Question

G let x be an exponentially distributed random variable with parameter λ = 1 / 2 . determine the probability distribution function of the random variable y = x 2 . what kind of distribution does y have?

Answer 1

$X$ has CDF

$F_X(x)=\mathbb P(X\le x)=\begin{cases}1-e^{-\lambda x/2}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

The CDF of $Y$ is then

$F_Y(y)=\mathbb P(Y\le y)=\mathbb P(X^2\le y)=\mathbb P(X\le\sqrt y)=F_X(\sqrt y)$
$\implies F_Y(y)=\begin{cases}1-e^{-\lambda\sqrt y/2}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$
$\implies F_Y(y)=\begin{cases}1-e^{-(y/(4/\lambda^2))^{1/2}}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$

which is the CDF of a Weibull distribution with shape parameter $\dfrac4{\lambda^2}$ and scale parameter $\dfrac12$.