Probability density of square

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 $$X^2$$?

• $P(X^{2}\leq x)=P(-x \leq X \leq x)=\int_{-x}^{x} f_X(t)dt$. Differentiate this. – Kavi Rama Murthy Nov 22 at 11:33
• Nearly. $P(X^{2}\leq x)=P(-\sqrt x \leq X \leq \sqrt x )=\int_{-\sqrt x}^{\sqrt x} f_X(t)dt$ – oskar szarowicz Nov 22 at 23:29

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 $$U=X^{2}$$.

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

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