Question

Acontinous random variable X has a pdf given by p(x) = (5x4 0≤x≤1 0, otherwise) Let Y=X3. Find the probability distribution function ​

Answer 1

I'll use the method of transformations.

If
`f_X(x)` denotes the PDF of
`X`, and
`y=g(x)=x^3 \iff x=g^(-1)(y) = y^(1/3)`, we have

`f_Y(y) = f_X\left(g^(-1)(y)\right) \left|(dg^(-1))/(dy)\right|`

`(dg^(-1))/(dy) = \frac13 y^(-2/3)`

`\implies f_Y(y) = f_X\left(y^(1/3)\right) \left|\frac13 y^(-2/3)\right| = \boxed{\begin{cases} \frac53 y^(2/3) & \text{if } 0 \le y \le 1 \\ 0 & \text{otherwise} \end{cases}}`