# Meaning of expected value?

Given probability space $(\Omega, \mathcal{F}, P)$ and $P$-measurable function $X: \Omega \to \mathbb{R}$ called a random variable, define the expectation of $X$ to be $E[X]=\int_\Omega X(\omega)dP(\omega) = \int_\mathbb{R} xd\mu_X(x)$ where $\mu_X$ is a measure on $\mathbb{R}$ defined as $\mu_X(B)=P(X^{-1}(B))$.

I'm trying to learn some rigorous probability theory and I was wondering how to interpret the definition of the expectation. Thanks!

• It is the average of your random variable, namely random function, with respect to the probability measure associated to it. Actually your integral is still over the sample space but doesnt matter. Feb 25, 2015 at 3:21
• I can't be the only one who finds it bizarre that someone would know what a measure is before they have some intuition about what expected value is, or know how to look that intuition up for themselves. Feb 25, 2015 at 3:59

Another option is to consider the law of large numbers. Suppose that $X$ is a random variable such that $E[X^2] < \infty$. Let $X_1,X_2,X_3,\ldots$ be i.i.d. samples of $X$. Then almost surely $$\lim_{n\to\infty} \frac{X_1+\cdots+X_n}{n} = E[X].$$ That is, if you take the average of many copies of $X$, then you get something very close to $E[X]$, and in the limit, you get exactly $E[X]$. So the expectation is some sort of average value of your random variable.
If the random variable $X$ is discrete, then there is a simple explanation for the formula. Suppose that $X$ attains value in the domain $D$. Its expectation is $$E[X] = \sum_{x \in D} \Pr[X = x]x.$$ What's the idea here? Suppose that you take $n$ samples of $X$. Roughly $\Pr[X=x]n$ of them are equal to $x$, and their contribution to the average is roughly $\Pr[X=x]x$. Summing over all $x$, we obtain the formula.