I am trying to evaluate the characteristic function of half-normal distribution (actually I was wondering whether if I take iid copies of half normal distribution, then defining $X=X_1-X_2$ would make $X$ normally distributed or not, I am yet to check that). The result I have read on Wikipedia page of folded normal distribution says that it is
$$\phi(t)=2e^{-\frac{\sigma^2t^2}{2}}\Phi(i\sigma t)$$
I am stuck on how to derive that. I am thinking of using indicator of whether the original normally distributed random variable (of which this one is modulus) is positive or negative, but to no avail.
Also, is there any method in general to find characteristic functions of modulus of random variables?
Also, can you please tell whether symmetrising the modulus of a random variable by taking difference of iid copies like above would give back original random variable times a constant or not.
Please help.
Best Answer
For a random variable $X$ having even pdf, $\varphi_{|X|}(t)=\varphi_X(t)+\mathbf{i}\,H\varphi_{X}(t)$, where $H\varphi$ is the Hiblert transform of $\varphi$ (see, e.g., this note). Thus, \begin{align} \varphi_{|X|}(t)&=e^{-\frac{(\sigma t)^2}{2}}-\mathbf{i}\left(\mathbf{i}\,e^{-\frac{(\sigma t)^2}{2}}\operatorname{erf}\!\left(\mathbf{i}\,\frac{\sigma t}{\sqrt{2}}\right)\!\right) \\ &=e^{-\frac{(\sigma t)^2}{2}}\left(1+\operatorname{erf}\!\left(\mathbf{i}\,\frac{\sigma t}{\sqrt{2}}\right)\!\right) \\[0.4em] &=2e^{-\frac{(\sigma t)^2}{2}}\Phi(\mathbf{i}\,\sigma t). \end{align} (the derivation of $H\varphi$ can be found in this note).
You can directly check that the difference between i.i.d. half-normal random variables $X_1$ and $X_2$ is not normally distributed (though, it's pdf is also even): $$ \varphi_{X_1-X_2}(t)=\varphi_{X_1}(t)\varphi_{X_2}(-t)=e^{-(\sigma t)^2}(1-[\operatorname{erf}(\mathbf{i}\, t/\sqrt{2})]^2), $$ which is not the c.f. of a normal distribution (with zero mean).
If you want to compute the c.f. of $|X|$ directly (assume, for simplicity, that $\sigma=1$), note that $$ \varphi_{|X|}(t)=\frac{1}{\sqrt{2\pi}}\int_0^{\infty}e^{\mathbf{i}\,tx}e^{-\frac{x^2}{2}}\,dx+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 e^{-\mathbf{i}\,tx}e^{-\frac{x^2}{2}}\,dx. $$ Consider the first integral (the second can be evaluated similarly): $$ \frac{1}{\sqrt{2\pi}}\int_0^{\infty}e^{\mathbf{i}\,tx}e^{-\frac{x^2}{2}}\,dx=\frac{e^{-t^2/2}}{\sqrt{2\pi}}\int_0^{\infty}e^{-\frac{(x-\mathbf{i}\, t)^2}{2}}\,dx=\frac{e^{-t^2/2}}{2}(1+\operatorname{erf}(\mathbf{i}\, t/\sqrt{2})). $$