# 连续型随机变量函数的分布

$f_X(x)=\begin{cases}\frac{x}{8},&0<x<4\\ 0, &\text{其它}\end{cases}$

\begin{align*}F_Y(y)&=P(Y\le y)=P(2X+8\le y)\\&=P\left(X\le \frac{y-8}{2}\right)=F_X\left(\frac{y-8}{2}\right)\end{align*}

\begin{align*}f_Y(y)&=F’_Y(y)=(F_x(\frac{y-8}{2}))’\\ &=f_X(\frac{y-8}{2})\cdot\frac{1}{2}\\ &=\begin{cases}\frac{1}{8}\cdot \frac{y-8}{2},&0<\frac{y-8}{2}<4\\ 0, &\text{其它}\end{cases}\\ &=\begin{cases}\frac{y-8}{16},&8<y<16\\ 0, &\text{其它}\end{cases}\end{align*}

$F_X(x)=\begin{cases}0,&x\le0\\ \frac{x^2}{16},&0<x<4\\ 1,&x\ge 4\end{cases}$

（1）$$Y=e^X$$，

\begin{align*}F_Y(y)&=P(Y\le y)=P(e^X\le y)=P(X\le \ln y)=F_X(\ln y)\end{align*}

\begin{align*}f_Y(y)&=F’_Y(y)=(F_X(\ln y))’\\ &=f_X(\ln y)\frac{1}{y}\\ &=\begin{cases}\frac{1}{y},&0< \ln y<1\\ 0,&\text{其它}\end{cases}\\ &=\begin{cases}\frac{1}{y},&1< y< e\\ 0,&\text{其它}\end{cases}\end{align*}

（2）$$Y=-2\ln X$$，

\begin{align*}F_Y(y)&=P(Y\le y)=P(-2\ln X\le y)\\ =P(X\ge e^{-\frac{y}{2}})=1-P(X< e^{-\frac{y}{2}})\\ &=1-F_X\left(e^{-\frac{y}{2}}\right)\end{align*}

\begin{align*}f_Y(y)&=F’_Y(y)=(1-F_X\left(e^{-\frac{y}{2}}\right))’\\ &=\frac{1}{2}e^{-\frac{y}{2}}f_X\left(e^{-\frac{y}{2}}\right)\\ &=\begin{cases}\frac{1}{2}e^{-\frac{y}{2}},&0< e^{-\frac{y}{2}}<1\\ 0,&\text{其它}\end{cases}\\ &=\begin{cases}\frac{1}{2}e^{-\frac{y}{2}},&y>0\\ 0,&y\le 0\end{cases}\end{align*}