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How do I make this statement slightly more rigorous?

If $X$ is a positive random variable (i.e. $X \geq a$ for some $a > 0$) and if $\mathbb{E}[X]= a$, then the "typical value of $X$ is $a$".

My question is the meaning of the phrase in parenthesis, is this the same as "almost always $X=a$" or "$P((X-a)^2 > \epsilon) = 0$ for every $\epsilon$"? If so how do I show that?

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    $\begingroup$ Do you mean for the value $a$ to be both the lower bound and the expectation? $\endgroup$
    – Joe
    Commented Aug 11, 2022 at 1:48
  • $\begingroup$ @Joe yes. Hm, is that not clear from the formulation of the statement? $\endgroup$
    – nervxxx
    Commented Aug 11, 2022 at 1:52
  • $\begingroup$ I just wasn’t sure $\endgroup$
    – Joe
    Commented Aug 11, 2022 at 1:52
  • $\begingroup$ @AlexOrtiz That's begging the question. $\endgroup$
    – nervxxx
    Commented Aug 11, 2022 at 1:56
  • $\begingroup$ @AlexOrtiz Comment to your edit that $X$ is in this case just a constant, non-random number. This isn't true I believe. I can imagine there can be events where $X$ takes a non-$a$ value. It's just that the measure of those events is 0. $\endgroup$
    – nervxxx
    Commented Aug 11, 2022 at 2:00

4 Answers 4

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There is no precise meaning for the phrase "the typical value of $X$ is $a$" if $X$ is a random variable and $a$ is a given fixed number. It can have a precise meaning only once it has been defined. In the situation we are dealing with in the OP, $EX$ is essentially the only value $X$ takes, which we can prove as follows.

Consider the events $A_n=\{X > EX+1/n\}$, for each $n\ge 1$. Using the definition of $A_n$, and the lower bound $X\ge EX$, we have \begin{align*} EX &= E[X;A_n] + E[X;A_n^c]\\ &\ge (EX+1/n)P(A_n)+(EX)P(A_n^c)\\ &= EX + P(A_n)/n\\ &\ge EX. \end{align*} Therefore, all the inequalities must actually be equalities, so $P(A_n) = 0$ for every $n$. Since $P(X>EX) = \lim_{n\to\infty}P(A_n)$, we conclude $P(X>EX) = 0$. Notice we do not require $EX > 0$ to reach this conclusion: if $X$ is any random variable which is greater than or equal to its mean almost surely, then $X = EX$ almost surely.


The notation $E[X;A]$ where $X$ is a random variable and $A$ is an event means the same thing as $\int_A X(\omega)P(d\omega)$.

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Yeah, almost always $X=a$ as in $P(X\neq a)=0$.

This follows because:

$$E[X]=a=E[X|X\leq a]P(X\leq a) + E[X|X>a]P(X>a) = $$ $$aP(X\leq a) + E[X|X>a]P(X>a) \implies P(X>a)=0$$

This follows from $X\geq a$ and that $E[X|X>a]>a$

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  • $\begingroup$ I think it isn't clear, at least it isn't to me, why what's written before the $\implies$ actually implies $P(X>a) = 0$. $\endgroup$
    – Alex Ortiz
    Commented Aug 11, 2022 at 2:46
  • $\begingroup$ @AlexOrtiz $E[X|X>a]>a$ and if $P(X>a)>0$ then $E[X]>a$, since $X=a,X>a$ partition the values of X. $\endgroup$
    – Annika
    Commented Aug 11, 2022 at 3:11
  • $\begingroup$ Part of the problem is that in the general case, we cannot simply partition the values of $X$. A priori, $X$ could be a continuous random variable, so the equality $E[X] = aP(X=a) + \dots$ doesn't necessarily make sense. $\endgroup$
    – Alex Ortiz
    Commented Aug 11, 2022 at 3:25
  • $\begingroup$ @AlexOrtiz I don't make any assumptions about continuity. $P(X=a)$ makes perfect sense for a continuous RV, it's just that most cases it is $0$ since $P_X$ is atomless. But here we cannot have any positive probability for $X>a$ or we shift the mean away from $a$, contradicting our assumptions. I'll revise if you have a counterexample. $\endgroup$
    – Annika
    Commented Aug 11, 2022 at 3:30
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    $\begingroup$ Of course, there isn't any counterexample to the claim, since it's a true claim, but I think I see the way your argument works now. I had to work through the technical apparatus of conditional expectation on my own to follow the manipulation of $E[X]$, though. $\endgroup$
    – Alex Ortiz
    Commented Aug 11, 2022 at 3:49
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We have $E(X - a) = 0$. Since $X - a$ is nonnegative, a standard result of measure theory implies $X - a = 0$ almost surely. Hence $X = a$ almost surely.

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Define $$ E(X)=\int_U x p(x)dx $$ where $U$ is the universe, then divide the universe into a part where $x=a$, say $G$, and $x>a$, say $H$, then $G\cup H=U$ since $p(x<a)=0$. Then $$ \begin{align} E(X) &=\int_G a p(x)dx+\int_H x p(x) dx\\ &=\int_Ua p(x)dx+\int_H(x-a)p(x)dx\\ \end{align} $$ But $$\int_U ap(x)dx=a\int_U p(x)dx=a,$$ and we are given $E(X)=a$ so subtracting $\int_H(x-a)p(x)dx=0$ and since $x>a$ for $x\in H$ and $p(x)\geq0$ that must mean that $\int_H p(x)dx=0$.

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