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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?

AnnieFrannie
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  • I think it should be about the inverse Fourier transform. – Yuta Sep 14 '18 at 09:07
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    This is the characteristic function of a random variable taking values $-1$ and $1$ with probability $\frac 1 2 $ each. There is no general procedure for finding the distribution from characteristic function (except for the inversion theorem which does not apply to this case!) – Kavi Rama Murthy Sep 14 '18 at 09:07
  • Characteristic functions of discrete distributions are of the type $\sum c_n e^{ita_n}$. You can compare your function with this to guess what the distribution is. – Kavi Rama Murthy Sep 14 '18 at 09:10

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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.

J.G.
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  • Thank you for your answer! I think I'm starting to get it. But from where did you get $$\frac{1}{2T}\int_{-T}^T\exp -itx\frac{\exp it+\exp -it}{2}dt$$? I thought I knew all about integrals, but apparently not (or maybe I've stared at this for too long). I'm not sure I understand the part where you wrote that the denominator "vanishes" at $x=\pm 1$ . – AnnieFrannie Sep 15 '18 at 11:38
  • I just rewrote $\cos t$ complex exponentials. What I'm saying is $1-x^2=0$ for those values of $x$. – J.G. Sep 15 '18 at 11:43