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Let $X$ be a random variable, then for any $n\in \mathbb{N}, \mathbb{E}^\mathbb{P}(X^n)<\infty.$ Then use Fubini's theorem to prove that

$$ \mathbb{E}^\mathbb{P}(X^n) = n \int_0^\infty t^{n-1}(1-F_X(t))dt - n \int_{-\infty}^0t^{n-1}F_X(t) dt$$

for CDF $F_X$ of $X$ and probability measure $\mathbb{P}$.

We know Fubini's theorem states for stochastic kernel $K$ that:

$$ \mathbb{E}^\mathbb{P}(X) = \int_{\Omega_1} \left(\int_{\Omega_2}X(\omega_1,\omega_2)K(\omega_1,d\omega_2) \right)\mathbb{P}_1(d{\omega_1})$$ where $\mathbb{P}_1$ is a probability measure on $\mathcal{F}_1$. So now, for our case:

$$ \mathbb{E}^\mathbb{P}(X^n) = \int_{\Omega_1} \left(\int_{\Omega_2}X^n(\omega_1,\omega_2)K(\omega_1,d\omega_2) \right)\mathbb{P}_1(d{\omega_1})$$

But i'm not too sure how to rewrite the kernel and get it in the form desired?

I know that $$ \mathbb{E}^\mathbb{P}(X^n) =\int_0^\infty x^nf(x) dx...$$ but i am not sure how to use the above theorem

Eric Wofsey
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Btzzzz
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1 Answers1

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Hint

Denote $(X^n)^+$ and $(X^n)^-$ the positive and negative part. One can prove that if $n$ is odd, then $(X^n)^+=(X^+)^n$ and $(X^n)^-=(X^-)^n$. And if $n$ is even, $(X^n)^+=X^n$ and $(X^n)^-=0$. Then,

\begin{align*} \mathbb E[(X^+)^n]&=\int_{\Omega } (X^+(\omega ))^n\mathbb P(\mathrm d \omega )\\ &=\int_\Omega \int_0^{X^+(\omega )}nx^{n-1}\,\mathrm d x \mathbb P(\mathrm d \omega )\\ &\underset{\text{Fubini}}{=}\int_{0}^\infty nx^{n-1}\int_\Omega \boldsymbol 1_{\{X(\omega )>x\}}\mathbb P(\mathrm d \omega )\,\mathrm d x\\ &=n\int_0^\infty x^{n-1}\mathbb P\{X(\omega )^+>x\}\,\mathrm d x. \end{align*}

I let you do the other cases and conclude.

Surb
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