The distribution of the sample means determines the distribution of the population.

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It is obvious that the distribution of the sample means is completely determined from the distribution of the population. I wonder if the reverse holds. In other words, if the distribution of the sample means is known, can we determine the distribution of the population from it?

I claim that this holds true, which is equivalent to the following.

Claim:

For two random variables $X$ and $Y$, let $X_1 , ..., X_n$ be random samples from $X$ and $Y_1 , ..., Y_n$ be those from $Y$. That is, $X_1 , ..., X_n$ are independent and identically distributed as $X$ and so do $Y_1 , ..., Y_n$ and $Y$. Suppose that the distributions of $\tilde{X} := X_1 + \cdots + X_n$ and $\tilde{Y}:=Y_1 + \cdots + Y_n$ are the same, then the distributions of $X$ and $Y$ also coincide.


I made the first proof of my claim with MGF(moment generating function)s.

Proof(when $X$ and $Y$ both have their MGFs):

Let $M_X (t):=\operatorname{E}\left(e^{tX}\right)$ and $M_Y (t):=\operatorname{E}\left(e^{tY}\right)$ be MGFs of $X$ and $Y$, respectively. Then the MGFs of $\tilde{X}$ and $\tilde{Y}$ are given by $M_{\tilde{X}} (t) = \left(M_X (t)\right)^n$ and $M_{\tilde{Y}} (t) = \left(M_Y (t)\right)^n$, which are the same by the assumtion. Since MGFs are real and positive, it follows that $M_X (t)=M_Y (t)$, implying that $X$ and $Y$ have the same distribution.

As one might notice from the title this is not applicable when $X$ or $Y$ doesn't have MGF near $0$.


So I tried with ch.f.(characteristic function)s of $X$ and $Y$ which always exist and proceeded in a similar way.

Proof(with ch.f. of $X$ and $Y$, not completed):

Let $\phi_X (t) := \operatorname{E}\left(e^{itX}\right)$ and $\phi_Y (t) := \operatorname{E}\left(e^{itY}\right)$ be ch.f.s of $X$ and $Y$, respectively. Then the ch.f.s of $\tilde{X}$ and $\tilde{Y}$ are given by $\phi_{\tilde{X}} (t) = \left(\phi_X (t)\right)^n$ and $\phi_{\tilde{Y}} (t) = \left(\phi_Y (t)\right)^n$, which are the same by the assumtion.

I would like to argue that $\phi_X (t)$ and $\phi_Y (t)$ are the same here, but since ch.f.s are no longer real, I can no longer progress.


I would appreciate if someone suggest a way to solve this (or show that it doesn't hold in some special case).

Thanks for your time:)