Let $X ∼ \operatorname{Uniform}(−1,1)$ and $Y=X^2$. Find the $\operatorname{CDF}$ and $\operatorname{PDF}$ of $Y$.

Question

I have some questions about the solution of Example 4.8 from Probability Course:

First, we note that $R_Y=[0,1]$. As usual, we start with the CDF. For y∈[0,1], we have

$F_Y(y) = P(Y≤y) = P(X^2≤y) = P(−√y≤X≤√y) $

$=\frac{√y−(−√y)}{1−(−1)}$, since $X∼\operatorname{Uniform}(−1,1)$

$=√y.$

...

\begin{equation} f_Y(y)=F'_Y(y)=\begin{cases} \frac{1}{2\sqrt{y}}, & for & \text{$0 \leq y \leq 1$}.\\ 0, & otherwise.\\ \end{cases} \end{equation}

My questions are:

1. Where does $=\frac{√y−(−√y)}{1−(−1)}$ comes from? Could it be $\int_{-\sqrt{y}}^{\sqrt{y}} dy$?
2. Why do we have $\geq 0$ for the CDF? Should $\frac{1}{2\sqrt 0}$ invalid?

Thank you so much!

Answer 1

For $(a,b) \subset (-1,1)$ we have $P(a\leq X\leq b)=\frac {b-a} 2$. Since $X$ is uniformly distributed in $(-1,1)$ the probability that $X$ takes values in $(a,b)$ is the length of the interval $(a,b)$ divided by the length of the entire interval $(-1,1)$ ( the latter being $(1)-(-1)=2$). For $a=-\sqrt y, b=\sqrt y$ this is also equal to $\frac 1 2 \int_{-\sqrt{y}}^{\sqrt{y}} dy$.

$Y=X^{2}$ is always non-negative and $Y$ takes values in $[0,1)$.