Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Do the mean and variance of a random variable define it completely? In particular,
if $X$ is a random variable, such that $E(X) = \operatorname{Var}(X) = u$,
does that imply that $X \sim \operatorname{Poisson}(u)$ ?
If so, then a proof that a RV is Poisson would require only to show the above relationship which is very useful. Otherwise, what measures can uniquely define a random variable, or a Poisson Random Variable?
No, for example you can take a standard normal distribution, multiply it by $\sqrt u$ and add $u$ to get a different distribution with mean and variance $u$. (You can do this trick with any non-constant starting distribution with finite mean and variance: take a linear transformation to get any mean and variance you want).
What is true is that if all the moments exist and match what they would be for a Poisson distribution (i.e. $\mathbb E(X^k)$ has the correct value for every $k$) then the distribution is Poisson.
[edit] The previous paragraph wouldn't necessarily be true if you were trying to determine whether $X$ was equal to some arbitrary distribution, but it is true for comparing with a Poisson distribution (and many other common distributions). Here is a thread on mathoverflow discussing when it is true.
A general random variable is completely defined by its distribution; if the support of a r.v. is an infinite set, finite number of point parameters like mean and variance can't define the type of a random variable.
