Mean and Variance of Continuous Random Variables

A2 · S2 · 9 min

The mean and variance of a continuous random variable answer the same questions as for a discrete one: where is the centre of the distribution, and how spread out is it? The only change is that sums become integrals. Every Paper 6 continuous random variable question asks for at least one of $E(X)$ and $\text{Var}(X)$ , and the formulas are in the formula booklet, so the marks are earned by setting up the integrals correctly and carrying out the integration accurately.

From sums to integrals

For a discrete random variable,

E(X) = \sum x\,P(X = x), \qquad \text{Var}(X) = \sum x^2 P(X = x) - [E(X)]^2.

Each value is weighted by its probability. For a continuous variable, a thin strip of width $\delta x$ around the value $x$ has probability approximately $f(x)\,\delta x$ . Weighting each $x$ by that and adding over all strips gives an integral.

Key result

For a continuous random variable $X$ with probability density function $f(x)$ :

E(X) = \int x\,f(x)\,dx

\text{Var}(X) = \int x^2 f(x)\,dx - [E(X)]^2 = E(X^2) - [E(X)]^2

The integrals are taken over the interval on which $f$ is defined. Both formulas are in MF19.

The mean $E(X)$ is the balance point of the region under the curve: if the region were cut out of card, it would balance on a pivot at $x = E(X)$ .

Mean and variance from a pdf

Multiply $f(x)$ by $x$ , expand, and integrate over the interval: this is $E(X)$ .
Multiply $f(x)$ by $x^2$ , expand, and integrate over the interval: this is $E(X^2)$ .
$\text{Var}(X) = E(X^2) - [E(X)]^2$ . Do not forget to subtract the square of the mean.
Check: the mean lies inside the interval, and the variance is positive. The standard deviation is $\sqrt{\text{Var}(X)}$ .

A routine mean and variance

The random variable $X$ has probability density function $f(x) = \dfrac{x^2}{9}$ for $0 \le x \le 3$ , and $0$ otherwise. Find $E(X)$ and $\text{Var}(X)$ .

Solution

E(X) = \int_0^3 x \cdot \frac{x^2}{9}\,dx = \frac{1}{9}\left[\frac{x^4}{4}\right]_0^3 = \frac{81}{36} = 2.25

E(X^2) = \int_0^3 x^2 \cdot \frac{x^2}{9}\,dx = \frac{1}{9}\left[\frac{x^5}{5}\right]_0^3 = \frac{243}{45} = 5.4

\text{Var}(X) = 5.4 - 2.25^2 = 5.4 - 5.0625 = 0.3375

The mean $2.25$ is towards the right of $[0, 3]$ , which makes sense: the density increases with $x$ , so more of the probability is at the upper end.

Symmetry: the mean for free

If the graph of $f$ is symmetric about the line $x = c$ , the balance point is $c$ , so $E(X) = c$ with no integration. Typical symmetric densities are $kx(a - x)$ on $[0, a]$ (symmetric about $\frac{a}{2}$ ), $k(a^2 - x^2)$ on $[-a, a]$ (symmetric about $0$ ), and constant densities.

If the question says "state" or "write down" the mean, symmetry is what is wanted. If it says "find", a sentence quoting the symmetry is still acceptable, but the variance must always be integrated.

Using symmetry

$X$ has probability density function $f(x) = \dfrac{3}{32}x(4 - x)$ for $0 \le x \le 4$ , and $0$ otherwise.

(a) Write down $E(X)$ .

(b) Find $\text{Var}(X)$ .

Solution

(a) The graph of $f$ is symmetric about $x = 2$ , so $E(X) = 2$ .

(b)

E(X^2) = \frac{3}{32}\int_0^4 (4x^3 - x^4)\,dx = \frac{3}{32}\left[x^4 - \frac{x^5}{5}\right]_0^4 = \frac{3}{32}\left(256 - \frac{1024}{5}\right) = \frac{3}{32} \times \frac{256}{5} = 4.8

\text{Var}(X) = 4.8 - 2^2 = 0.8

Infinite intervals and exponentials

The same formulas work on infinite intervals, using improper integrals. Exponential densities need integration by parts from Pure 3, which the syllabus assumes.

An infinite interval

$X$ has probability density function $f(x) = \dfrac{3}{x^4}$ for $x \ge 1$ , and $0$ otherwise. Find the mean and the standard deviation of $X$ .

Solution

E(X) = \int_1^\infty x \cdot 3x^{-4}\,dx = \int_1^\infty 3x^{-3}\,dx = \left[-\frac{3}{2x^2}\right]_1^\infty = 0 + \frac{3}{2} = 1.5

E(X^2) = \int_1^\infty 3x^{-2}\,dx = \left[-\frac{3}{x}\right]_1^\infty = 3

\text{Var}(X) = 3 - 1.5^2 = 0.75, \qquad \text{standard deviation} = \sqrt{0.75} = 0.866

An exponential density

The waiting time $T$ minutes has probability density function $f(t) = \tfrac{1}{2}e^{-t/2}$ for $t \ge 0$ , and $0$ otherwise. Find $E(T)$ and $\text{Var}(T)$ .

Solution

By parts, with $u = t$ and $\dfrac{dv}{dt} = \tfrac{1}{2}e^{-t/2}$ , so $v = -e^{-t/2}$ :

E(T) = \int_0^\infty t \cdot \tfrac{1}{2}e^{-t/2}\,dt = \left[-te^{-t/2}\right]_0^\infty + \int_0^\infty e^{-t/2}\,dt = 0 + \left[-2e^{-t/2}\right]_0^\infty = 2

The term $te^{-t/2} \to 0$ as $t \to \infty$ because the exponential decays faster than $t$ grows.

For $E(T^2)$ , with $u = t^2$ , $v = -e^{-t/2}$ :

E(T^2) = \left[-t^2e^{-t/2}\right]_0^\infty + \int_0^\infty 2te^{-t/2}\,dt = 0 + 4\int_0^\infty t \cdot \tfrac{1}{2}e^{-t/2}\,dt = 4E(T) = 8

\text{Var}(T) = 8 - 2^2 = 4

Linear transformations

The expectation and variance rules from linear combinations of random variables apply to continuous variables too:

E(aX + b) = aE(X) + b, \qquad \text{Var}(aX + b) = a^2\,\text{Var}(X).

So once you have $E(X)$ and $\text{Var}(X)$ , you never need to integrate again for a scaled or shifted version.

More generally, the mean of any function of $X$ is found by weighting that function by the density: $E(g(X)) = \displaystyle\int g(x)f(x)\,dx$ . $E(X^2)$ is the special case $g(x) = x^2$ . You will occasionally need, for example, $E\left(\frac{1}{X}\right)$ , but this is rare.

Unknown constants from the mean

If a density has two unknown constants and you are given the mean, you get two equations: total area $= 1$ and $\int x f(x)\,dx =$ given mean.

An exam-style question

The random variable $X$ has probability density function $f(x) = a + bx$ for $0 \le x \le 1$ , and $0$ otherwise. It is given that $E(X) = \dfrac{7}{12}$ .

(a) Find the values of $a$ and $b$ .

(b) Find $\text{Var}(X)$ .

(d) Find $P(X > E(X))$ .

Solution

(a) Total area:

\int_0^1 (a + bx)\,dx = a + \frac{b}{2} = 1

Mean:

\int_0^1 (ax + bx^2)\,dx = \frac{a}{2} + \frac{b}{3} = \frac{7}{12}

From the first, $a = 1 - \dfrac{b}{2}$ . Substituting: $\dfrac{1}{2} - \dfrac{b}{4} + \dfrac{b}{3} = \dfrac{7}{12}$ , so $\dfrac{b}{12} = \dfrac{1}{12}$ , giving $b = 1$ and $a = \dfrac{1}{2}$ .

(b)

E(X^2) = \int_0^1 \left(\tfrac{1}{2}x^2 + x^3\right)dx = \frac{1}{6} + \frac{1}{4} = \frac{5}{12}

\text{Var}(X) = \frac{5}{12} - \left(\frac{7}{12}\right)^2 = \frac{60}{144} - \frac{49}{144} = \frac{11}{144} = 0.0764

(c)

E(3X + 2) = 3 \times \frac{7}{12} + 2 = 3.75, \qquad \text{Var}(3X + 2) = 9 \times \frac{11}{144} = \frac{11}{16} = 0.6875

(d)

P\left(X > \tfrac{7}{12}\right) = \int_{7/12}^1 \left(\tfrac{1}{2} + x\right)dx = \left[\frac{x}{2} + \frac{x^2}{2}\right]_{7/12}^1 = 1 - \left(\frac{7}{24} + \frac{49}{288}\right) = \frac{155}{288} = 0.538

The probability of exceeding the mean is not $0.5$ . The density rises to the right, so the mean and the median are not the same point. This contrast is explored in median and percentiles.

Common mistakes

Forgetting to square the mean. $\text{Var}(X) = E(X^2) - [E(X)]^2$ , not $E(X^2) - E(X)$ .
Integrating $f(x)$ instead of $xf(x)$ . $\int f(x)\,dx = 1$ always; that is not the mean.
Multiplying only one term by $x$ . For $f(x) = \frac{3}{32}(4x - x^2)$ , $xf(x) = \frac{3}{32}(4x^2 - x^3)$ . Expand first, then integrate.
Writing $E(X^2) = [E(X)]^2$ . These are different; their difference is the variance.
Losing the constant. Keep $\frac{3}{32}$ (or $k$ ) outside the integral and remember to multiply by it at the end.
A negative variance. This always means an arithmetic or setup error. Check it before moving on.

Exam tip

The formulas are given in MF19, so the marks are for the integral with the correct integrand and limits, the integration, and the arithmetic. Write $\displaystyle\int_0^3 x \cdot \frac{x^2}{9}\,dx$ before you integrate.
Exact fractions ( $\frac{11}{144}$ ) are ideal. If you give a decimal, give at least 3 significant figures.
"Show that $E(X) = \dots$ " questions need every step, including substituting limits.
If the question says "state the value of $E(X)$ ", the answer is by symmetry: give a reason ("by symmetry of the graph about $x = 2$ ").
Expect the next part to ask for a median or a probability involving the mean, such as $P(X > \mu)$ . Use the unrounded mean.

Summary

$E(X) = \int x f(x)\,dx$ and $E(X^2) = \int x^2 f(x)\,dx$ , over the interval where $f$ is defined.
$\text{Var}(X) = E(X^2) - [E(X)]^2$ .
Symmetry about $x = c$ gives $E(X) = c$ immediately.
Infinite intervals and exponential densities: improper integrals and integration by parts.
$E(aX + b) = aE(X) + b$ ; $\text{Var}(aX + b) = a^2\text{Var}(X)$ .
Two unknown constants: use total area $= 1$ together with a given mean or probability.

Practice questions

Question

$f(x) = \tfrac{1}{4}(x + 1)$ for $0 \le x \le 2$ , and $0$ otherwise. Find $E(X)$ and $\text{Var}(X)$ .
$f(x) = \dfrac{4}{x^5}$ for $x \ge 1$ , and $0$ otherwise. Find $E(X)$ and $\text{Var}(X)$ .
$f(x) = \tfrac{1}{5}$ for $2 \le x \le 7$ , and $0$ otherwise. Write down $E(X)$ and find $\text{Var}(X)$ .
$f(x) = \tfrac{3}{8}x^2$ for $0 \le x \le 2$ , and $0$ otherwise. Find $E(X)$ and $\text{Var}(X)$ , and hence find $E(4X - 1)$ and $\text{Var}(4X - 1)$ .
$f(x) = \tfrac{3}{4}(1 - x^2)$ for $-1 \le x \le 1$ , and $0$ otherwise. State $E(X)$ and find the standard deviation of $X$ .
$f(x) = 2e^{-2x}$ for $x \ge 0$ , and $0$ otherwise. Find $E(X)$ and $\text{Var}(X)$ .
$f(x) = ax + bx^2$ for $0 \le x \le 1$ , and $0$ otherwise, where $a$ and $b$ are constants. Given that $E(X) = 0.7$ , find $a$ and $b$ , and find $\text{Var}(X)$ .
$f(x) = \cos x$ for $0 \le x \le \frac{\pi}{2}$ , and $0$ otherwise. Show that $E(X) = \frac{\pi}{2} - 1$ and that $\text{Var}(X) = \pi - 3$ .

Answers

$E(X) = \tfrac{1}{4}\displaystyle\int_0^2 (x^2 + x)\,dx = \tfrac{1}{4}\left(\tfrac{8}{3} + 2\right) = \tfrac{7}{6}$ . $E(X^2) = \tfrac{1}{4}\displaystyle\int_0^2 (x^3 + x^2)\,dx = \tfrac{1}{4}\left(4 + \tfrac{8}{3}\right) = \tfrac{5}{3}$ . $\text{Var}(X) = \tfrac{5}{3} - \tfrac{49}{36} = \tfrac{11}{36} = 0.306$ .
$E(X) = \displaystyle\int_1^\infty 4x^{-4}\,dx = \left[-\tfrac{4}{3x^3}\right]_1^\infty = \tfrac{4}{3}$ . $E(X^2) = \displaystyle\int_1^\infty 4x^{-3}\,dx = \left[-\tfrac{2}{x^2}\right]_1^\infty = 2$ . $\text{Var}(X) = 2 - \tfrac{16}{9} = \tfrac{2}{9}$ .
By symmetry $E(X) = 4.5$ . $E(X^2) = \tfrac{1}{5}\left[\tfrac{x^3}{3}\right]_2^7 = \tfrac{1}{15}(343 - 8) = \tfrac{67}{3}$ . $\text{Var}(X) = \tfrac{67}{3} - 4.5^2 = \tfrac{25}{12} = 2.08$ .
$E(X) = \tfrac{3}{8}\left[\tfrac{x^4}{4}\right]_0^2 = 1.5$ ; $E(X^2) = \tfrac{3}{8}\left[\tfrac{x^5}{5}\right]_0^2 = \tfrac{3}{8} \times \tfrac{32}{5} = 2.4$ ; $\text{Var}(X) = 2.4 - 2.25 = 0.15$ . $E(4X - 1) = 5$ ; $\text{Var}(4X - 1) = 16 \times 0.15 = 2.4$ .
By symmetry about $x = 0$ , $E(X) = 0$ . $E(X^2) = \tfrac{3}{4}\displaystyle\int_{-1}^1 (x^2 - x^4)\,dx = \tfrac{3}{4}\left(\tfrac{2}{3} - \tfrac{2}{5}\right) = \tfrac{1}{5}$ , so $\text{Var}(X) = \tfrac{1}{5}$ and the standard deviation is $\sqrt{0.2} = 0.447$ .
By parts, $E(X) = \left[-xe^{-2x}\right]_0^\infty + \displaystyle\int_0^\infty e^{-2x}\,dx = 0 + \tfrac{1}{2} = \tfrac{1}{2}$ . $E(X^2) = \left[-x^2e^{-2x}\right]_0^\infty + \displaystyle\int_0^\infty 2xe^{-2x}\,dx = 0 + E(X) = \tfrac{1}{2}$ . $\text{Var}(X) = \tfrac{1}{2} - \tfrac{1}{4} = \tfrac{1}{4}$ .
Area: $\tfrac{a}{2} + \tfrac{b}{3} = 1$ . Mean: $\tfrac{a}{3} + \tfrac{b}{4} = 0.7$ . Multiply the first by $6$ : $3a + 2b = 6$ . Multiply the second by $12$ : $4a + 3b = 8.4$ . Solving: $a = 1.2$ , $b = 1.2$ . (Check: $f(x) = 1.2x + 1.2x^2 \ge 0$ on $[0, 1]$ .) $E(X^2) = \tfrac{a}{4} + \tfrac{b}{5} = 0.3 + 0.24 = 0.54$ ; $\text{Var}(X) = 0.54 - 0.49 = 0.05$ .
$E(X) = \displaystyle\int_0^{\pi/2} x\cos x\,dx = [x\sin x]_0^{\pi/2} - \displaystyle\int_0^{\pi/2}\sin x\,dx = \tfrac{\pi}{2} - [-\cos x]_0^{\pi/2} = \tfrac{\pi}{2} - 1$ . $E(X^2) = \displaystyle\int_0^{\pi/2} x^2\cos x\,dx = [x^2\sin x]_0^{\pi/2} - \displaystyle\int_0^{\pi/2} 2x\sin x\,dx$ . $\displaystyle\int_0^{\pi/2} 2x\sin x\,dx = [-2x\cos x]_0^{\pi/2} + \displaystyle\int_0^{\pi/2} 2\cos x\,dx = 0 + 2 = 2$ , so $E(X^2) = \tfrac{\pi^2}{4} - 2$ . $\text{Var}(X) = \tfrac{\pi^2}{4} - 2 - \left(\tfrac{\pi}{2} - 1\right)^2 = \tfrac{\pi^2}{4} - 2 - \tfrac{\pi^2}{4} + \pi - 1 = \pi - 3$ .

From sums to integrals

Symmetry: the mean for free

Infinite intervals and exponentials

Linear transformations

Unknown constants from the mean

Practice questions

Console