For random variable X, is there any general result for satisfying $E[X] = E[X^k]$ for $k>0, k \neq 1, k \in \mathbb{R}$? For random variable $X$, is  there any general result for satisfying $E[X] = E[X^k]$ for $k>0$, $k \neq 1$, $k \in \mathbb R$ and where $k$ is given? 
I do not assume any distribution here, but if there is no general result for all distributions, normal distribution case is fine.
 A: It will certainly depend on the distribution.  In the normal distribution with mean $\mu$ and standard deviation $\sigma$, $E[X] = \mu$ and $E[X^k]$ is a rather complicated function of $\mu$, $\sigma$ and $k$: in the case where $k$ is a positive integer it is the coefficient of $x^k$ in the Maclaurin series of
$\exp(\mu x + \sigma^2 x^2/2)$, and this is
$$ \left(-i \sigma\right)^k H_k\left(\dfrac{i \mu}{\sigma}\right)$$
where $H_k$ is the "probabilists'" $k$'th Hermite polynomial.
A: A random variable that takes values in $\{0,1\}$ will satisfy $\Bbb E[X] = \Bbb E[X^k]$ for any $k>1$. A random variable that takes values in $\{-1,0,1\}$ will satisfy $\Bbb E[X] = \Bbb E[X^k]$ for $k=3,5,7,\ldots$. I don't think there's a general result, aside from (by definition)
$$ \int_{\Bbb R} x\,\mathrm dF(x) = \int_{\Bbb R} x^k\,\mathrm dF(x). $$
where $F$ is the distribution function of $X$. For a continuous random variable this reduces to $$\int_{\Bbb R}x\,f(x)\mathrm dx = \int_{\Bbb R}x^k\,f(x)\mathrm dx $$
where $f$ is a probability density function of $X$. 
