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Let $X$ be a non-negative random variable. Why does the following hold:

$$\int_0^\infty \boldsymbol{1}_{X > c} \, dc = \int_0^X dc = X \quad\text{(?)} $$

I am confused because I thought that $\int_0^\infty \boldsymbol{1}_{X > c} \, dc$ gives me something like $\sum_{c \in [0,\infty)}\lambda(\{X > c\})$.

EDIT: This was the original equation: $\mathbb E(\int_0^\infty \boldsymbol{1}_{X > c}) \, dc =\mathbb E(\int_0^X dc) = \mathbb E(X)$. Maybe I read this wrong!

Pazu
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    The integral in question has nothing to do with randomness of $X$. – Sangchul Lee Jul 17 '17 at 20:50
  • This is part of $\int\limits_0^{\infty} (1-F_X(c) dc = \mathbb E(X)$ – Pazu Jul 17 '17 at 20:52
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    Let me elaborate my comment: the integral in question is a direct consequence of the following deterministic result: for any $a \geq 0$, $$ \int_{0}^{\infty} \mathbf{1}{{ c < a }} , dc = \int{0}^{a} dc = a. $$ Since this is true for any $a \geq 0$, you can even plug $a = X(\omega)$ which depends on some variable $\omega$ that is completely irrelevant to the computation of the integral above. – Sangchul Lee Jul 17 '17 at 20:58
  • @MichaelHardy You are kinda right, I had: $\mathbb E(\int_0^\infty \boldsymbol{1}_{X > c}) , dc =\mathbb E(\int_0^X dc) = \mathbb E(X)$ – Pazu Jul 17 '17 at 21:04
  • @Pazu : My earlier comment was mistaken. I made the crude mistake of neglecting the fact that $c$ is a bound variable. The answer by user365239 is correct. – Michael Hardy Jul 17 '17 at 21:08
  • This was posted countlessly many times on the site (by several users, including me). Is your question about a specific post on the site? – Did Jul 17 '17 at 21:13
  • @Did : I think the thing that's been posted repeatedly involved the proposition that $$ \int_0^\infty \Pr(X>c),dc = \operatorname{E}(X) \text{ if }\Pr(X>0) = 1.$$ But this is simpler than that.; it just says $$ \int_0^\infty \boldsymbol{1}_{X,>,c}(c) ,dc = X. $$ See the answer by user365239. Nothing in that answer, nor in the question, depends on $X$ being a random variable; that comes in afterward. – Michael Hardy Jul 17 '17 at 21:21
  • @Pazu : I now suspect that you read that $$ \int_0^\infty \Pr(X>c),dc = \operatorname{E}(X) \text{ if }\Pr(X>0) = 1 $$ and then in a proposed proof of that you found $$ \int_0^\infty \boldsymbol{1}_{X,>,c}(c) ,dc = \int_0^X , dc = X. $$ – Michael Hardy Jul 17 '17 at 21:25
  • @MichaelHardy No, I am referring to the identity between random variables. – Did Jul 17 '17 at 21:26
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    @Did : ok, But the fact that it is a random variable doesn't really have anything to do with the proof of that identity. $\qquad$ – Michael Hardy Jul 17 '17 at 21:27
  • @Did I understand if you want to close the thread. I think I got everything together and the proof was indeed discussed several times on this site. – Pazu Jul 17 '17 at 21:31
  • @MichaelHardy It has not, but there are definitely two identities here, one being the integrated form of the other, and I am referring to the other (if the mention "random variable" irks you, replace it by "function"). – Did Jul 17 '17 at 21:32
  • Possible duplicate of $E(X)=\int_{0}^{\infty}P(X>y)dy$ for non-negative r.v. $X$ – Pazu Jul 17 '17 at 21:39
  • One thing is missing here: If $\Pr(X>0)= 1$ then $$ \begin{align} X & = \int_0^\infty \boldsymbol{1}{X,>,c}(c) , dc \ \ \text{and therefore } \operatorname{E}(X) & = \operatorname{E}\int_0^\infty \boldsymbol{1}{X,>,c} (c) ,dc \ \ & = \int_0^\infty \operatorname{E}(\boldsymbol{1}_{X,>,c} (c)) , dc \longleftarrow \text{How is this step justified?} \ \ & = \int_0^\infty \Pr(X>c),dc. \end{align} $$ To justify the step mentioned above, I would cite Tonelli's theorem,$,\ldots\qquad$ – Michael Hardy Jul 17 '17 at 22:08
  • $\ldots,$which says that if a function being integrated is non-negative, then the double integral equals the iterated integral in either of the two orders regardless of whether that is finite or $+\infty. \qquad$ – Michael Hardy Jul 17 '17 at 22:09
  • @MichaelHardy Sorry but the notation $$\mathbf 1_{X>c}(c)$$ is absurd. One can either write $$X=\int_0^\infty\mathbf 1_{X>c},dc$$ or $$X(\omega)=\int_0^\infty\mathbf 1_{X>c}(\omega),dc$$ or even $$X(\omega)=\int_0^\infty\mathbf 1_{X(\omega)>c},dc$$ – Did Jul 18 '17 at 05:54
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    Even if this poster initially intended this to be a question with content other than what it's turned out to be, and that other content turned out to duplicate another question (and it does) nonetheless what was intended is not what happened, and this is not an exact duplicate, and it actually gives a slightly interesting way of looking at the matter that is not necessarily present in that older question. – Michael Hardy Jul 18 '17 at 15:37

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$$ \int_0^\infty \mathbf{1}_{X>c} \; dc = \int_0^X 1 \; dc + \int_X^\infty 0 \; dc = \int_0^X 1 \; dc = X $$

user365239
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  • Thanks you are of course totally right. I think I missed that $\int_0^\infty \boldsymbol{1}_{X > c} dc$ defines a random variable itself. – Pazu Jul 17 '17 at 21:37