Difference Between Mean, Average, Expected Value for Calculus

Question

Lately, I am having trouble understanding the difference between what is average, what is expected value, and what is mean when using Calculus. I am under the impression now that they are the same thing which confuses me. Why would expected value, the $E[X]$ symbol, have the form $E[X]=\int_a^bx\cdot f(x)dx$, rather than $E[X]=\frac{1}{b-a}\int_a^bf(x)dx$. I thought the second equation was the definition of average from a calculus stand point. I am wanting to know the difference between using these three terms and when to use their corresponding equations appropriately.

Answer 1

Note that $$\int_a^b f(x) dx=1$$ so $$ \frac{1}{b-a} \int_a^b f(x) dx=\frac{1}{b-a} $$ is in some sense "an average probability used" (see example below).

You want to say that expectation is the weighted average of value points, where the weights are given by the distribution, so this is like finding a center of mass, with density given by $f(x)$, and the result is exactly $$ \mathbb{E}[X] = \frac{\int_a^b x f(x) dx}{\int_a^b f(x) dx} = \int_a^b x f(x) dx $$ because the integral in the denominator is 1.

Example

It is like, if I have 0.1 kg at -1, 0.8 kg at 0, and 0.1 kg at 1, you see the average (center of mass) should be at 0 by symmetry, but your suggestion would give an average weight of $$ \frac{0.1+0.8+0.1}{1 - (-1)} = \frac{1}{2}. $$

But the expected value is not the average weight used, but the location of the center of mass, here $$ (-1) \cdot 0.1 + 0 \cdot 0.8 + 1 \cdot 0.1 = 0, $$ as intuitively expected by symmetry.