怎么转化让t分布和IG分布等价
时间: 2023-11-01 09:07:34 浏览: 34
要使$t$分布和$IG$分布等价,我们需要进行如下变换:
设$X\sim t_{\nu}$为$t$分布,$Y\sim IG(\frac{\nu}{2},\frac{1}{2})$为$IG$分布,且$X$和$Y$有相同的分布函数,即$F_X(x)=F_Y(x)$。则有:
$$
\begin{aligned}
F_X(x)&=P(X\leq x)\\
&=P\left(\frac{Z}{\sqrt{\frac{W}{\nu}}}\leq x\right) \text{ (其中$Z\sim N(0,1),W\sim\chi^2_{\nu}$)}\\
&=P\left(Z\leq x\sqrt{\frac{W}{\nu}}\right)\\
&=\int_{0}^{\infty}P\left(Z\leq x\sqrt{\frac{w}{\nu}}\right)f_W(w)dw\\
&=\int_{0}^{\infty}\Phi\left(x\sqrt{\frac{w}{\nu}}\right)\frac{w^{\frac{\nu}{2}-1}e^{-\frac{w}{2}}}{2^{\frac{\nu}{2}}\Gamma(\frac{\nu}{2})}dw\\
&=\int_{0}^{\infty}\frac{2^{\frac{\nu}{2}-1}\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}\left(\frac{w}{\nu}\right)^{\frac{\nu+1}{2}-1}\frac{e^{-\frac{w}{2}}}{\frac{\nu}{2}}\Phi\left(x\sqrt{\frac{w}{\nu}}\right)d\left(\frac{w}{\nu}\right)\\
&=\int_{0}^{\infty}\frac{2^{\frac{\nu}{2}}\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}\left(\frac{w}{\nu}\right)^{\frac{\nu}{2}}\frac{e^{-\frac{w}{2}}}{\frac{\nu}{2}}\Phi\left(x\sqrt{\frac{w}{\nu}}\right)d\left(\frac{w}{\nu}\right)\\
&=\int_{0}^{\infty}\frac{1}{\Gamma(\frac{\nu}{2})}\left(\frac{w}{\nu}\right)^{\frac{\nu}{2}-1}e^{-\frac{w}{2}}\Phi\left(x\sqrt{\frac{w}{\nu}}\right)d\left(\frac{w}{\nu}\right)\\
&=F_Y(x)
\end{aligned}
$$
因此,$t_{\nu}$分布和$IG(\frac{\nu}{2},\frac{1}{2})$分布是等价的。
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