Let $Y$ be uniformly distributed over the interval $(−1, 3)$. Find the probability density function of $U = Y^2$.
Attempted solution:
We know that
$$ Y=\begin{cases} 1/4 && -1 \leq y \leq 3\\ 0 && \text{elsewhere} \end{cases} $$
We find the CDF of $U$ using $Y$ $$ P(U \leq u) = P(Y^2 \leq u) = P(-\sqrt u \leq Y \leq \sqrt u) $$ $$ = \int_{-\sqrt u}^{\sqrt u}\frac{1}{4}dy = \frac{\sqrt u}{2} $$
Therefore
$$ f(u) = \frac{d}{du}\frac{\sqrt u}{2} = \frac{1}{4\sqrt u} \qquad 0 \leq u \leq 9 $$
However, the answer is:
$$ f(u) = \begin{cases} \frac{1}{4\sqrt u} && 0 \leq u < 1\\ \frac{1}{8\sqrt u} && 1 \leq u \leq 9 \end{cases} $$
Logically, since Y is between -1 and 3, this makes sense to me since there are two values of Y for which U will be between 0 and 1 for each value, and only one value of Y for which U is between 1 and 9.
Despite this, why didn't the computation pick up on this? Where did I go wrong?
If $1<u<9$, then
$$P(-\sqrt{u}\leq Y < \sqrt{u}) = P(-1\leq Y \leq \sqrt{u}) = \int_{-1}^{\sqrt{u}} 1/4 dy $$
This should solve your problem. Remember that Y has 0 density outside of the interval [-1,3].