Let $f(x)=1/3$ if $-1<x<2$ and zero elsewhere, be the PDF of X. Find the CDF and PDF of $Y=X^2$.

Question

This is problem 1.7.24 Hogg, McKean, and Craig. I understand the process for finding such a CDF and PDF. Here is what I have,

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

So I would think the answer would be

$$F_Y(y)=\int_{-\sqrt{y}}^\sqrt{y}\frac{1}{3}dx$$ on $0\leq y<4$. But the solutions manual says

$$F_Y(y)=\int_{-\sqrt{y}}^\sqrt{y}\frac{1}{3}dx$$ on $0\leq y<1$ and

$$F_Y(y)=\int_{-1}^\sqrt{y}\frac{1}{3}dx$$ on $1\leq y<4$.

I don't understand how the reasoning behind using these bounds. Any explanation would be much appreciated.

Answer 1

We are given $$ f(x) = \begin{cases} \frac{1}{3}, & -1<x<2\\ 0, & x\leq -1 \text{ or } x\geq 2 \end{cases}. $$ We then have $$ F_Y(y) = \int_{-\sqrt{y}}^{\sqrt{y}} f(x) \,dx. $$ Note that we cannot just insert $\frac{1}{3}$ into the integral here, because $f$ is not equal to $\frac{1}{3}$ on its whole domain.

If $0 \leq y < 1,$ then $(-\sqrt{y}, \sqrt{y})\subset (-1,2)$, so $f$ is equal to $\frac{1}{3}$ on $(-\sqrt{y}, \sqrt{y})$ and we may indeed just insert $\frac{1}{3}$ into the integral.

However, when $y$ passes $1$, our lower bound stops changing, because when $1 \leq y < 4,$ we have $$ \int_{-\sqrt{y}}^{\sqrt{y}} f(x) \,dx = \int_{-\sqrt{y}}^{-1} f(x) \,dx + \int_{-1}^{\sqrt{y}} f(x) \,dx = 0 + \int_{-1}^{\sqrt{y}} \frac{1}{3} \,dx, $$ where the first integral is $0$ because $f$ is zero on $(-\sqrt{y},-1).$ A similar thing happens when $y$ passes $4$.

Answer 2

As given $f(x)$ is $\frac{1}{3}$ only when $-1<x<2$ other wise it is 0

Thus $$-1\le\sqrt y\le2$$

But $\sqrt y$is always greater than 0

$$ 0\le \sqrt y$$

Thus $$ 0 \le y \le 4$$

also, $\sqrt y $ is defined only when $y\ge 0 $

$$-\sqrt y \ge -1$$ $$\sqrt y \le 1$$ but $0\le \sqrt y$

$$0\le y\le 1$$