Transformation of Uniform Distribution using $Y = X^2$

Question

I know this has already been answered somewhere but I can't seem to find where I am going wrong.

$X$ has a uniform distribution on $[-1,1]$

Let $Y = X^2$

P.d.f of X:

$f_X(x) = \frac{1}{2}$ for $-1 \le x \le 1$

Then using $f_Y(y) = f_X(g^{-1}( y))$ $|\frac{d}{dy}[g^{-1}( y)]|$

My new range of $y$ is $0 \le y \le 1$

Then $|\frac{d}{dy}[g^{-1}( y)]|$ = $\frac{1}{2\sqrt{y}}$

Subbing in I get that there is a an extra factor of one half stopping the p.d.f summing to 1 over the interval. Where have I gone wrong?

Answer 1

You can use the definition of CDF

$$F_Y(y)=P(Y\leq y)=P(-\sqrt{y}\leq X\leq \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})=$$

$$=\frac{\sqrt{y}+1}{2}-\frac{-\sqrt{y}+1}{2}=\sqrt{y}$$

Thus derivating you get your density

$$f_Y(y)=\frac{1}{2\sqrt{y}}\mathbb{1}_{(0;1]}(y)$$