So we are given a measure $\mathbb{P}$ such that $\mathbb{P}(X\in\mathbb{Z})=1$. And we are asked to prove $$\mathbb{P}(X=n)=\frac1{2\pi}\int_0^{2\pi}\exp(-itn)\varphi_X(t)\ dt$$

where $\varphi_X(t)$ is a characteristic function of X.

I can believe the statement is true as I can spot inverse fourier trasform of $\varphi_X(t)$ which in fact is a fourier transform of $X$. What I did is I simply substituded $\varphi_X(t)$ from the definition and try to play with it but I get:

$RHS=\frac1{2\pi}\int_0^{2\pi}\exp(-itn)\sum_{k\in\mathbb{Z}}\exp(ikt)\mathbb{P}(t=k)\ dt$

I tried some algebraic manipulations it took me nowhere closer to the result and what's more I am not even sure if I can swap the order of the limits.

I would like to get some hint on that one and explanation why I can swap the order of the limits we are taking (if this is the right way)


Hint: First, $\varphi_X(t) = \sum_{k \in \mathbb{Z}} e^{ik t} \mathbb{P}(X = k)$, so you should correct those typos. Then, what is $$\frac{1}{2\pi}\int_0^{2\pi} e^{it(k-n)}\,dt$$ equal to?

Hint pt 2: To justify changing the order, show that $$\int_0^{2\pi} \sum_{k \in \mathbb{Z}} \left| e^{i t(k - n)} \mathbb{P}(X = k) \right| < \infty$$ and then use Fubini's theorem.

  • $\begingroup$ I solved the problem already, however not entirely because I did not give an argument for why I can change the limit of integration and summation. So that's all that is left. $\endgroup$ – Kran Dec 5 '19 at 16:32
  • $\begingroup$ @Kran, I added a hint for how to justify changing the order. $\endgroup$ – Marcus M Dec 5 '19 at 17:30
  • $\begingroup$ oooohh right! I just fixed myself on Lebesgue theorems..... $\endgroup$ – Kran Dec 5 '19 at 17:45
  • $\begingroup$ can you also take a look at my solution of this: math.stackexchange.com/questions/3464384/… Sorry for imposing this on you haha, but it seems no one besides me is interested in this problem. $\endgroup$ – Kran Dec 5 '19 at 17:47

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