I'm new to characteristic functions and I would really appriciate some help with the following question:
"Give the distribution which has characteristic function $\varphi(t)=cos(t)$."
I've tried to go backwards by the definition of a characteristic function, but can't get it right. Any suggestions on how I should think to solve this?
It seems that the distribution is a symmetric Bernoulli, but
in my book there's a theorem that says:
"If the distribution of X is discrete, then
$P(X=x)=\displaystyle{\lim_{T \to \infty}} \frac{1}{2T} \int_{-T}^{T} e^{-itx} \cdot \varphi(t) dt$."
The problem is that I get a really weird result when trying to integrate this (even though I used the (I + I)-trick):
$\frac{(e^{-iTx}+e^{iTx})sin(T)-ix(e^{-iTx}-e^{iTx})cos(T)}{(1-x^2)}$
This doesn't look very "Bernoulli". What am I doing wrong?
Best Answer
Careful! Your $1-x^2$ denominator vanishes at $x=\pm 1$, which as you've noted are the places where $P(X=x)\ne 0$. Here's another approach, using the fact that $\lim_{T\to\infty}\int_{-T}^T e^{-ity}dt=0$ for all $y\ne 0$ (proof is an exercise; you should find the integral is $\frac{\sin Ty}{Ty}$). Since$$\frac{1}{2T}\int_{-T}^T\exp -itx\frac{\exp it+\exp -it}{2}dt=\frac{1}{4T}\int_{-T}^T(\exp -it(x-1)+\exp -it(x+1))dt,$$if $x\ne \pm 1$ the integral vanishes as $T\to\infty$, so that dividing out a multiple of $T$ gets $0$. But when $x=1$, what we get instead is$$\lim_{T\to\infty}\frac{1}{4T}\int_{-T}^T (1+\exp -2it)dt=\frac{1}{2}.$$The result $P(X=-1)=\frac{1}{2}$ follows similarly.