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
`\frac4{\lambda^2}` and scale parameter
`\frac12`.