Finding the cumulative distribution function of a transformed random variable

Question

Question

Let $X$ have cumulative distribution function

$$F_X(x) = \begin{cases}
0\qquad \mathrm{if}\ x < 0,\\
\frac 1 8 x^3\quad \mathrm{if}\ 0 \leq x \leq 2,\\
1\qquad \mathrm{if}\ x > 2.\\
\end{cases}$$

Find the cumulative distribution function of $Z = X^2$.

My working

I attempted to use to the CDF method to compute the CDF of $Z$ as

$$\begin{aligned}
F_Z(z) & = \mathbb{P}(Z \leq z)\\
& = \mathbb{P}(X^2 \leq z)\\
& = \mathbb{P}(-\sqrt z \leq X \leq \sqrt z)\\
& = F_X(\sqrt z) – F_X(-\sqrt z)\\
& = \frac 1 8 (\sqrt z)^3 – \frac 1 8 (-\sqrt z)^3\\
& = \frac 1 4 (\sqrt z)^3\ \forall\ z \in [0, 4]
\end{aligned}$$

However, I know this is incorrect as $F_Z(4)$ gives me $2$, but I am not sure where I have gone wrong.

Any intuitive explanations will be greatly appreciated 🙂

Edit

As mentioned in the comments, I was careless in that I did not realise that $F_X(-\sqrt z) = 0$!

Answer 1

The transformation function $Z=X^2$ is monotone in X - support thus

$$F_Y(y)=F_X(\sqrt{z})=\frac{1}{8}(\sqrt{z})^3\cdot\mathbb{1}_{(0;4)}(z)$$