n}\ln \left ( \dfrac {x_{i}}{{\theta }^{2}}{e}^{-\dfrac {{x_{i}}^{2}}{2{\theta }^{2}}} \right )$
$=\sum _{i=1}^{n}\left ( \ln x_{i}-2\ln \theta -\dfrac {{x_{i}}^{2}}{2{\theta }^{2}} \right )$
步骤 3:求对数似然函数关于θ的导数
为了找到极大似然估计值,我们需要求对数似然函数关于θ的导数,并令其等于0:
$\dfrac {\partial l(\theta )}{\partial \theta }=\sum _{i=1}^{n}\left ( -\dfrac {2}{\theta }+\dfrac {{x_{i}}^{2}}{{\theta }^{3}} \right )$
令导数等于0,得到:
$\sum _{i=1}^{n}\left ( -\dfrac {2}{\theta }+\dfrac {{x_{i}}^{2}}{{\theta }^{3}} \right )=0$
步骤 4:求解θ的极大似然估计值
将导数等于0的方程化简,得到:
$\sum _{i=1}^{n}\dfrac {{x_{i}}^{2}}{{\theta }^{3}}=\sum _{i=1}^{n}\dfrac {2}{\theta }$
$\sum _{i=1}^{n}{x_{i}}^{2}=2n{\theta }^{2}$
$\theta ^{2}=\dfrac {\sum _{i=1}^{n}{x_{i}}^{2}}{2n}$
$\theta =\sqrt {\dfrac {\sum _{i=1}^{n}{x_{i}}^{2}}{2n}}$