$X$ is distributed beta$(a,b)$. $Y=X^k$, where $k$ is a positive integer. I want to find the expectation of $Y$.
$$ f_X(x) = \frac{1}{\beta(a,b)} x^{a-1} (1-x)^{b-1} $$
Using chain rule, I have $$ f_Y(y) = \frac{1}{\beta(a,b)} (y^{1/k})^{a-1} (1-y^{1/k})^{b-1} \frac{1}{k} y^\frac{1-k}{k} \\ $$
Is there another form of this that's some other beta pdf? How can I get it into such a form?
$$\begin{align*} \operatorname{E}[X^k] &= \int_{x=0}^1 x^k \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1} \, dx \\ &= \frac{\Gamma(a+b)\Gamma(a+k)}{\Gamma(a)\Gamma(a+k+b)} \int_{x=0}^1 \frac{\Gamma(a+k+b)}{\Gamma(a+k)\Gamma(b)} x^{a+k-1} (1-x)^{b-1} \, dx \\ &= \frac{\Gamma(a+b)\Gamma(a+k)}{\Gamma(a)\Gamma(a+k+b)}. \end{align*}$$
You can see that it holds for any $k \in \mathbb R$ with $k>-a$.
