Variation of PDF

Question

Suppose I have the PDF $$ f_X(x) = \frac{12x^2}{7} $$ with $-1 \leq x \leq 1$.

What would be the PDF of $Y = X^2$?

Would it be $$ f_Y(y) = \frac{12x^{1/2}}{49},\quad 0 \leq y \leq 1 ? $$

Answer 1

Note that yours is not a valid density, i.e. $f_X(x)$ doesn't integrate to 1. Indeed

$$ \int_{-1}^1 \frac{12x^2}{7}\,dx = \frac{8}{7}. $$

The correct density should thus be $f_X(x)= \frac{3x^2}{2}.$ For $Y = X^2$ we have

\begin{align*} F_Y(y) = P(Y \leq y) = P(X^2\leq y) = P(-\sqrt{y} \leq X \leq \sqrt{y}).\tag{*} \end{align*}

To obtain $f_Y$ compute the above integral in (*) and differentiate w.r.t $y$.