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?
 A: 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 $E[X|X>a]>a$
A: 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 $\int_A X(\omega)P(d\omega)$.
A: 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.
A: 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 $\int_H p(x)dx=0$.
