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Given: $F_X(x)$ is a CDF and: $E[X] = \int\limits_0^\infty (1-F_X (x))\, dx\ $

How do I prove: $E[X^n] = \int\limits_0^\infty nx^{n-1}(1-F_X(x))dx $

Alex Pozo
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Thomas Sowell
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    Note that you need $X\ge 0$ for the first equation to hold. In this regime, the function $x \mapsto x^n$ is monotone. Can you then work out the cdf of $X^n$ in terms of that of $X$? If so, simply apply the first equality to $X^n$ and perform a substitution. – stochasticboy321 Mar 16 '17 at 00:17
  • This process doesn't seem to get me the correct answer. – Thomas Sowell Mar 16 '17 at 00:33
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    Byron's answer is of course correct, but it's curious that you didn't get the answer using the above. Are you sure you did the calculations correctly? Let $Y=X^n$. Unpacking the definition of $F_Y$, we have $F_Y(y) = P(X^n \le y) = P(X \le y^{1/n}) = F_X(y^{1/n})$ (here we're using that $X\ge 0$ a.s.). Thus, $$\mathbb{E}[X^n] = \int_0^\infty (1-F_Y(y)),\mathrm{d} y = \int_0^\infty (1-F_X(y^{1/n})) ,\mathrm{d} y .$$ Now use the substitution $u = y^{1/n}$ in the final integral, and the correct expression pops right out. – stochasticboy321 Mar 16 '17 at 07:20

2 Answers2

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The usual trick is to write $1-F_X(x)$ as an integral over the probability space $\Omega$ and use Fubini's theorem to exchange the order of integration. \begin{eqnarray} \int_0^\infty n x^{n-1}P(X>x)\,dx&=&\int_0^\infty n x^{n-1}\int_\Omega {\bf 1}_{X(\omega)>x}\,P(d\omega)\,dx\\[5pt] &=&\int_\Omega\int_0^\infty n x^{n-1} {\bf 1}_{X(\omega)>x}\,dx\,P(d\omega)\\[5pt] &=&\int_\Omega\int_0^{X(\omega)} n x^{n-1} \,dx\,P(d\omega)\\[5pt] &=&\int_\Omega X(\omega)^n \,P(d\omega)\\[5pt] &=&E(X^n). \end{eqnarray}

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Another way is recall that for a nonnegative random variable Y, $E[Y] = \int_{0}^{\infty} P[Y > t] dt.$

Then, for nonnegative random variable X and $n \in \mathbb{N}$, we have

\begin{align} E[X^n] &= \int_{0}^{\infty} P[X^n > t] dt\\ &= \int_{0}^{\infty} P[X^n > x^n] nx^{n-1}dx\\ &= \int_{0}^{\infty} nx^{n-1} P[X > x] dx\\ \end{align}

Note: We have used the change of variable $t=x^n$, so $dt=nx^{n-1}dx$.

Q.E.D.

Bayesian guy
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