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Let $X$ be a real non-negative random variable on the probability space $(\Omega,\mathcal{F}, \mathbb{P})$. Given that $$ E[X]=\int_\Omega \int_0^\infty \chi_{t<X}\,dt\,d\mathbb{P}=\int_0^\infty \mathbb{P}[ X\geq t]\,dt, $$ show that, for all $\epsilon>0$ $$ E[X]\leq \sum_{n=0}^\infty \epsilon\mathbb{P}[X\geq n\epsilon]\leq E[X]+\epsilon. $$ Tried to use some Fubini combined with rewriting stuff as countable sums (like $\mathbb{P}[X\geq t]=\mathbb{P}\left[\bigcup_{n=1}^\infty \{X\geq nt\}\right]$) but I am a bit lost. Some intuition is also highly appreciated (I think I am beaten by the misunderstanding of notation).

Lee David Chung Lin
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Rgkpdx
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1 Answers1

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Notice that the summation in the middle is the Riemann sum of the decreasing function $f(t) = P(X>t).$ The monotonicity makes it easy to estimate the difference between the sum and the integral. Now, the question is: is the function Riemann integrable, and if not, does it matter?

Did
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Igor Rivin
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  • I would say that we do not know whether the probability measure of $X$ is Riemann integrable (is this what you are talking about?). However calling it a Riemann sum isn't a stronger assumption given that the trick can be done within the Lebesgue integration (doesn't it?). – Rgkpdx Jan 27 '14 at 23:15
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    @Ton It is actually NOT the probability measure, but the CDF. Part of the point is that this is a monotonic function and therefore IS Riemann-integrable, so... – Igor Rivin Jan 28 '14 at 01:06