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Suppose that $X$ is a random variable with $E(X) = Var(X)$. Then the distribution of $X$

  1. is necessarily Poisson,

  2. is necessarily Exponential,

  3. is necessarily Normal,

  4. Cannot be identified from the given data.

For Poisson distribution we have mean = variance = $\mu$. For Exponential and Normal this equality doesn't hold. So can we conclude that option 1 is correct?

Thank you.

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No, you cannot make that conclusion. It is not correct that the equality always doesn't hold for a normal distribution. One can take a normal distribution with any (finite) variance and mean. E.g. see the wikipedia article.

Edit thanks to joriki: Note that you are mistaken about the facts as pointed out in the main answer, but also your proposed logical deduction doesn't hold. You said that you "know" that the equation holds in case (1) (Poisson) and doesn't hold in cases (2) and (3) (exponential and normal). This doesn't allow us to conclude the distribution is necessarily Poisson. There could be another distribution that is neither Poisson nor exponential nor normal for which this equality holds. The only way you could conclude (1) is if Poisson were the unique distribution for which this relation held. I showed that that this is not the case, hence you must deduce (4).

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    $\begingroup$ It's perhaps worth mentioning that even if it were true that this equality never holds for exponential and normal distributions, one could still not conclude that the distribution is necessarily Poisson. $\endgroup$
    – joriki
    Apr 18, 2020 at 17:54

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