Beyond expectations

Does there exist a random variable such that the probability of it exceeding or equaling its expected value is 0?

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1 solution

Alex Li
Jul 1, 2015

Relevant wiki: Expected Value - Problem Solving

Consider a variable x = 1 n x=\frac{1}{n} , where n n is randomly chosen from the interval ( 0 , 1 ) (0,1) . A simple integration can show that the expected value of x x is \infty , which is obviously unattainable.

This is not real satisfying. "Expected values" typically have real life meaning, so what would be a real life example of this counterexample? Are there any counterexamples that do not appeal to the infinite?

Michael Mendrin - 5 years, 11 months ago

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  1. infinite expectation are possible in reality (see @Agnishom Chattopadhyay ’s reply).
  2. no: there is no example of a r. v. with a non-infinite expected value, with the property in the original problem (see my post below).

R Mathe - 3 years ago

oh wow...but i don't really know calculus. can you give me a non calculus explanation? thanks!!!

Willia Chang - 5 years ago

Poorly worded problem. Can't make simplistic remarks about equaling or exceeding infinity.

Aditya Dua - 3 years, 11 months ago

This is rubbish, one may only talk of ‘the’ expected value of a real-valued r. v., X X , if and only if it is in L 1 L^{1} , which requires < E [ X ] < -\infty<\mathbb{E}[X]<\infty . Now, suppose P [ X m ] = 0 \mathbb{P}[X\geq m]=0 , where m : = E [ X ] m:=\mathbb{E}[X] and the expected value exists. Then m X > 0 m-X>0 a.e. By basic measure theory, this implies E [ m X ] > 0 \mathbb{E}[m-X]>0 . But E [ m X ] = m E [ X ] = 0 \mathbb{E}[m-X]=m-\mathbb{E}[X]=0 . This is a contradiction. Hence the assumption was wrong, and P [ X m ] 0 \mathbb{P}[X\geq m]\neq 0 .

R Mathe - 3 years ago

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