Determine the density function for $Y=3X^2$

Question

The following function is the density function for the random variable $X$:

$$f(x)=\begin{cases}k\,(1-x^2)&\text{if}&\lvert x\rvert\leq 1\\[1ex]0&\text{if}&\lvert x\rvert\gt 0\end{cases}$$

Determine the density function for $Y=3X^2$

Answer 1

Since $X$ has even PDF and $P(|X|>1)=0$, $P(Y\le y)$ is $0$ for $y<0$, $1$ for $y>3$, and is otherwise$$P(|X|\le\sqrt{y/3})=\int_{-\sqrt{y/3}}^{\sqrt{y/3}}k(1-x^2)dx=2k[x-x^3/3]_0^{\sqrt{y/3}}=2k\sqrt{y/3}(1-y/9).$$This needs to be equal to $1$ for $y=3$, so$$1=4k/3\implies k=\frac34\implies\forall y\in[0,\,3]P(Y\le y)=\sqrt{3y/4}(1-y/9).$$Differentiating, the density is $0$ outside this range, but is otherwise $y^{-1/2}(3-y)/(4\sqrt{3})$.

Answer 2

Hint: for $0\leq y \leq 3$

$$F_Y(y)=P(Y\leq y)=P(3X^2\leq y)=P(X^2\leq \frac{y}{3})=P(|X|\leq \sqrt{\frac{y}{3}})=P(-\sqrt{\frac{y}{3}} \leq X \leq \sqrt{\frac{y}{3}})$$

after that

$$f_Y(y)=\frac{d}{dy}F_Y(y)$$