Expected value, variance and standard deviation – Random Variables (Senior 6 Mathematics) | Ganzaa

Random Variables

Lessons

Senior 6 · Mathematics

Random Variables

5 lessons

Lesson 2 of 5

Expected value, variance and standard deviation

35 min

The long-run average

An insurance company does not know which of its clients will claim this year, but it can predict quite closely how much it will pay out on average per client, and it sets its premiums above that figure. A national lottery works the same way in the other direction. The average value of a random variable over many repetitions is its expected value, and the typical distance of its values from that average is measured by the variance and the standard deviation. You met the mean and standard deviation of a data set in earlier years; here they are computed from probabilities instead of frequencies.

Definition

For a discrete random variable $X$ with values $x_{i}$ and pmf $p (x_{i})$ :

the expected value (or mean) is $E(X) = \mu = \sum_i x_i \, p(x_i)$ ;
the variance is $\operatorname{Var}(X) = \sigma^2 = E\big[(X - \mu)^2\big] = \sum_i (x_i - \mu)^2 \, p(x_i)$ ;
the standard deviation is $\sigma = \sqrt{\operatorname{Var}(X)}$ , in the same units as $X$ .

More generally, for a function $g$ , $E\big(g(X)\big) = \sum_i g(x_i) \, p(x_i)$ . Example: for a fair coin with $X = 1$ for heads and 0 for tails, $E(X) = 0 \times \frac{1}{2} + 1 \times \frac{1}{2} = \frac{1}{2}$ .

Important

Computing formula for the variance

\operatorname{Var}(X) = E(X^2) - \mu^2 .

Proof. $\sum (x_i - \mu)^2 p_i = \sum x_i^2 p_i - 2\mu \sum x_i p_i + \mu^2 \sum p_i = E(X^2) - 2\mu^2 + \mu^2$ , since $\sum x_i p_i = \mu$ and $\sum p_i = 1$ .

Linear change of variable. For constants $a$ and $b$ :

E(aX + b) = aE(X) + b , \qquad \operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X) , \qquad \sigma_{aX + b} = |a| \, \sigma_X .

Reason. $E(aX + b) = \sum (ax_i + b)p_i = a\sum x_ip_i + b\sum p_i$ . Adding $b$ shifts every value and the mean by the same amount, so the deviations $X - \mu$ do not change; multiplying by $a$ multiplies every deviation by $a$ and every squared deviation by $a^{2}$ .

The expected value is the balance point: weights

p (x)

placed at the positions

x

balance on a pivot at

\mu = E(X)

Worked example 1: three coins again

Find the mean, variance and standard deviation of the number of heads $X$ in three tosses of a fair coin (lesson 1).

Solution.

E(X) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} = \frac{12}{8} = 1.5 , \qquad E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = 3 .

So $\operatorname{Var}(X) = 3 - 1.5^2 = 0.75$ and $\sigma = \sqrt{0.75} \approx 0.866$ . The mean 1.5 is not a possible value of $X$ : it is a long-run average, not a prediction for one toss of three coins.

Worked example 2: a linear change

$P(X = x) = \frac{x}{10}$ for $x = 1, 2, 3, 4$ (lesson 1, worked example 2). Find $E (X)$ and $\operatorname{Var}(X)$ , then the mean and standard deviation of $Y = 3 X - 2$ .

Solution. $E(X) = \frac{1 + 4 + 9 + 16}{10} = 3$ and $E(X^2) = \frac{1 + 8 + 27 + 64}{10} = 10$ , so $\operatorname{Var}(X) = 10 - 9 = 1$ . Then $E (Y) = 3 (3) - 2 = 7$ , $\operatorname{Var}(Y) = 3^2 \times 1 = 9$ and $\sigma_Y = 3$ . The figure shows the pmf of $X$ balancing at 3.

Worked example 3: is the game fair?

At a school fair (an illustrative game), you pay 500 Frw to roll a fair die. A 6 wins 2000 Frw, a 4 or a 5 wins 500 Frw, and anything else wins nothing. Let $G$ be your net gain. Find $E (G)$ and the standard deviation of $G$ .

Solution.

Score	6	4 or 5	1, 2 or 3
$g$ (Frw)	$2000 - 500 = 1500$	$500 - 500 = 0$	$- 500$
$P (G = g)$	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{3}{6}$

$E(G) = \frac{1500 + 0 - 1500}{6} = 0$ : the game is fair (neither side gains in the long run). $E(G^2) = \frac{1500^2 + 3 \times 500^2}{6} = \frac{3\,000\,000}{6} = 500\,000$ , so $\operatorname{Var}(G) = 500\,000$ and $\sigma \approx 707$ Frw: a fair game can still swing widely from one play to the next.

Warning

$E (X^{2})$ is not $\big(E(X)\big)^2$ . In worked example 1, $E (X^{2}) = 3$ but $\big(E(X)\big)^2 = 2.25$ ; their difference is the variance, which can never be negative.

Constants do not add to a variance. $\operatorname{Var}(3X - 2) = 9\operatorname{Var}(X)$ , not $9\operatorname{Var}(X) - 2$ and not $3\operatorname{Var}(X)$ .

Tip

Set the work out as a table with rows $x$ , $p$ , $x p$ and $x^{2} p$ . The sums of the last two rows are $E (X)$ and $E (X^{2})$ , and one subtraction gives the variance.

jbstatistics: Expected Value and Variance of Discrete Random Variables (about 8 min). Watch the table method for E(X) and E(X²), and the shortcut Var(X) = E(X²) − μ², used exactly as in this lesson.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Find the mean, variance and standard deviation of the score on one roll of a fair die.
$X$ takes the values 0, 1, 2, 3 with probabilities 0.1, 0.3, 0.4, 0.2. Find $E (X)$ , $\operatorname{Var}(X)$ and $\sigma$ .
$E (X) = 4$ and $\operatorname{Var}(X) = 2.25$ . Find $E (2 X + 5)$ , $\operatorname{Var}(2X + 5)$ and the standard deviation of $5 - X$ .
In an illustrative raffle, 500 tickets are sold at 200 Frw each. There is one prize of 20 000 Frw and two prizes of 5000 Frw. Find the expected net gain of a person who buys one ticket.
$X$ takes the values 1, 2, 3 with probabilities $a$ , $b$ and 0.3, and $E (X) = 2.1$ . Find $a$ , $b$ and $\operatorname{Var}(X)$ .
Using the pmf of $R$ found in lesson 1, practice question 6 (red balls drawn), find $E (R)$ and $\operatorname{Var}(R)$ .
Prove that $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$ .

Show answersHide answers

$\mu = 3.5$ ; $E(X^2) = \frac{91}{6}$ , $\operatorname{Var}(X) = \frac{91}{6} - \frac{49}{4} = \frac{35}{12} \approx 2.92$ ; $\sigma \approx 1.71$ .
$E (X) = 1.7$ , $E (X^{2}) = 3.7$ , $\operatorname{Var}(X) = 0.81$ , $\sigma = 0.9$ .
13, 9 and 1.5.
Expected prize $\frac{20\,000 + 2 \times 5000}{500} = 60$ Frw, so the expected net gain is $60 - 200 = - 140$ Frw.
$a + b = 0.7$ and $a + 2 b + 0.9 = 2.1$ : $b = 0.5$ , $a = 0.2$ ; $E (X^{2}) = 4.9$ , $\operatorname{Var}(X) = 0.49$ .
$E (R) = 0.6 + 0.6 = 1.2$ ; $E (R^{2}) = 0.6 + 1.2 = 1.8$ ; $\operatorname{Var}(R) = 0.36$ .
$E(aX + b) = a\mu + b$ , so $aX + b - E(aX + b) = a(X - \mu)$ and $\operatorname{Var}(aX + b) = E\big[a^2(X - \mu)^2\big] = a^2 \operatorname{Var}(X)$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagram, the worked examples, the practice items and the quiz questions. The fair game and the raffle are illustrative.
Video: Expected Value and Variance of Discrete Random Variables, jbstatistics (Jeremy Balka), embedded from YouTube.

1The long-run average

The long-run average