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According to Wikipedia, a characteristic function completely determines the properties of a probability distribution. This means it must be unique. However, the definition given is:

$$ \text{Char of }X (t)=E[e^itX]$$

Now $e^{iz}$ repeats for every $2 \pi$ increase in $z$. So how can it be unique?

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I thought at first you were asking how the characteristic function can be unique. There's no issue here, because e^ix is a well-defined function: for a given value of x, there is a unique value of e^ix. And so the expectation (an integral) has a unique value as well.

On second thought, it appears you're asking how it's possible, considering that e^ix is not injective (i.e., multiple x can have the same value of e^ix), that the original distribution can be completely recovered from the characteristic function. The answer to that is that you're probably missing the "t" in the expression: the characteristic function is a function of t, and although for a given value of t (e.g. t=1), the random variables $X$ and $X+2k\pi/t$ would have the same value of the characteristic function at that point, they would have different values at other t. So the characteristic functions are different, and yes, the distribution can be completely recovered from the characteristic function.

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Okay, so thats why my reasoning is wrong. I don't suppose you know how to prove that no two probabilities distributions have the same characteristic function? – Casebash Jul 30 '10 at 23:33
As a first step, note that you can recover all the moments from the characteristic function: the derivative of the characteristic function E[e^(itX)] is E[iXe^(itX)]=iE[X] at t=0, the second derivative gives -E[X^2], etc. And you can also recover the actual distribution itself, see the section on inversion formulas at… – ShreevatsaR Jul 30 '10 at 23:48
BTW: when the distribution has a density function, the characteristic function is just the Fourier transform and vice-versa, so for most continuous random variables you encounter, a simpler reason is: take the [inverse] Fourier transform of the characteristic function and you get the pdf. – ShreevatsaR Jul 31 '10 at 1:53

It's just a Fourier transform. E(x) is an integral over the probability distribution. The function is unique; if you focus on the value inside the expectation integral, that's not, but so what?

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