I am trying to follow a proof where $Y$ is a continuous random variable with probability density function $f_{Y}$.

What I don't get is how or why changing order of integration of the following

  1. $\int_0^{\infty} P(Y \gt a) dy = \int_0^{\infty}\int_{a}^{\infty} f_{Y}(x) dx\ dy $ leads to

  2. $ = \int_0^{\infty}(\int_{0}^{x}\ dy) f_{Y}(x)\ dx$

What are the explicit algebraic or calculus manipulations which take us from 1 to 2?

  • $\begingroup$ They are applying Fubini's theorem. What exactly do you want to know? I am not really understanding what is your question. Are you having trouble with the integral bounds? $\endgroup$ – Patrick Da Silva Oct 26 '13 at 1:30
  • $\begingroup$ By the way, there is a little problem with $$ \int_0^{\infty} P(Y > a) dy $$ because $y$ does not appear in the integrand. There is also a problem with $2.$ because there $a$ has disappeared. $\endgroup$ – Patrick Da Silva Oct 26 '13 at 1:32
  • $\begingroup$ what's the problem? $a$ is just some constant. The integral $\int_0^{\infty}$ just happens to contain the expression $P(Y \gt a)$. $\endgroup$ – T. Webster Oct 26 '13 at 1:33
  • $\begingroup$ @PatrickDaSilva I realize they are applying Fubini's theorem, but what is the algebra that takes us from the equation in 1 to the one in 2? $\endgroup$ – T. Webster Oct 26 '13 at 1:34
  • $\begingroup$ The lower bound of the inner integral should be $y$ not $a$. $\endgroup$ – Mhenni Benghorbal Oct 26 '13 at 1:36

I will guess what would've been the correct expression. I am guessing $$ \int_0^{\infty} P(Y > y) dy $$ is the integral you want to compute (there is a theorem in which this expression appears...). $$ \int_0^{\infty} P(Y > y) dy = \int_0^{\infty} \int_y^{\infty} f_Y(x) \, dx \, dy. $$ Now you want to start integrating this integral with respect to $y$ instead of $x$ first. The region of the $xy$-plane you are integrating over is $$ \{ (x,y) \, | \,0 \le y \le x \} $$ because $y$ has no constraint except being positive, but $x$ has the constraint of being greater than $y$. If you swap these conditions around, $x$ has no constraint, but $y$ has the constraint of being smaller than $y$. This means that by Fubini's theorem, $$ \int_0^{\infty} \int_y^{\infty} f_Y(x) \, dx \, dy = \int_0^{\infty} \int_0^x f_Y(x) \, dy \, dx. $$ Note that the order of integration has changed, so that since $f_Y(x)$ does not depend on $y$, the first integral can be factored : $$ \int_0^{\infty} \int_0^x f_Y(x) \, dy \, dx = \int_0^{\infty} \left( \int_0^x \, dy \right) f_Y(x) \, dx = \int_0^{\infty} x \, f_Y(x) \, dx = \mathbb E [ Y ], $$ which I'm guessing was the point of this computation.

Hope that helps,

  • $\begingroup$ Yes that helps and you guess correct. The reason I took $a$ to be the place of $y$ was just confusion over application of e.g. $P(Y \le a) = F_y(a) = \int_0^a f_y(a) dy $. Here the proof specifically intends to prove $\mathrm{E}[Y] = \int_0^{\infty} P(Y > y) dy$ $\endgroup$ – T. Webster Oct 26 '13 at 21:46
  • $\begingroup$ Did you mean $y$ has the constraint of being smaller than $x$? $\endgroup$ – T. Webster Oct 26 '13 at 22:00
  • $\begingroup$ Well essentially if you first integrate over $y$ then over $x$, then $y$ is free and $x \ge y$. But if you integrate first over $x$ and then over $y$, then $x$ is free and $0 \le y \le x$. But of course this is just two different ways of saying that $0 \le y \le x$. $\endgroup$ – Patrick Da Silva Oct 27 '13 at 0:58

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