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Suppose I know the probability density of a one-dimensional random variable $X$. Towards what direction do I need to think in order to calculate the probability density of $X^2$?

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    $\begingroup$ $P(X^{2}\leq x)=P(-x \leq X \leq x)=\int_{-x}^{x} f_X(t)dt$. Differentiate this. $\endgroup$ – Kavi Rama Murthy Nov 22 at 11:33
  • $\begingroup$ Nearly. $P(X^{2}\leq x)=P(-\sqrt x \leq X \leq \sqrt x )=\int_{-\sqrt x}^{\sqrt x} f_X(t)dt$ $\endgroup$ – oskar szarowicz Nov 22 at 23:29
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If you want to find the probability distribution of a function of random variables, you can use the nexts methods:

  1. The method of distribution function.

  2. The method of transformation.

  3. The method of moment generating functions.

Now, I'm going to explain each method:

  1. The method of distribution function: Let, $U$ be a function of the random variables $Y_{1}, Y_{2},\ldots, Y_{n}$. 1.1 Find the region $U=u$ in the $(y_{1},y_{2},\ldots,y_{n})$ space.
    1.2. Find the region $U\leq u$.
    1.3. Find the $F_{U}(u)=\mathbb{P}[U\leq u]$ by integrating $f(y_{1},y_{2},\ldots,y_{n})$ over the region $U\leq u$.
    1.4. Find the density function $f_{U}(u)$ by differenting $F_{U}(u)$. Thus, $\displaystyle f_{U}(u)=\frac{d F_{U}(u)}{du}$.

Note: For example, in your problem $U=X^{2}$, so you need to find $\displaystyle \mathbb{P}[X^{2}\leq x]$.

  1. The method of transformation: Let $U=h(Y)$, where $h(Y)$ is either an increasing or decreasing function of $y$ for all $y$ such that $f_{Y}(y)>0$. So,
    2.1. Find the inverse function, $y=h^{-1}(u)$.
    2.2. Evaluate $$\frac{dh^{-1}}{du}=\frac{dh^{-1}(u)}{du}$$ 2.3. Find $f_{U}(u)$ by $$f_{U}(u)=f_{Y}\circ h^{-1}(u) \left|\frac{dh^{-1}}{du} \right|$$

Note: In your problem, you can take $U=h(X)=X^{2}$, and use this method if possible.

  1. The method of moment generating functions: Let $U$ be a function of the random variables $Y_{1},Y_{2},\ldots, Y_{n}$.
    3.1. Find the moment generating function for $U$, $m_{U}(t)$.
    3.2. Compare $m_{U}(t)$ with the other well-known moment generating function. If $m_{U}(t)=m_{V}(t)$ for all values of $t$, so $U$ and $V$ have a identical distribution.

Note: In your problem, you could take $U=X^{2}$.

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Suppose $f_X$ is PDF of $X$ and $F_X$ is CDF of $X$. Then

$$F_{X^2}(t) = P(X^2 < t) = P(-\sqrt{t} < X < \sqrt{t}) = F_{X}(\sqrt{t}) - F_{X}(- \sqrt{t})$$

And if we differentiate both sides by $t$, we get

$$f_{X^2}(t) = \frac{f_X(\sqrt{t}) + f_X(-\sqrt{t})}{2\sqrt{t}}$$

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