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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.

user157012
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    Hint: If E(X)=E(X^2)=1 then E(X^2)=E(X)^2 hence this puts some very heavy restriction on the distribution of X... (Anecdotally, no Poisson distribution fits these conditions.) – Did Jun 14 '14 at 08:33

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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 $n$.

Vladimir
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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, $E(X)=E(X^2)=1\Rightarrow k=k(k+1)=1\Rightarrow k=0=1!!$

Samrat Mukhopadhyay
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  • It's likely that OP is (mistakenly) guessing that it's Poisson or Gamma, having confused $\text{E}(X) = \text{E}(X^2)$ with $\text{E}(X) = \text{V}(X)$. I don't think that's part of the question. – M. Vinay Jun 14 '14 at 09:44