Transformation of Random Variable $Y = X^2$

Question

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 1

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:

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

Answer 2

If the square of a number is between $0$ and $1$ then the number itself has to be between $-1$ and $1$. Like, the squares of $-0.5$ and $0.5$ are both $0.25$. However, no real number will produce negative squares. This is independent of the nature of the number, it can be a fixed one or a randomly selected one. So,

$$P(X^2

If $y\geq 1$ then the square of every number between $-1$ and $\sqrt y$ will be less than $ y$. So

$$P(X^2

$$P(X^2

This is why

$$P(X^2

The rest is given by integrating the pdf over the respective domains.