Given the probability distribution of X, whats the PDF of X²?

Question

Let's say we have a random variable $X$ with a certain probability density function $f_x(x)$.

1) How should I find out the PDF of the random variable $X^2$?

Problem background:

$X_1 = s_1 + W$, where $W \sim \mathcal(0,\sigma^2)$ and $s_1 = A \cos(\phi)$ with $\phi \sim unif[-\pi,\pi]$ and $A$ a positive constant value. The pdf of $s_1$ is than $f_{s_1}(s_{1}) = \frac{1}{\pi \sqrt{A^2-s_1^2}}$. The pdf of $X_1$ can be described as:

\begin{equation} f_{x_1}(x_1) = f_{s_1}(s_1)*f_w(w) \end{equation}

Note the assumption that $s_1$ and $W$ are independent.

2) Any ideas for the analytical expression of $f_{x_1}(x_1)$?

By the way, the PDF of $X_1$ is attached below.

Let a second random variable be $X_2 = s_2 + W$, where $W \sim \mathcal(0,\sigma^2)$ and $s_2 = A \sin(\phi)$ with $\phi \sim unif[-\pi,\pi]$ and $A$ the same constant value as with $X_1$. The pdf of $s_2$ is than $f_{s_2}(s_{2}) = \frac{1}{\pi \sqrt{A^2-s_2^2}}$. The pdf of $X_2$ can be described the same way as before:

\begin{equation} f_{x_2}(x_2) = f_{s_2}(s_2)*f_w(w) \end{equation}

3) Any ideas for the PDF of $T = \sqrt{{X_1}^2+{X_2}^2}$?

Below is a plot of the PDFs of $X_1^2$ (blue) and $X_2^2$ (red).

Surprisingly, the response to 3) seems to be a normal distribution... but why? Even the variable ${T}^2 = {X_1}^2+{X_2}^2$ seems to follow also a normal distribution. The delay between both PDFs of $X_1$ and $X_2$ seems to be really important, since $\sqrt{2{X_1}^2} = \sqrt{2} X_1$ does not follow a normal distribution. The result of 3) is plotted below.

Really appreciate your help, regards.

Answer 1

On 1):

For $x>0$ we have: $$F_{X^{2}}\left(x\right)=F_{X}\left(x^{\frac{1}{2}}\right)-F_{X}\left(-x^{\frac{1}{2}}\right)$$

Taking the derivative on both sides we find:

$$f_{X^{2}}\left(x\right)=\frac{1}{2}x^{-\frac{1}{2}}\left[f_{X}\left(x^{\frac{1}{2}}\right)+f_{X}\left(-x^{\frac{1}{2}}\right)\right]$$

Answer 2

It seems that a random variable $Y = X^2$ would have a PDF

$$ f_Y(y) = \frac{f_X(\sqrt{y})+f_X(-\sqrt{y})}{2\sqrt{y}} \qquad y \geq 0 $$

but I have to give that some thought to verify/disprove it.

Off the top of my head, I can't think of a short cut for the PDF of $Z = \sqrt{X_1^2+X_2^2}$ where both of $X_1, X_2$ independently share the PDF $f_X(x)$. (Meaning integration is the path forward.) But I'll also give that some thought, too.