10C

Mean and variance of a continuous distribution

The mean or expected value, variance and standard deviation of a continuous random variable found by integration, the uniform distribution, and linear changes aX + b.

The formulas

These match the discrete formulas, with the sum replaced by an integral.

Mean and variance

$$\mu = E(X) = \int_{-\infty}^{\infty} x f(x)\,dx \qquad \operatorname{Var}(X) = \int_{-\infty}^{\infty} x^2 f(x)\,dx - \mu^2$$

The standard deviation is $\sigma = \sqrt{\operatorname{Var}(X)}$. As before, $E(aX + b) = aE(X) + b$ and $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$.

Example 1●●

$f(x) = 2x$ for $0 \le x \le 1$. Find the mean, variance and standard deviation.

$\mu = \displaystyle\int_0^1 2x^2\,dx = \tfrac23$.

$E(X^2) = \displaystyle\int_0^1 2x^3\,dx = \tfrac12$, so $\operatorname{Var}(X) = \tfrac12 - \tfrac49 = \tfrac{1}{18}$.

$\sigma = \sqrt{\tfrac{1}{18}} \approx 0.24$.

The uniform distribution

Uniform on a ≤ x ≤ b

$$\mu = \frac{a + b}{2} \qquad \operatorname{Var}(X) = \frac{(b - a)^2}{12}$$

Example 2●●

$X$ is uniform on $0 \le x \le 6$. Find the mean and standard deviation.

$\mu = 3$ by symmetry.

$E(X^2) = \displaystyle\int_0^6 \tfrac16 x^2\,dx = 12$, so $\operatorname{Var}(X) = 12 - 9 = 3$ and $\sigma = \sqrt3 \approx 1.73$. This agrees with $\tfrac{6^2}{12} = 3$.

Flashcards

$E(X)$ for a PDF

$\displaystyle\int x f(x)\,dx$

$operatorname{Var}(X)$ for a PDF

$\displaystyle\int x^2 f(x)\,dx - \mu^2$

Uniform variance

$\dfrac{(b - a)^2}{12}$

Practice

$f(x) = 2x$ on $0 \le x \le 1$. Find $E(X)$.
For the same PDF, find $E(X^2)$.
For the same PDF, find $\operatorname{Var}(X)$.
$X$ is uniform on $0 \le x \le 6$. Find $E(X)$.
$X$ is uniform on $0 \le x \le 6$. Find $\operatorname{Var}(X)$.
$X$ is uniform on $4 \le x \le 10$. Find $E(X)$.
$f(x) = 3x^2$ on $0 \le x \le 1$. Find $E(X)$.
$\operatorname{Var}(X) = 3$. Find $\operatorname{Var}(2X + 5)$.
A PDF is symmetric about $x = 4$. Its mean is: