# Power (Square) of Different Distributions?

If there is a PDF $$\text{f}(x)$$, and $$x$$ follows different distributions, for example, Rayleigh, Rice, Nakagami, and Weibull. Then what will be the distribution of $$|x|^2$$ for these distributions? Please refer me to any book, research paper or any article which helps me with this.

You have some random variable $$X$$ with density function $$f$$ and CDF $$F$$, and you want to know something about the density and/or CDF of the random variable $$Y := X^2$$. You can do that by considering first the CDF of $$Y$$.
Let $$a$$ be a fixed nonnegative number. Then \begin{align*} \mathbb P(Y \leq a) &= \mathbb P(X^2 \leq a) \\ &= \mathbb P(-\sqrt a \leq X \leq \sqrt a) \\ &= F(\sqrt a) - F(-\sqrt a) \end{align*} which gives an explicit formula for the CDF of $$Y$$ in terms of that of $$X$$. Since the density is the derivative of the CDF, we can leverage this to get the density function of $$Y$$, which we'll call $$g$$: \begin{align*} g(a) &= \frac{\mathrm d}{\mathrm da} \mathbb P(Y \leq a) \\ &= \frac{\textrm d}{\textrm d a} \left[ F(\sqrt a) - F(-\sqrt a) \right] \\ &= f(\sqrt a) \cdot \frac{1}{2 \sqrt a} - f(-\sqrt a) \cdot \frac{-1}{2 \sqrt a} \\ &= \frac{f(\sqrt a) + f(-\sqrt a)}{2 \sqrt a} \\ \end{align*} and in particular if $$f$$ is symmetric, then we have $$g(a) = \frac{f(\sqrt a)}{\sqrt a}.$$
In general, if $$X$$ has cdf $$F(x) = P(X \leq x)$$ and density function $$f(x)$$, the cdf for $$X^2$$ will be $$P(X^2 \leq x) = P(-\sqrt{x} \leq X \leq \sqrt{x}) = F(\sqrt{x}) - F(-\sqrt{x}).$$ Then assuming $$F$$ is differentiable, the density function for $$X^2$$ is $$\frac{d}{dx}F(\sqrt{x}) - F(-\sqrt{x}) = f(\sqrt{x}) / (2 \sqrt{x}) + f(-\sqrt{x}) / (2 \sqrt{x}).$$ See also this question and the very detailed answer given there.