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\frac 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\frac y2\right)=F_X\left(\ln\frac y2\right)`


`F_X(x)=1-e^(-4x)`

`\implies F_Y(y)=1-e^(-4\ln\frac y2)=1-(16)/(y^4)`


`\implies f_Y(y)=(\mathrm dF_Y(y))/(\mathrm dy)=(64)/(y^5)`