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I have difficulties with an old exam problem :

Let $X$ be a positive random variable defined on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$. Show that
$$\int_0^\infty t^k \mathbf{P}(X\geq t) dt = \int_0^\infty \int_{\Omega} t^k\int_{\{X(\omega)\geq t\}}dt d\mathbf{P}(\omega)$$ Infer from this the integral expression of $\mathbb{E}(X^{k+1})$ (where $\mathbb{E}$ is the expectation)

We have Fubini theorem, which we can apply to a $\mathbb{B}(\mathbb{R})\otimes\mathcal{F}$-measurable function because the Lebesgue measure is $\sigma$-finite and $\mathbf{P}$ is also $\sigma$-finite because it is a probability. I think we can write $\mathbf{P}(X\geq t)$ as $\int_{\{ X(w)\geq t\}} d\mathbf{P}(\omega)$ but I don't know how to proceed next. Especially I don't see how to introduce the $\int_{\Omega}$.

Edit

From the comments, there must be an error in the description of the exam problem. It should have been the following :

Let $X$ be a positive random variable defined on a probability space $(\Omega, \mathcal{F}, \mathbf{P})$. Show that
$$\int_0^\infty t^k \mathbf{P}(X\geq t) dt = \int_0^\infty \int_{\Omega} t^k\mathbf{1}_{\{X(\omega)\geq t\}}dt d\mathbf{P}(\omega)$$ Where $\mathbf{1}_{\{X(\omega)\geq t\}}$ is the characteristic function of $\{ X(\omega)\geq t\}$

Infer from this the integral expression of $\mathbb{E}(X^{k+1})$ (where $\mathbb{E}$ is the expectation)

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    $\begingroup$ Something seems wrong with the statement. Are there really supposed to be three $\int$ signs on the right-hand side? $\endgroup$ – Math1000 May 28 '15 at 17:23
  • $\begingroup$ Egads. This is an example of how measure theory can be used to make simple things difficult. I agree that the $\int_{\Omega}$ is likely a typo. For intuition, you can assume $X$ has a density $f_X(x)$ and use integration by parts to get: $$ \int_0^{\infty} t^k (1-F_X(t))dt = \left.\left(\frac{t^{k+1}}{k+1}(1-F_X(t))\right)\right|_0^{\infty} +\frac{1}{k+1} \int_0^{\infty} t^{k+1}f_X(t)dt = \frac{1}{k+1}E[X^{k+1}]$$ where you need to justify the limit $\lim_{t\rightarrow\infty} t^{k+1}(1-F_X(t))=0$. $\endgroup$ – Michael May 28 '15 at 17:28
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    $\begingroup$ The right-hand side doesn't make sense. First of all, there are three integrals there (I suspect the innermost integral should have been an indicator function). Second, the outermost integral is over $[0,\infty)$ but you're integrating with respect to $P$ which is a measure on $(\Omega,\mathcal{F})$. @Michael: That is essentially making things harder even at the cost of generality. $\endgroup$ – Stefan Hansen May 28 '15 at 17:31
  • $\begingroup$ @Math1000 Well, that is what it is stated on the exam sheet. Do you think that's an error? $\endgroup$ – Luc M May 28 '15 at 17:32
  • $\begingroup$ @StefanHansen , how does it make it more difficult? I agree it is not as general (assuming a density). Yet, one thing I have learned over the years is that "understandability" trumps "generality." In this case, the measure theory notation is so daunting that most people would ignore it, never getting the main idea that you can use integration by parts. $\endgroup$ – Michael May 28 '15 at 17:37
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By Fubini's theorem, we have

$$\begin{align*} \int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt&= \int_0^\infty t^k \mathbb E\left[1_{\{\omega:X(\omega)\geqslant t\}} \right]\mathsf dt\\ &=\int_0^\infty t^k\int_{\Omega} 1_{\{\omega:X(\omega)\geqslant t\}}\mathsf d\mathbb P\;\mathsf dt\\ &=\int_{\Omega}\int_0^{X(\omega)}t^k \mathsf dt\; \mathsf d\mathbb P\\ &=\int_{\Omega} \frac1{k+1}X^{k+1}(\omega)\mathsf d\mathbb P(\omega)\\ &= \frac1{k+1}\mathbb E[X^{k+1}]. \end{align*}$$

Hence $$\mathbb E[X^{k+1}] = (k+1)\int_0^\infty t^k\mathbb P(X\geqslant t)\mathsf dt.$$

The crucial part here is that $$1_{\{\omega : X(\omega) \geqslant t\}}(\omega) = 1_{\{t: t\leqslant X(\omega)\}}(t). $$

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    $\begingroup$ We submitted answers at the same time. You have a typo: A factor of $1/(k+1)$ is missing. $\endgroup$ – Michael May 28 '15 at 18:39
  • $\begingroup$ Actually I think you need to use Tonelli for switching integrals of non-negative functions. Else, more work is needed to justify conditions for when Fubini can be applied, and justifying $\infty = \infty$ in cases when Fubini does not work. $\endgroup$ – Michael May 28 '15 at 19:11
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    $\begingroup$ By "Fubini's theorem" I really mean the Fubini-Tonelli theorem ;) $\endgroup$ – Math1000 May 28 '15 at 19:23
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    $\begingroup$ Well, just as long as the original asker understands there is an issue here, and that non-negativity is what makes things work. I have seen several papers and books (mainly in engineering) that magically invoke theorems (like "Lebesgue dominated convergence") when conditions required for those theorems do not apply at all! =) $\endgroup$ – Michael May 28 '15 at 20:20
  • $\begingroup$ @Michael I think what I called Fubini theorem was also Fubini-Tonelli (I'm not sure we have seen each theorem separately in class). Thanks to you both for your answers. I'm a bit reassured to see there really was a error in the description of the exercise, I'll edit my question to make that clear. $\endgroup$ – Luc M May 29 '15 at 13:41
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Since I dissed measure theory, I feel obligated to give a measure-theory answer. This is essentially the method suggested by Stefan Hansen (using indicator function). The correct thing we want to prove is:

$$ \int_0^{\infty} t^k P[X\geq t]dt = \int_{\omega \in \Omega} \left[\int_0^{\infty} 1\{X(\omega) \geq t\} t^k dt\right] dP(\omega) $$ where $1\{X(\omega)\geq t\}$ is an indicator function that is 1 if $X(\omega)\geq t$, and 0 else.

To do this, we can use the Tonelli theorem about double integrals of non-negative functions. The main steps (you can fill in details) are:

\begin{align} \int_0^{\infty} t^k P[X\geq t] dt &= \int_0^{\infty} t^k \left[\int_{\omega: X(\omega)\geq t} dP(\omega) \right] dt\\ &= \int_0^{\infty} t^k\left[ \int_{\omega\in\Omega} 1\{X(\omega)\geq t\} dP(\omega)\right]dt \end{align}

and then use Tonelli to switch the order of integration.

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