Power of a random variable and related notion of moment I am trying to grasp the idea of a power of a random variable, defined as a function $\Omega \to E$. In understand how a function can self-compose with itself, but I am unable to relate this to the power of random variables. I haven't been able to verify online that the referred to power is indeed the functional power, and I'm uncertain to assume so, since, as far as I understand, a self-composing function $f:X \to Y$ would require that $Y ⊆ X$, which does not hold with a RV. I am self-teaching, and so getting by on bits and pieces, thus even though I make a sincere effort not to, there might be an important thing I'm missing for which I'd greatly appreciate any reference and perhaps explanation. 
A related question I have concerns the moments of a RV's distribution. How can we assume an expected value for the kth power of a random variable X $E[X^k]$, or better, how can we know the specific underlying distribution of the kth powers of X, for which its first moment (expected value) would be given with $E[X^k]$? Where does this multitude arise? 
 A: I assume you mean "power" like the random variable $X^k$? It's not a composition. $X$ is a function $\Omega\to \mathbb R.$ $X^k $ is just a function $\Omega\to \mathbb R$ given by $X^k(\omega) = (X(\omega))^k.$ 
In other words it's just like $\sin^2(x)$ is a power of $\sin(x).$
If you know the distribution of $X$ then the distribution of $X^k$ can be obtained by a change of variables. For instance if $X$ is a non-negative continuous RV with CDF $F_X(x) = P(X\le x),$ then the CDF of $X^k$ is $$ F_{X^k}(y) = P(X^k\le y) = P(X\le y^{1/k}) = F_X(y^{1/k}).$$ Since $X$ is continuous, so is $X^k$ and if we want the PDF we differentiate the CDF like $$ f_{X^k}(y) = \frac{d}{dy}F_{X^k}(y) = \frac{1}{k}y^{1/k-1} F_X'(y^{1/k}) = \frac{1}{k}y^{1/k-1}f_X(y^{1/k})$$ where we used the chain rule and the fact that the PDF of $X$ is the derivative of the CDF, i.e. $f_X(x)=F_X'(x).$
(Equivalently, you can use the change of variables formula to compute $f_{X^k}(y).$ It amounts to the same thing.)
We could then compute $E(X^k)$ in two ways. Either $$ E(X^k)=\int_0^\infty yf_{X^k}(y)dy$$ or $$ E(X^k) = \int_0^\infty x^kf_X(x)dx$$ and both will give the same answer.
