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Ok, according to some notes I have, the following is true for a random variable $X$ that can only take on positive values, i.e $P(X<0=0)$

$\int_0^{\infty}(1-F_X(x))dx=\int_0^{\infty}P(X>x)dx$

$=\int_0^{\infty}\int_x^{\infty}f_X(y)dydx$

$=\int_0^{\infty}\int_0^{y}dxf_X(y)dy$

$=\int_0^{\infty}yf_X(y)dy=E(X)$

I'm not seeing the steps here clearly. The first line is obvious and the second makes sense to me, as we are using the fact that the probability of a random variable being greater than a given value is just the density evaluated from that value to infinity.

Where I'm lost is why: $=\int_x^{\infty}f_X(y)dy=\int_0^{y}f_X(y)dy$

Also, doesn't the last line equal E(Y) and not E(X)?

How would we extend this to the discrete case, where the pmf is defined only for values of X in the non-negative integers?

Thank you

Justin
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  • See the answers to this question. The proof for the discrete case is, in my opinion, easier for beginners to understand. – Dilip Sarwate Sep 07 '13 at 21:00
  • It's not

    $$ \int_x^\infty f_X(y),dy = \int_0^y f_X(y),dy.$$

    The order of integration is changed (allowed per Fubini's theorem), you have

    $$\int_0^\infty \left(\int_x^\infty f_X(y),dy \right), dx = \int_0^\infty \left(\int_0^ydx \right),f_X(y) dy.$$

     – Daniel Fischer Sep 07 '13 at 21:03
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    You are integrating a function of two variables over the region ${(x,y) \colon x \leq y < \infty, 0 \leq x < \leq \infty}$ by letting $y$ vary from a fixed $x$ all the way to $\infty$ (inner integral) and then letting $x$ vary from $0$ to $\infty$ (outer integral). An alternative way is to fix $y$ and so $x$ varies from $0$ to $y$ (inner integral) and then let $y$ vary from $0$ to $\infty$ (outer integral). Since the integrand is not a function of $x$ at all, the inner integral just gives $y$. – Dilip Sarwate Sep 07 '13 at 21:08
  • @DilipSarwate, thank you for the explanation. It does make sense now. – Justin Sep 07 '13 at 22:34
  • Even though there's already a good link provided by Dilip Sarwate, I’d still like to make the links to the current choice of target posts for continuous and discrete. Please see the meta post for (abstract) duplicates. – Lee David Chung Lin Mar 10 '19 at 21:08

1 Answers1

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The region of integration for the double integral is $x,y \geq 0$ and $y \geq x$. If you express this integral by first integrating with respect to $y$, then the region of integration for $y$ is $[x, \infty)$. However if you exchange the order of integration and first integrate with respect to $x$, then the region of integration for $x$ is $[0,y]$. The reason why you get $E(X)$ and not something like $E(Y)$ is that $y$ is just a dummy variable of integration, whereas $X$ is the actual random variable that defines $f_X$.

user2566092
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