问题及解答

Weibull 分布是由瑞典物理学家 Waloddi Weibull 于1939年引进的.

他的1951年的一篇文章 "A Statistical Distribution Function of Wide Applicability"(J. Applied Mechanics, vol. 18: 293--297) 中讨论了一些应用.

定义: 

A random variable $X$ is said to have a Weilbull distribution with parameters $\alpha$ and $\beta$ ($\alpha > 0$, $\beta > 0$) if the pdf of $X$ is

\[
f(x;\alpha,\beta)=\begin{cases}
\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0
\end{cases}
\]

与之十分相近的是 $\Gamma$-分布(Gamma distribution). 回顾其 pdf 定义为

\[
f_2(x;\alpha,\beta)=\begin{cases}
\frac{1}{\beta^{\alpha}\Gamma{\alpha}}x^{\alpha-1}e^{-\frac{x}{\beta}}, & x\geqslant 0,\\
0, & x < 0
\end{cases}
\]

令 Weibull 分布和 $\Gamma$-分布中的参数 $\alpha=1$, $\lambda=\frac{1}{\beta}$, 我们都可以得到指数分布(exponential distribution)

\[
f(x;\lambda)=\begin{cases}
\lambda e^{-\lambda x}, & x\geqslant 0,\\
0, & x < 0.
\end{cases}
\]

也就是说, 指数分布同是 $\Gamma$-分布和 Weibull 分布的特殊情形. 但是, 存在 $\Gamma$-分布, 其不属于 Weibull 分布; 也存在 Weibull 分布, 不属于 $\Gamma$-分布.

对于所定义的 Weibull 分布, 证明其 cdf 为

\[
F(x;\alpha,\beta)=\begin{cases}
1-e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0.\\
\end{cases}
\]

Prop. 若 $X$ 是服从 Weibull 分布的随机变量, 则其均值和方差为

\[
\begin{eqnarray}
\mu=E(X)&=&\beta\cdot\Gamma(1+\frac{1}{\alpha})\\
\sigma^2=V(X)&=&\beta^2\cdot\biggl[\Gamma(1+\frac{2}{\alpha})-\Bigl(\Gamma(1+\frac{1}{\alpha})\Bigr)^2\biggr]
\end{eqnarray}
\]

\[
\begin{split}
F(x;\alpha,\beta)&=\int_{0}^{x}f(t;\alpha,\beta)dt=\int_{0}^{x}\frac{\alpha}{\beta^{\alpha}}\cdot t^{\alpha-1}e^{-(\frac{t}{\beta})^{\alpha}}dt\\
&=\int_{0}^{x}\frac{\alpha}{\beta}(\frac{t}{\beta})^{\alpha-1}e^{-(\frac{t}{\beta})^{\alpha}}dt=\int_{0}^{x}\alpha\cdot(\frac{t}{\beta})^{\alpha-1}e^{-(\frac{t}{\beta})^{\alpha}}d(\frac{t}{\beta})\\
&=\int_{0}^{\frac{x}{\beta}}\alpha\cdot u^{\alpha-1}e^{-u^{\alpha}}du=\int_{0}^{\frac{x}{\beta}}u^{\alpha-1}e^{-u^{\alpha}}du^{\alpha}\\
&=\int_{0}^{(\frac{x}{\beta})^{\alpha}}e^{-s}ds=-e^{-s}\biggr|_{0}^{(\frac{x}{\beta})^{\alpha}}\\
&=1-e^{-(\frac{x}{\beta})^{\alpha}}
\end{split}
\]

\[
\begin{split}
\mu=E(X)&=\int_{0}^{+\infty}xf(x;\alpha,\beta)dx=\int_{0}^{+\infty}x\cdot\frac{\alpha}{\beta^{\alpha}}\cdot x^{\alpha-1}\cdot e^{-(\frac{x}{\beta})^{\alpha}}dx\\
&=\int_{0}^{+\infty}\alpha\cdot(\frac{x}{\beta})^{\alpha}\cdot e^{-(\frac{x}{\beta})^{\alpha}}dx
\end{split}
\]

令 $u=(\frac{x}{\beta})^{\alpha}$, 则 $x=\beta\cdot u^{\frac{1}{\alpha}}$, $dx=\frac{\beta}{\alpha}u^{\frac{1}{\alpha}-1}du$. 于是

\[
\begin{split}
\mu&=\int_{0}^{+\infty}\alpha ue^{-u}\cdot\frac{\beta}{\alpha}u^{\frac{1}{\alpha}-1}du\\
&=\int_{0}^{+\infty}\beta u^{\frac{1}{\alpha}}e^{-u}du\\
&=\beta\cdot\int_{0}^{+\infty}u^{1+\frac{1}{\alpha}-1}e^{-u}du\\
&=\beta\cdot\Gamma(1+\frac{1}{\alpha})
\end{split}
\]

下面计算方差

\[
\sigma^2=V(X)=E(X^2)-[E(X)]^2=\int_{0}^{+\infty}x^2\cdot f(x;\alpha,\beta)dx-\biggl(\beta\cdot\Gamma(1+\frac{1}{\alpha})\biggr)^2
\]

其中

\[
\begin{split}
E(X^2)&=\int_{0}^{+\infty}x^2\cdot f(x;\alpha,\beta)dx\\
&=\int_{0}^{+\infty}x^2\cdot\frac{\alpha}{\beta^{\alpha}}\cdot x^{\alpha-1}\cdot e^{-(\frac{x}{\beta})^{\alpha}}dx\\
&=\int_{0}^{+\infty}\alpha\beta\cdot(\frac{x}{\beta})^{\alpha+1}\cdot e^{-(\frac{x}{\beta})^{\alpha}}dx
\end{split}
\]

令 $t=\frac{x}{\beta}$, 则 $x=\beta t$, $dx=\beta dt$, 于是

\[
\begin{split}
E(X^2)&=\int_{0}^{+\infty}\alpha\beta\cdot t^{\alpha+1}e^{-t^{\alpha}}\cdot\beta dt\\
&=\alpha\beta^2\cdot\int_{0}^{+\infty}t^{\alpha+1}e^{-t^{\alpha}}dt
\end{split}
\]

令 $u=t^{\alpha}$, 则 $t=u^{\frac{1}{\alpha}}$, $dt=du^{\frac{1}{\alpha}}=\frac{1}{\alpha}u^{\frac{1}{\alpha}-1}du$. 于是

\[
\begin{split}
E(X^2)&=\alpha\beta^2\cdot\int_{0}^{+\infty}u^{1+\frac{1}{\alpha}}\cdot e^{-u}\cdot\frac{1}{\alpha}u^{\frac{1}{\alpha}-1}du\\
&=\beta^2\cdot\int_{0}^{+\infty}u^{\frac{2}{\alpha}}\cdot e^{-u}du\\
&=\beta^2\cdot\int_{0}^{+\infty}u^{1+\frac{2}{\alpha}-1}\cdot e^{-u}du\\
&=\beta^2\cdot\Gamma(1+\frac{2}{\alpha})
\end{split}
\]

因此,

\[
\begin{split}
\sigma^2=V(X)&=\beta^2\cdot\Gamma(1+\frac{2}{\alpha})-\biggl(\beta\cdot\Gamma(1+\frac{1}{\alpha})\biggr)^2\\
&=\beta^2\cdot\biggl[\Gamma(1+\frac{2}{\alpha})-\Bigl(\Gamma(1+\frac{1}{\alpha})\Bigr)^2\biggr]
\end{split}
\]