Expectation and Variance for Continuous Random Variables

Expectation

For a continuous random variable $X$ with PDF $f$:

$E[X] = \int_{-\infty}^{\infty} x f(x) \, dx$

This is the continuous analogue of the discrete sum: integration replaces summation.

Intuition

The expectation is still a weighted average, but now with an integral. The density $f(x)$ weights each infinitesimal contribution.

Example: Uniform on [0, 2]

$E[X] = \int_0^2 x \cdot \frac{1}{2} \, dx = \frac{1}{2} \left[ \frac{x^2}{2} \right]_0^2 = \frac{1}{2} \cdot \frac{4}{2} = 1$

The centre of $[0, 2]$ is 1.

Example: Exponential

For $X \sim \text{Exp}(\lambda)$:

$E[X] = \int_0^{\infty} x \lambda e^{-\lambda x} \, dx$

Integration by parts gives:

$E[X] = \frac{1}{\lambda}$

Expectation of Functions (LOTUS)

For any function $g$:

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

Example: $E[X^2]$ for Uniform [0, 2]

$E[X^2] = \int_0^2 x^2 \cdot \frac{1}{2} \, dx = \frac{1}{2} \left[ \frac{x^3}{3} \right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$

Variance

$\text{Var}(X) = E[(X - E[X])^2]$

$\text{Var}(X) = E[X^2] - E[X]^2$

The computational formula works exactly as for discrete variables.

Example: Uniform on [0, 2]

$E[X] = 1, \quad E[X^2] = \frac{4}{3}$

$\text{Var}(X) = \frac{4}{3} - 1 = \frac{1}{3}$

Example: Exponential

$E[X] = \frac{1}{\lambda}, \quad E[X^2] = \frac{2}{\lambda^2}$

$\text{Var}(X) = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$

Standard deviation: $\sigma_X = \frac{1}{\lambda}$

Properties Remain Unchanged

All properties of expectation and variance carry over:

Property	Continuous
Linearity	$E[aX + b] = aE[X] + b$
Additivity	$E[X + Y] = E[X] + E[Y]$
Scaling	$\text{Var}(aX) = a^2 \text{Var}(X)$
Shift	$\text{Var}(X + c) = \text{Var}(X)$

Standard Deviation

$\sigma_X = \sqrt{\text{Var}(X)}$

This returns to the original units.

Example: Uniform on [a, b]

For $X \sim \text{Unif}[a, b]$:

$E[X] = \frac{a + b}{2}$

$\text{Var}(X) = \frac{(b-a)^2}{12}$

$\sigma_X = \frac{b-a}{\sqrt{12}}$

Quantiles

The $p$th quantile $x_p$ satisfies:

$F(x_p) = p$

Or: $\int_{-\infty}^{x_p} f(x) \, dx = p$.

Example: Exponential

Find $x_{0.5}$ (median) for $X \sim \text{Exp}(\lambda)$.

$P(X \leq x_{0.5}) = 1 - e^{-\lambda x_{0.5}} = 0.5$

$e^{-\lambda x_{0.5}} = 0.5$

$x_{0.5} = \frac{\ln 2}{\lambda}$

Note: $x_{0.5} \neq E[X] = \frac{1}{\lambda}$. For exponential, median $\approx 0.693/\lambda$, mean $= 1/\lambda$.

Summary

Quantity	Discrete	Continuous
Expectation	$\sum x p(x)$	$\int x f(x) dx$
$E[g(X)]$	$\sum g(x) p(x)$	$\int g(x) f(x) dx$
Variance	$E[X^2] - E[X]^2$	Same formula