Is this a correct $CDF$

Question

For a uniformly distributed r. v. $X$ on $[0, 3]$ find $CDF$ of $Y = X^2$.

Here's what I do:

1. $$p\left(x\right)=\begin{cases} \dfrac{1}{3},x\in\left[0;3\right]\\ 0,x\notin\left[0;3\right] \end{cases}$$

2. If $x\in [0, 3]$, then $y\in [0, 9]$.

3. $$F_{Y}\left(Y\le y\right)=F_{Y}\left(X^{2}\le y\right)=F_{Y}\left(X^{2}\le y\right)=F_{Y}\left(-\sqrt{y}\le X\le\sqrt{y}\right)$$

4. $$F_{Y}\left(Y\le y\right)=\frac{1}{3}x\bigg|_{-\sqrt{y}}^{\sqrt{y}}=\frac{2\sqrt{y}}{3}$$

5. Thus, $$F_{Y}\left(y\right)=\begin{cases} 0,y<0\\ \frac{2\sqrt{y}}{3},y\in\left[0;9\right]\\ 1,y>9 \end{cases}$$

Is this a correct solution? I'm a bit confused that at $y=9$ the $F_y$ is greater than $1$.

Answer 1

$F_Y(y)$ certainly cannot be $\frac23\sqrt y$ for $y$ close to $9$, because $\lim_{y\to 9^-}\frac23\sqrt y=2>1$.

If you want to calculate $P(-\sqrt{y}\le X\le \sqrt y)$ using the pdf of $X$, which you can do, then you must integrate the pdf of $X$. The pdf of $X$ is the function $p_X(t)=\begin{cases}0&\text{if }t\le 0\lor x\ge 3\\ \frac13&\text{if }0<t<3\end{cases}$ and therefore, for $y\ge 0$, $$P(-\sqrt y\le X\le y)=\int_{-\sqrt y}^{\sqrt y}p_X(x)\,dx=\int_0^{\min\{\sqrt y,\,3\}} \frac13\,dx=\frac13\min\{\sqrt y,\,3\}=\\=\begin{cases}1&\text{if }y\ge9\\ \frac13\sqrt y&\text{if }0\le y<9\end{cases}$$

Answer 2

Note that $p$ is $0$ on the interval $[- \sqrt{y}, 0),$ so actually $\mathbb{P}(Y \leq y) = \frac{1}{3} x \vert_0^{\sqrt{y}} = \frac{\sqrt{y}}{3}.$ Then it is immediate that this is indeed a cdf. I hope this helps. :)