Characteristic function of a square of normally distributed RV

Question

Let's assume that $ X \sim \mathcal{N}(0,1) $. I'm supposed to compute the characteristic function of $ X^2 $.

As far as I got is that the density of $ X^2 $ is $ g(y) = \frac{1}{2\sqrt{2\pi}} e^{-y/2} \frac{1}{\sqrt{y}} \textbf{1}_{[0,\infty)}$ and therefore

$$ \psi_{X^2}(t) = \frac{1}{2\sqrt{2\pi}}\int\limits_{0}^{\infty} e^{itx} e^{-x/2}\frac{1}{\sqrt{x}} dx$$

The problem is I'm not acquainted with computing comples integrals and don't know how to continue from this point. Is there a way around / an easy non-residue method of calculating that?

Answer 1

You messed things up when you used the density of $X^2$ and wrote $\mathrm e^{\mathrm itx^2}$ in the integral. Either use the density of $X$ and write $\mathrm e^{\mathrm itx^2}$ in the integral, or use the density of $Y=X^2$ and write $\mathrm e^{\mathrm ity}$ in the integral.

Personally, I find the former option more systematic and less error-prone than the latter, here the latter option yields $$\varphi_Y(t)=E(\mathrm e^{\mathrm itY})=E(\mathrm e^{\mathrm itX^2})=\int_\mathbb R\mathrm e^{\mathrm itx^2}\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\int_\mathbb R\frac1{\sqrt{2\pi}}\mathrm e^{-zx^2/2}\mathrm dx,$$ with $z=1-2\mathrm it$ such that $\Re(z)\gt0$. Maybe you can finish this.

Answer 2

A normally distributed r.v. has pdf $\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$, and therefore the characteristic function of $X^2$ where $X\sim\mathcal{N}(0,1)$ is $$ \mathbb{E} e^{itX^2} = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{it x^2}e^{-\frac{x^2}{2}} dx = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{(it-\frac12) x^2} dx= \frac{2}{\sqrt{2\pi}}\int_{0}^\infty e^{(it-\frac12) x^2} dx. $$ Now, it boils down to computing $\int_0^\infty e^{ax^2}dx$, for $a\in\mathbb{C}$ with $\operatorname{Re}(a) < 0$.

Answer 3

I think I'm fine with it now. We calculate:

$ \int\limits_{0}^{\infty} e^{\frac{x}{2}(2it - 1)} \frac{1}{\sqrt{x}} dx $ by substituting $ s^2 = x(-2it + 1), ds = \sqrt{1-2it}\frac{1}{2\sqrt{x}}dx $ and we get that it equals

$ \int\limits_{0}^{\infty} e^{-s^2/2} \frac{2}{\sqrt{1 - 2it}}ds = 2\sqrt{2\pi}/\sqrt{1-2it} $

And the characteristic function:

$ \phi_{X^2}(t) = \frac{1}{\sqrt{1-2it}} $