It is true that $P(X= Y)=1 \Rightarrow E(X)= E(Y)$ and $P(X\ge Y)=1 \Rightarrow E(X)\ge E(Y)$, but is it true that $P(X > Y)=1 \Rightarrow E(X) > E(Y)$ ?

The proofs for the first two don't quite carry over so I'm not sure if this is true or not.

  • $\begingroup$ Are you familiar with measure theory or are you exclusively using random variable notation? $\endgroup$ – j4GGy Apr 23 '15 at 12:52
  • $\begingroup$ @GenericNickname I'm taking an undergrad course on probability theory so I'm sort of familiar with the introductory parts of measure theory. $\endgroup$ – simonzack Apr 23 '15 at 12:54
  • $\begingroup$ You need the expectations to exist for the first two statements to be meaningful. For the third statement, it helps if at least one is finite. $\endgroup$ – Henry Apr 23 '15 at 13:36
  • $\begingroup$ @Henry Doesnt the third one require expectations to exist as well? Or do you mean the first two work for $X,Y\ge 0$ as well? $\endgroup$ – simonzack Apr 23 '15 at 13:54
  • $\begingroup$ On the third statement, they again need to exist, but if one is finite then the other will exist (even if possibly infinite) $\endgroup$ – Henry Apr 23 '15 at 13:58

Let $P\{Z>0\}=1$ and realize that $\{Z>0\}=\bigcup_{n=1}^{\infty}\{Z\geq\frac1{n}\}$.

That implies that $P\{Z\geq\frac1{n}\}$ converges to $P\{Z>0\}=1$ so that $P\{Z\geq\frac1{n}\}>\frac12$ for some $n$.

Consequently $\mathbb EZ\geq\frac1{n}\frac12>0$.

Apply this on $Z=X-Y$.


Alternatively $1=P\left\{ Z>0\right\} \leq\sum_{n=1}^{\infty}P\left\{ Z\geq\frac{1}{n}\right\} $ by subadditivity.

Then $P\left\{ Z\geq\frac{1}{n}\right\} >0$ for some $n$ implying the existence of an $\epsilon$ with $P\left\{ Z\geq\frac{1}{n}\right\} \geq\epsilon>0$.

Then $Z\geq\frac{1}{n}1_{\left\{ Z\geq\frac{1}{n}\right\} }$ a.s. so that $\mathbb{E}Z\geq\mathbb{E}\frac{1}{n}1_{\left\{ Z\geq\frac{1}{n}\right\} }=\frac{1}{n}\mathbb{P}\left\{ Z\geq\frac{1}{n}\right\} \geq\frac{1}{n}\epsilon>0$.

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  • $\begingroup$ I can see the other steps but how do you arrive at the "consequently" part? $\endgroup$ – simonzack Apr 23 '15 at 13:17
  • $\begingroup$ $Z\geq\frac{1}{n}1_{\left\{ Z\geq\frac{1}{n}\right\} }$ a.s, so that $\mathbb{E}Z\geq\mathbb{E}\frac{1}{n}1_{\left\{ Z\geq\frac{1}{2}\right\} }=\frac{1}{n}\mathbb{P}\left\{ Z\geq\frac{1}{2}\right\} >0$ $\endgroup$ – drhab Apr 23 '15 at 13:19

Yes. The proof for the general case is a bit messier than the other two, because as you know we usually only retain nonstrict inequalities under a limit process. The idea is that the set $X>Y$ can be broken up into a union of sets $X>Y+1/n$. If any of these has positive probability, and $X<Y$ has zero probability, then $E[X]>E[Y]$ (why?). The way we show that one of them has positive probability is using continuity of measure. We write:

$$\{ X>Y \}=\bigcup_{n=1}^\infty \{ X > Y + 1/n \}.$$

Now the union on the right is an increasing union, so we get

$$P(X>Y) = \lim_{n \to \infty} P(X>Y+1/n).$$

The definition of limit then implies that we have some $n$ with $P(X>Y+1/n)>1/2$, say. That finishes the proof. (Note that I have used the shorthand notation which is common in probability theory.)

The slightly tricky part is proving continuity of measure; this can be found in essentially any measure theory text, but let me know if you need help with it.

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  • $\begingroup$ Thanks we did do the continuity of measure. $\endgroup$ – simonzack Apr 23 '15 at 13:21

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