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In a proof I read recently, I saw the author conclude that $X$ is an Exp(1) random variable after finding $E(X^{k}) = k!$. Why and when is this allowed (i.e. when can I conclude that some r.v. follows a distribution whose expectation I recognize)? Is it not possible to construct another random variable with expectation $k!$ that isn't an Exp(1) random variable (maybe I just missed something earlier on)?

I haven't been able to find much online about the injectivity of the expectation function, so maybe someone can clarify for me.

  • The constant variable $k!$ also has that expectation. As does the variable which is $k!\pm 1$ each with probability $\frac 12$. You can easily find infinitely many other variables with the same expectation. – lulu Nov 03 '19 at 19:27
  • Then how can they conclude $X$ follows $\text{Exp}(1)$? –  Nov 03 '19 at 19:29
  • They must have more information. – lulu Nov 03 '19 at 19:29
  • In that case there is an enormous amount of information regarding the distribution. Not least, the fact that you have the desired relation for all $k$, not just one. But, if you have a question about that solution, you should ask the user who posted it. – lulu Nov 03 '19 at 19:33

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The claim is, that given $EX^k=k!$ for each $k=0,1,2,\dots$ implies $X$ has the indicated $\mathrm{Exp}(1)$ distribution. Most of the comments to your question ignore the statement that the $\mu_k=k!$ condition holds for all $k$, even though it is clear in the post you cite.

Read all about this kind of problem in general here and here. In your special case, since the series $$f(z)=\sum_{k\ge0} EX^k\frac {z^k}{k!}=\sum_{k\ge0}z^k = \frac 1{1-z}$$ converges for all complex $z$ in a neighborhood of $0$, there is only one probability distribution with the indicated moments.

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