probability - Transformation of Random Variable $Y = X^2$

I'm learning probability, specifically transformations of random variables, and need help to understand the solution to the following exercise:

Consider the continuous random variable $X$ with probability density function $$f(x) = \begin{cases} \frac{1}{3}x^2 \quad -1 \leq x \leq 2, \\ 0 \quad \quad \text{elsewhere}. \end{cases}$$ Find the cumulative distribution function of the random variable $Y = X^2$.

The author gives the following solution:

For $0 \leq y \leq 1: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-\sqrt y \leq X \leq \sqrt y) = \int_{-\sqrt y}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{2}{9}y\sqrt y.$

For $1 \leq y \leq 4: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-1 \leq X \leq \sqrt y) = \int_{-1}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{1}{9} + \frac{1}{9}y\sqrt y.$

For $y > 4: F_{Y}(y) = 1.$

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Previous to this exercise, I've managed to follow the solutions of two similar (obviously simpler) problems for a strictly increasing and strictly decreasing function of $X$, respectively. However in this problem, I don't understand the computations being done, specifically:

• How does the three intervals $0 \leq y \leq 1$, $1 \leq y \leq 4$ and $y > 4$ are determined? In the two previous problems I've encountered, we only considered one interval which was identical to the interval where $f(x)$ was non-zero.


• In the case where $0 \leq y \leq 1$, why does $P(X^2 \leq y) = P(-\sqrt y \leq X \leq \sqrt y)$ and not $P(X \leq \sqrt y)$? I have put question marks above the equalities that I don't understand.

I think I have not understand the theory well enough. I'm looking for an answer that will make me understand the solution to this problem and possibly make the theory clearer.

Answer

Let's start by seeing what the density function $f_X$ of $X$ tells us about the cumulative distribution function $F_X$ of $X$. Since $f_X(x) = 0$ for $-\infty < x < -1$, we see that
$$F_X(x) = \int_{-\infty}^x f_X(t) \, dt \equiv 0 $$
in this range. Similarly, since $f_X(x) = 0$ in the range $2 < x < \infty$, we see that

$$F_X(x) = \int_{-\infty}^x f_X(t) \, dt = \int_{-\infty}^{\infty} f_X(t) \, dt \equiv 1$$
in this range. In other words, the random variable is "supported on the interval $[-1,2]$" in the sense that $P(X \notin [-1,2]) = 0$.

Now let us consider $Y = X^2$. This variable is clearly non-negative and since $X$ is supported on $[-1,2]$, we must have that $Y$ is supported on $[0, \max((-1)^2,2^2)] = [0,4]$. This is intuitively clear because the variable $X$ (with probability $1$) takes values in [-1,2] and so $X^2$ takes values in $[0,\max((-1)^2,(2)^2)]$. So we only need to understand $F_Y(y)$ in the range $y \in [0,4]$. Now, we always have

$$ F_Y(y) = P(Y < y) = P(X^2 < y) = P(-\sqrt{y} < X < \sqrt{y}) = \int_{-\sqrt{y}}^{\sqrt{y}} f_X(t) \, dt $$

but since $f_X$ is defined piecewise, to proceed at this point we need to analyze several cases. We already know that $F_Y(y) = 0$ if $y \leq 0$ and $F_Y(y) = 1$ if $y \geq 4$.

If $0 \leq y \leq 1$ then $[-\sqrt{y},\sqrt{y}]$ is contained in $[-1,1]$ and on $[-1,1]$ the density function is $f_X(x) = \frac{1}{3}x^2$ so we can write

$$ F_Y(y) = \int_{-\sqrt{y}}^{\sqrt{y}} \frac{1}{3} t^2 \, dt. $$

However, if $1 < y \leq 4$ then $-\sqrt{y} < -1$ and so the interval of integration splits as $[-\sqrt{y}, -1] \cup [-1,\sqrt{y}]$. Over the left $[-\sqrt{y},-1]$ part, the density function is zero so the integal will be zero and we are left only with calculating the integral over the right part:

$$ F_Y(y) = \int_{-\sqrt{y}}^{-1} f_X(t) \, dt + \int_{-1}^{\sqrt{y}} f_X(t) \, dt = \int_{-1}^{\sqrt{y}} \frac{1}{3}t^2 \, dt. $$