Show that $\mathbb E[Y^p]=\int_0^\infty py^{p-1}\mathbb P\{Y\geq y\}\mathrm d y.$

Question

Let $Y\geq 0$ a r.v. Show that $$\mathbb E[Y^p]=\int_0^\infty py^{p-1}\mathbb P\{Y\geq y\}\mathrm d y.$$

I have that $$\mathbb E[Y^p]=\int_0^\infty Y^p\mathrm d \mathbb P,$$ how can I continue ?

Answer 1

For $k\geqslant 1$ and $t\geqslant 0$, the map $(t,\omega)\mapsto kt^{k-1}\mathsf 1_{\{\omega:X(\omega)\geqslant t\}}$ is measurable as the product of measurable functions, and because it is nonnegative, from Tonelli's theorem we have \begin{align} \int_{[0,\infty)\times\Omega} kt^{k-1}\mathsf 1_{\{\omega:X(\omega)\geqslant t\}}\mathsf d(t\times\mathbb P(\omega)) &= \int_0^\infty kt^{k-1} \int_\Omega \mathsf 1_{\{\omega:X(\omega)\geqslant t\}}\,\mathsf d\mathbb P(\omega)\,\mathsf dt\\ &= \int_0^\infty kt^{k-1}\mathbb E\left[\mathsf 1_{\{\omega:X(\omega)\geqslant t\}}\right]\,\mathsf dt\\ &= \int_0^\infty kt^{k-1}\mathbb P(X\geqslant t)\,\mathsf dt, \end{align} and also \begin{align} \int_{[0,\infty)\times\Omega} kt^{k-1}\mathsf 1_{\{\omega:X(\omega)\geqslant t\}}\,\mathsf d(t\times\mathbb P(\omega)) &= \int_\Omega \int_0^\infty kt^{k-1}\mathsf 1_{\{\omega:X(\omega)\geqslant t\}}\,\mathsf dt\,\mathsf d\mathbb P(\omega) \\ &= \int_\Omega \int_0^{X(\omega)}kt^{k-1}\,\mathsf dt\,\mathsf d\mathbb P(\omega)\\ &= \int_\Omega X(\omega)^k\,\mathsf d\mathbb P(\omega)\\ &= \mathbb E\left[X^k\right]. \end{align} It follows that $$\mathbb E\left[X^k\right] = \int_0^\infty kt^{k-1}\mathbb P(X\geqslant t)\,\mathsf dt. $$