In basic probability courses, we learnt that if $$X \sim Poisson(\lambda)$$ , then

$$E(X) = V(X) = \lambda$$

So obviously, the expectation and variance of a Poisson random variable are the same.

Now, is there other probability distribution which has same expectation and variance?

If yes, then by using what method could we find that probability distribution(s)?

  • 2
    $\begingroup$ I think you could come up with a lot if you consider the 1st and 2nd moments. If $f$ is the moment generating function, then to have equal mean and variance you would have $f'(0)=f''(0)-f'(0)^2$. $\endgroup$ – gd1035 Oct 9 '18 at 0:59


For any random variable $X$ with $\Var[X] < \infty$, the random variable $$Y = X - \mathbb{E}[X] + \Var[X]$$ has $$\Var[Y] = \Var[X] \qquad \mathrm{and} \qquad \mathbb{E}[Y] = \Var[X]\,.$$

In other words, for any random variable with finite variance, you can shift it so that the mean and variance are the same.

  • $\begingroup$ Wow this is interesting!! Then we should have many examples other than the Poisson distribution. $\endgroup$ – son520804 Oct 9 '18 at 1:42

Define a distribution with the same mean and variance as having property S.

Obtaining a two-valued (discrete) distribution with property S is easy. Let $P(x=m+d)=P(x=m-d)=\frac12$ with $m>0$. The mean is obviously $m$ and the variance is $d^2$, so $d=\sqrt m$. Similar computations lead to more-valued discrete distributions with property S.

For continuous distributions, there is obviously the normal $\mathcal N(m,m)$ distribution, which indeed is sometimes used to approximate the Poisson distribution. The uniform distribution can be made to have property S; fixing the lower bound at 0 we must have $b/2=b^2/12$ for the upper bound $b$, leading to $b=6$.

The most general method of obtaining distributions with property S would be through the moment generating function, as suggested in comments.


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