Post Test 2023-05-13 16:05 本文约 字 Hello world!!! 中文 tex equatiaon test a=b+cb=c+d(1)a = b+c \\ b = c+d \tag{1} a=b+cb=c+d(1) fY(y)=fX[h(y)]∣h′(y)∣=fX[h(y)]h′(y)=1θe−xθ[dxdy(−θln(1−y))]=1θe−−θln(1−y)θθ1−y=1θeln(1−y)θ1−y=1−yθθ1−y=1\begin{aligned} f_Y(y) & = f_X[h(y)]|h'(y)| \\[2ex] & = f_X[h(y)]h'(y) \\[2ex] & = \frac{1}{\theta}e^{-\frac{x}{\theta}}[\frac{dx}{dy}(-\frac{\theta}{ln(1-y)})] \\[2ex] & = \frac{1}{\theta}e^{-\frac{-\frac{\theta}{ln(1-y)}}{\theta}}\frac{\theta}{1-y} \\[2ex] & = \frac{1}{\theta}e^{ln(1-y)}\frac{\theta}{1-y} \\[2ex] & = \frac{1-y}{\theta}\frac{\theta}{1-y} \\[2ex] & = 1 \end{aligned} fY(y)=fX[h(y)]∣h′(y)∣=fX[h(y)]h′(y)=θ1e−θx[dydx(−ln(1−y)θ)]=θ1e−θ−ln(1−y)θ1−yθ=θ1eln(1−y)1−yθ=θ1−y1−yθ=1 inline equatiaon ab\frac{a}{b}ba Hera is a figure.
