Mathematical Expectation

Discrete case

If $X$ is a random variable and $f (x)$ is the probability distribution of $X$ at x, the mean/expected value of $X$ is

E (x) = \sum_{x} x \cdot P (X = x) = \sum_{x} x \cdot f (x)

Joint Probability Mass function

E [g (X, Y)] = \sum_{x} \sum_{y} g (x, y) \cdot f_{X, Y} (x, y)

Continuous case

If $X$ is a random variable and $f (x)$ is the probability density of $X$ at x, the mean or expected value of $X$ is

E (X) = \int_{\infty}^{\infty} x \cdot f (x) d x

if the integral is not finite, $E (X)$ does not exist

Joint Probability Density Function

E [g (X, Y)] = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} g (x, y) \cdot f_{X, Y} (x, y) d x

Properties

Expected value of a function of $X$

If the series is absolutely convergent / the integral is finite

E [g (X)] = \int_{- \infty}^{\infty} g (x) f (x) d x

E [g (X)] = \sum_{x} g (x) f (x)

Otherwise, the expectation doesn't exist

If $g (X)$ is something simple, $E [g (X)]$ could be calculated using $E [X]$ , but when $g (X)$ has different behaviours depending on $X$ (piecewise, etc), the integral with both functions is necessary, possibly also splitting up the bounds

Simplifying the base definition:

E [a X + b] = a E [X] + b

Linear Combination

E [\sum_{i = 1}^{n} c_{i} g_{i} (X)] = \sum_{i = 1}^{n} c_{i} E [g_{i} (x)]

Expansion with the binomial theorem

E [(a X + b)^{n}] = \sum_{i = 0}^{n} (\begin{matrix} n \\ i \end{matrix}) a^{n - i} b^{i} E (X^{n - 1})

Darth Vader Rule

E (X) = \int_{0}^{\infty} [1 - F (X)] d x