# Typical value of random variable

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?

• Do you mean for the value $a$ to be both the lower bound and the expectation?
– Joe
Aug 11 at 1:48
• @Joe yes. Hm, is that not clear from the formulation of the statement? Aug 11 at 1:52
• I just wasn’t sure
– Joe
Aug 11 at 1:52
• @AlexOrtiz That's begging the question. Aug 11 at 1:56
• @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. Aug 11 at 2:00

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 $$\int_A X(\omega)P(d\omega)$$.

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 $$E[X|X>a]>a$$

• I think it isn't clear, at least it isn't to me, why what's written before the $\implies$ actually implies $P(X>a) = 0$. Aug 11 at 2:46
• @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.
– Bey
Aug 11 at 3:11
• 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. Aug 11 at 3:25
• @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.
– Bey
Aug 11 at 3:30
• 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. Aug 11 at 3:49

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.

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 $$p(x. Then \begin{align} E(X) &=\int_G a p(x)dx+\int_H x p(x) dx\\ &=\int_Ua p(x)dx+\int_H(x-a)p(x)dx\\ \end{align} But $$\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$$.