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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)

Luc M
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    Something seems wrong with the statement. Are there really supposed to be three $\int$ signs on the right-hand side? – Math1000 May 28 '15 at 17:23
  • 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 $\lim{t\rightarrow\infty} t^{k+1}(1-F_X(t))=0$. – Michael May 28 '15 at 17:28
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    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. – Stefan Hansen May 28 '15 at 17:31
  • @Math1000 Well, that is what it is stated on the exam sheet. Do you think that's an error? – Luc M May 28 '15 at 17:32
  • @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. – Michael May 28 '15 at 17:37
  • @Michael Why is integration by parts the main idea? There is no need for integration by parts, even in the general case - it's a straight-forward application of Fubini's theorem. – saz May 28 '15 at 17:44
  • @Michael: For one, you would have to show that the limit in your expression is in fact 0. I think the measure-theoretic approach is mathematically much neater and essentially the exercise is just a simple application of Fubini. – Stefan Hansen May 28 '15 at 17:47
  • @saz , Well, it seems the measure theory notation was scary enough so that it was not immediately obvious there was even a typo in the question. Perhaps we can agree that the argument on 1-dimensional integrals uses basic calculus (understandable by a larger crowd), while the other requires familiarity with measure theory. Obviously measure theory is a foundation, and a measure theory class is about measure theory. But it is often useful to get other (perhaps simpler) perspectives. – Michael May 28 '15 at 17:51
  • @StefanHansen , You cannot use Fubini without justifying the integral of the absolute value is finite (which is not always true), a proper justification seems as difficult as considering the cases when $t^{k+1}(1-F_X(t))\rightarrow 0$ (consider the tail of the integral for $E[X^{k+1}]$ when this is a finite value). However, the Tonelli theorem let's us switch integrals of non-negative functions without worrying about this issue (as it gives $\infty = \infty$ when that arises). – Michael May 28 '15 at 19:10
  • @Michael: People usually mean Tonelli when the function is non-negative, so there's no need for further justification here. – Stefan Hansen May 28 '15 at 19:48
  • @StefanHansen , *laugh. I guess I get no love then, hey? =) – Michael May 28 '15 at 20:08
  • @Michael: Sure, I just didn't agree with what you were saying. My comments weren't meant in any harsh way :) – Stefan Hansen May 28 '15 at 20:45
  • This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post. Also see a measure-theoretic treatment. – Lee David Chung Lin Nov 13 '18 at 13:48

2 Answers2

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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 $$1_{\{\omega : X(\omega) \geqslant t\}}(\omega) = 1_{\{t: t\leqslant X(\omega)\}}(t). $$

Math1000
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    We submitted answers at the same time. You have a typo: A factor of $1/(k+1)$ is missing. – Michael May 28 '15 at 18:39
  • 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. – Michael May 28 '15 at 19:11
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    By "Fubini's theorem" I really mean the Fubini-Tonelli theorem ;) – Math1000 May 28 '15 at 19:23
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    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! =) – Michael May 28 '15 at 20:20
  • @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. – Luc M May 29 '15 at 13:41
  • The third equality here is not obvious at all to me – Trajan Feb 09 '24 at 14:26
  • @Trajan By "Fubini" (the author said (s)he meant "Fubini-Tonelli") $\int_0^\infty \int_{\Omega} f(t,\omega)\mathsf d\mathbb P;\mathsf dt=\int_{\Omega}\int_0^\infty f(t,\omega)\mathsf dt;\mathsf d\mathbb P$, where $f(t,\omega)=t^k$ if $X(\omega) \geqslant t$, and $0$ else. – Anne Bauval Feb 09 '24 at 14:39
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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.

Michael
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