If $X$ is distributed $\mathcal{N}\sim(\mu,1)$ then what is the PDF of $X^2$?

Question

If $X$ is distributed $\mathcal{N}\sim(\mu,1)$ then what is the distribution of $X^2$? My try: \begin{align*} P(X^2<x)\\ P(-\sqrt{x}<X<\sqrt{x})\\ \int_{-\sqrt{x}}^{\sqrt{x}} f(x) \,dx \end{align*} Then: \begin{align*} f_{X^{2}}(x)=\frac{d}{dx}\int_{\sqrt{-x}}^{\sqrt{x}} f(x) \,dx\\ =f_X(\sqrt{x})\frac{1}{2\sqrt{x}}+f_X(-\sqrt{x})\frac{1}{2\sqrt{x}}\\ =\frac{1}{\sqrt{2\pi x}} e^{\frac{-(\sqrt{x}-\mu)^2}{2}} \end{align*} How should I continue?

Answer 1

$X^2$ is a member of the family of noncentral chi-squared distributions. There are a few problems with your calculations as others have pointed out. Notably, you should not use the same variable to represent both the limits of integration and the variable of integration. You should write instead $$\begin{align} f_{X^2}(x) &= \frac{d}{dx} \int_{t=-\sqrt{x}}^\sqrt{x} f_X(t) \, dt \\ &= \frac{f_X(\sqrt{x})}{2\sqrt{x}} + \frac{f_X(-\sqrt{x})}{2\sqrt{x}} \\ &= \frac{1}{2 \sqrt{2 \pi x}} \left( e^{-(\sqrt{x}-\mu)^2/2} + e^{-(\sqrt{x} + \mu)^2/2} \right). \end{align}$$ Your main computational error is in assuming that $f_X(\sqrt{x}) = f_X(-\sqrt{x})$. This is not true unless $\mu = 0$.

We can stop here, or we can perform additional algebraic manipulations: note $$(\sqrt{x} \pm \mu)^2 = x \pm 2\mu \sqrt{x} + \mu^2,$$ therefore $$e^{-(\sqrt{x} - \mu)^2/2} + e^{-(\sqrt{x} + \mu)^2/2} = e^{-x/2} e^{\mu \sqrt{x}} e^{-\mu^2/2} + e^{-x/2} e^{-\mu \sqrt{x}} e^{-\mu^2/2} = e^{-(x + \mu^2)/2} (e^{\mu \sqrt{x}} + e^{-\mu \sqrt{x}}).$$ Since $e^z + e^{-z} = 2 \cosh z$, we can write $$f_{X^2}(x) = \frac{e^{-(x+\mu^2)/2} \cosh (\mu \sqrt{x})}{\sqrt{2\pi x}}, \quad x \ge 0.$$

This density corresponds to a noncentral chi-squared distribution with $1$ degree of freedom and noncentrality parameter $\mu^2$.