What if a random variable is (surely/almost surely) greater than (or even less than) its mean? $X \ge \mu$ ($X \le E[X]$)

integrationmeasure-theoryprobabilityprobability theory

Assume that $X$ is integrable in probability space $(\Omega, \mathscr F, \mathbb P)$, i.e. $X \in \mathscr L^1 (\Omega, \mathscr F, \mathbb P)$.

  1. What does it mean if a random variable is (surely/almost surely) greater than its expected value? $X \ge E[X]$ I think this means $X$ is, at least almost surely, constant. (I'm not sure $X$ is surely constant even if $X \ge E[X]$ surely.) How do we prove this though? What I've done so far:

    • 1.1. I can prove this for $X$ indicator and nonnegative simple (and nonnegative discrete).
    • 1.2. I didn't bother anymore to try for nonnegative integrable and general integrable because I'm hoping for some simple proof I might've over looked like…
    • 1.3. …like prove that $P(X > E[X]) = 0$ through something like this. Maybe consider $E[X1_A]$ where $A=\{X > E[X]\}$ or something.
    • 1.4. If standard machine is really the way to go about this, then I'm stuck: For nonnegative integrable, probably monotone convergence theorem, but not really sure how. But since we're still in nonnegative, I'm guessing we'll have $X=0$. For general integrable, ok this part I remember is actually not just simple but also easy, so I must really be over looking something.

  2. Does the same conclusion in (1) (I mean whatever is the correct conclusion and not necessarily what I have stated) hold if $X$ is instead (surely/almost surely) less than its expected value? $X \le E[X]$

  3. Elementary/basic probability theory: If $X$ is a continuous random variable, then how do we show it is impossible that $X \ge E[X]$ surely (and also $X \le E[X]$ surely) (and also almost surely, but you know, it's still elementary/basic)? (I guess ignore this part if you can answer the above without measure theory.)

  4. If answering any of the above is easier if we assume $X$ is square integrable, then please tell me how (eg somehow we can say $Var[X]=0$).

Best Answer

This can be shown through measure theory. Let's use your $A=\{X>\mu\}$. Then on $A$, $X-\mu$ is positive. On its complement $A^c$, $X$ equals $\mu$ so $X-\mu$ is zero.

Hence:

$$\mu=E[X]= \mu+ \int_A X-\mu dP + \int_{A^c} X-\mu dP = \mu + \int_A X-\mu dP.$$

Now note that the final term has to be zero and the integrand is strictly positive on $A$. Hence, $A$ has to be a null-set.

You can try to follow a similar proof strategy for the case of a continuous random variable with a pdf.

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