Show $E(X^k) = \int^\infty_0 (1-F_X(x))h(x) dx$

Question

I've been asked to show that for the positive continuous random variable $X$, $E(X^k) = \int^\infty_0 (1-F_X(x))h(x) dx$ for some $h(x)$ which I must find. The previous question was to show that $E(X) = \int^\infty_0 (1-F_X(x)) dx$. Could someone give me a hint on how to use this result for this question?

Thanks!

Answer 1

$E(X^k)=\int_0^\infty (1-F_{X^k}(x))dx$ using the result that for any positive r.v. $Y$, $E(Y)=\int_0^\infty (1-F_Y(y))dy$.

Now $F_{X^k}(x)=P(X^k\leq x)=P(X\leq x^{1/k})=F_X(x^{1/k})$.

Substitute this into the integrand and change the variable $x\mapsto y^k$ to get $h(x)=kx^{k-1}$.

Answer 2

This is another way to do it.

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Note that \begin{align} X = \int_0^Xds \end{align}

Using this observation, we have \begin{align} X^k = \int_0^X ks^{k-1}ds = \int_0^{\infty} 1_{\{s<X\}}ks^{k-1}ds \end{align} Taking expectations, we obtain \begin{align} \mathbb{E}[X^k]&=\mathbb{E}\left[\int_0^{\infty} 1_{\{s<X\}}ks^{k-1}ds\right]\\ &=\int_0^{\infty} \mathbb{E}[1_{\{s<X\}}]ks^{k-1}ds\\ &=\int_0^{\infty} \mathbb{P}(s<X)ks^{k-1}ds\\ &=\int_0^{\infty} [1-F_X(s)]ks^{k-1}ds \end{align}