Question

(1 pt) let x be an exponential random variable with parameter \lambda = 4, and let y be the random variable defined by y = 2 e^x. compute the probability density function of y:

Answer 1

$Y=2e^X\iff X=\ln\dfrac Y2$

Note that $x\in[0,\infty)$, so that $y\in[2,\infty)$.

$F_Y(y)=\mathbb P(Y\le y)=\mathbb P(2e^X\le y)=\mathbb P\left(X\le\ln\dfrac y2\right)=F_X\left(\ln\dfrac y2\right)$

$F_X(x)=1-e^{-4x}$
$\implies F_Y(y)=1-e^{-4\ln\frac y2}=1-\dfrac{16}{y^4}$

$\implies f_Y(y)=\dfrac{\mathrm dF_Y(y)}{\mathrm dy}=\dfrac{64}{y^5}$