expectation of random variable

Question

I have this question :

What is the expected value of $E(X^{100})$ if X is a random variable such that $E(X)=E(X^2)=1$?

I am very confused as $X$ could be a poisson or gamma variate.

Answer 1

The condition $EX=EX^2=1$ seems to be wrong if the RV has to be Poisson or Gamma distributed.

As @Did said in the comments, $X$ can not be distributed as a Poisson random variable,because for Poisson RV's $Var(X)={E}X$. So, if it is a gamma RV, i.e if $$f_{X}(x)=\dfrac{x^{k-1}e^{- x}}{\Gamma(k)},\ x\ge 0$$ Then, $E(X)=E(X^2)=1\Rightarrow k=k(k+1)=1\Rightarrow k=0=1!!$

Answer 2

We have $$ E(X-E(X))^2=E(X^2)-E(X)^2=0 $$ Hence $X$ has zero variance and we have $E(X^n)=E(X)^n$ for all $n$. Indeed, we know this for $n=1,2$, and for $n\ge 2$ we have $$ E((X-E(X))^n)=\int_{-\infty}^\infty (x-E(X))^{n-2} (x-E(X))^2 d\mu(x), $$ where $d\mu(x)$ is the probability measure of $X$. But we already know that this is zero for $n=2$ and hence the measure $(x-E(X))^2d\mu(x)$ being nonnegative is identically zero. Now for $0=E((X-E(X))^n)$ we use the binomial formula and prove that $E(X^n)=E(X)^n$ by induction on $n$.