Variance of a Discrete Random Variable

The expected value $E(X)$ tells you the long-run average of a random variable, but not how much the results vary. Two games can both have an expected payout of $2, one paying exactly $2 every time and the other paying $0 or $20. The variance and standard deviation measure that spread. You’ll also see how they change when every value is scaled or shifted, for example when a prize is doubled or an entry fee is subtracted.

Key ideas

Variance and standard deviation

For a discrete random variable $X$ with mean $\mu = E(X)$ , the variance is the expected squared distance from the mean:

\text{Var}(X) = E\big((X - \mu)^2\big) = \sum (x - \mu)^2 \, P(X = x)

The standard deviation is its square root, $\sigma = \sqrt{\text{Var}(X)}$ . It’s in the same units as $X$ , which makes it easier to interpret.

The calculation formula

Expanding the square gives a formula that’s usually quicker:

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

In words: “the mean of the squares minus the square of the mean.” To use it, add two rows to the probability table, $xP(X = x)$ and $x^2P(X = x)$ , and add each row.

Since a variance can’t be negative, $E(X^2)$ is always at least $[E(X)]^2$ . If you get a negative variance, you’ve made an error.

Using your GDC

Put the values of $x$ in one list and the probabilities in another, then run one-variable statistics with the probabilities as the frequency list. The mean $\bar{x}$ is $E(X)$ , and the population standard deviation $\sigma_x$ is the standard deviation of $X$ (don’t use $s_x$ ). Square $\sigma_x$ for the variance.

Linear transformations

If every value of $X$ is multiplied by $a$ and then $b$ is added, the new random variable is $aX + b$ . Then

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

so the standard deviation of $aX + b$ is $|a|\sigma$ .

Why? Adding $b$ slides every value along by the same amount: the centre moves, but the spread doesn’t change. Multiplying by $a$ stretches every distance from the mean by a factor of $|a|$ , so squared distances are multiplied by $a^2$ .

Change to $X$	Effect on $E$	Effect on Var	Effect on SD
add $b$	add $b$	none	none
multiply by $a$	multiply by $a$	multiply by $a^2$	multiply by $\lvert a \rvert$

More general expectations

For any function $g$ , $E\big(g(X)\big) = \sum g(x)P(X = x)$ . That’s how $E(X^2)$ is defined. Be careful: in general $E(X^2) \ne [E(X)]^2$ , and their difference is exactly the variance.

Worked examples

Example 1: Rolling a die

Let $X$ be the score on a fair six-sided die. Find $E(X)$ , $\text{Var}(X)$ and the standard deviation.

Solution. Each value has probability $\tfrac{1}{6}$ .

E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5

E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6}

\text{Var}(X) = \frac{91}{6} - 3.5^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}

The variance is $\dfrac{35}{12} \approx 2.92$ and the standard deviation is $\sqrt{35/12} \approx 1.71$ (3 s.f.).

Example 2: A table with unknowns

The random variable $X$ has this distribution, and $E(X) = 1.7$ .

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$0.1$	$p$	$q$	$0.2$

Find $p$ and $q$ , then find $\text{Var}(X)$ and the standard deviation of $X$ .

Solution. The probabilities add to $1$ : $0.1 + p + q + 0.2 = 1$ , so $p + q = 0.7$ .

The mean is $1.7$ : $0(0.1) + 1p + 2q + 3(0.2) = 1.7$ , so $p + 2q = 1.1$ .

Subtracting the equations: $q = 0.4$ , and then $p = 0.3$ .

E(X^2) = 0^2(0.1) + 1^2(0.3) + 2^2(0.4) + 3^2(0.2) = 0 + 0.3 + 1.6 + 1.8 = 3.7

\text{Var}(X) = 3.7 - 1.7^2 = 3.7 - 2.89 = 0.81

The standard deviation is $\sqrt{0.81} = 0.9$ .

Example 3: A game with a fee

In a fairground game, $X$ is the number of targets a player hits, with the distribution from Example 2. The player pays $8 to play and wins $5 per target. Let $Y$ be the player’s profit in dollars.

(a) Write $Y$ in terms of $X$ .
(b) Find $E(Y)$ and the standard deviation of $Y$ .
(c) Is the game fair?

Solution.

(a) $Y = 5X - 8$ .

(b) Using $E(X) = 1.7$ and $\text{Var}(X) = 0.81$ :

E(Y) = 5(1.7) - 8 = 0.5 \qquad \text{Var}(Y) = 5^2(0.81) = 20.25

The standard deviation of $Y$ is $\sqrt{20.25} = 4.5$ . (Or directly: $5 \times 0.9 = 4.5$ .)

The expected profit is 50 cents per game, with a standard deviation of $4.50.

Example 4: Exact values from a formula

The probability distribution of $X$ is $P(X = x) = \dfrac{k}{x}$ for $x = 1, 2, 3$ . Find the exact values of $k$ , $E(X)$ and $\text{Var}(X)$ .

Solution. The probabilities add to $1$ :

k\left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{11k}{6} = 1 \quad\Rightarrow\quad k = \frac{6}{11}

Notice that $x \cdot \dfrac{k}{x} = k$ , so each term of $E(X)$ is just $k$ :

E(X) = k + k + k = 3k = \frac{18}{11}

E(X^2) = \sum x^2 \cdot \frac{k}{x} = k(1 + 2 + 3) = 6k = \frac{36}{11}

\text{Var}(X) = \frac{36}{11} - \left(\frac{18}{11}\right)^2 = \frac{396}{121} - \frac{324}{121} = \frac{72}{121}

(The standard deviation is $\dfrac{6\sqrt{2}}{11} \approx 0.771$ .)

Common mistakes

Squaring the mean instead of finding the mean of the squares. $E(X^2)$ is $\sum x^2P(X = x)$ , not $[E(X)]^2$ . In Example 2, $E(X^2) = 3.7$ but $[E(X)]^2 = 2.89$ .

Forgetting to subtract. $E(X^2)$ on its own is not the variance. Finish with $- [E(X)]^2$ .

Writing Var(aX + b) = aVar(X) + b. Adding a constant doesn’t change the spread, and multiplying by $a$ multiplies the variance by $a^2$ . For $Y = 5X - 8$ , $\text{Var}(Y) = 25\,\text{Var}(X)$ , not $5\,\text{Var}(X) - 8$ .

A negative variance or standard deviation. $\text{Var}(-2X) = 4\,\text{Var}(X)$ , which is positive, and the standard deviation is $2\sigma$ , not $-2\sigma$ . Square $a$ for the variance; use $|a|$ for the standard deviation.

Mixing up variance and standard deviation. Read which one the question asks for. If you’re given $\sigma = 3$ , then $\text{Var}(X) = 9$ .

Using the sample standard deviation on a GDC. For a probability distribution, use $\sigma_x$ , not $s_x$ .

Practice

1. (Warm-up) Find $E(X)$ , $\text{Var}(X)$ and the standard deviation of $X$ .

$x$	$1$	$2$	$3$
$P(X = x)$	$0.2$	$0.5$	$0.3$

Solution

E(X) = 0.2 + 1.0 + 0.9 = 2.1 \qquad E(X^2) = 0.2 + 2.0 + 2.7 = 4.9

\text{Var}(X) = 4.9 - 2.1^2 = 4.9 - 4.41 = 0.49

The standard deviation is $\sqrt{0.49} = 0.7$ .

2. (Warm-up) $E(X) = 4$ and $\text{Var}(X) = 9$ . Find:

(a) $E(3X - 2)$ and $\text{Var}(3X - 2)$ ;
(b) $E(5 - X)$ and $\text{Var}(5 - X)$ .

Solution

(a) $E(3X - 2) = 3(4) - 2 = 10$ . $\text{Var}(3X - 2) = 3^2(9) = 81$ .

(b) $5 - X = -X + 5$ , so $a = -1$ , $b = 5$ . $E(5 - X) = -4 + 5 = 1$ . $\text{Var}(5 - X) = (-1)^2(9) = 9$ .

3. (Core) $P(X = x) = k(x + 1)$ for $x = 0, 1, 2, 3$ . Find $k$ , $E(X)$ and $\text{Var}(X)$ .

Solution

$k(1 + 2 + 3 + 4) = 10k = 1$ , so $k = 0.1$ . The probabilities are $0.1, 0.2, 0.3, 0.4$ .

E(X) = 0 + 0.2 + 0.6 + 1.2 = 2 \qquad E(X^2) = 0 + 0.2 + 1.2 + 3.6 = 5

\text{Var}(X) = 5 - 2^2 = 1

4. (Core) A fair coin is tossed three times, and $X$ is the number of heads. Write down the probability distribution of $X$ , and find $E(X)$ and $\text{Var}(X)$ .

Solution

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\tfrac{1}{8}$	$\tfrac{3}{8}$	$\tfrac{3}{8}$	$\tfrac{1}{8}$

E(X) = \frac{0 + 3 + 6 + 3}{8} = \frac{12}{8} = 1.5 \qquad E(X^2) = \frac{0 + 3 + 12 + 9}{8} = \frac{24}{8} = 3

\text{Var}(X) = 3 - 1.5^2 = 0.75

Check: $X \sim B(3, 0.5)$ , and the binomial formulas give $np = 1.5$ and $np(1 - p) = 0.75$ . ✓

5. (Core) A game costs $8 to play. You roll a fair die and win twice the score in dollars. Let $W$ be your profit in dollars.

(a) Write $W$ in terms of the score $X$ , and find $E(W)$ and the standard deviation of $W$ .
(b) What should the game cost for it to be fair?

Solution

(a) $W = 2X - 8$ . From Example 1, $E(X) = 3.5$ and $\text{Var}(X) = \dfrac{35}{12}$ .

E(W) = 2(3.5) - 8 = -1 \qquad \text{SD}(W) = 2\sqrt{\frac{35}{12}} = 3.42 \text{ (3 s.f.)}

On average you lose $1 per game.

(b) With a cost of $c$ dollars, $E(2X - c) = 7 - c$ . This is $0$ when $c = 7$ , so the game is fair if it costs $7.

6. (Core) The random variable $X$ has $E(X) = 0.5$ and this distribution.

$x$	$-1$	$0$	$2$	$5$
$P(X = x)$	$0.3$	$p$	$q$	$0.1$

Find $p$ and $q$ , and the standard deviation of $X$ .

Solution

Mean: $-0.3 + 0 + 2q + 0.5 = 0.5$ , so $2q = 0.3$ and $q = 0.15$ . Total: $0.3 + p + 0.15 + 0.1 = 1$ , so $p = 0.45$ .

E(X^2) = 1(0.3) + 0 + 4(0.15) + 25(0.1) = 0.3 + 0.6 + 2.5 = 3.4

\text{Var}(X) = 3.4 - 0.5^2 = 3.15 \qquad \sigma = \sqrt{3.15} = 1.77 \text{ (3 s.f.)}

7. (Core) The daily maximum temperature $C$ ( $^\circ\text{C}$ ) in a city in July has mean $22$ and standard deviation $4$ . The temperature in degrees Fahrenheit is $F = 1.8C + 32$ . Find the mean, standard deviation and variance of $F$ .

Solution

E(F) = 1.8(22) + 32 = 71.6 \qquad \text{SD}(F) = 1.8(4) = 7.2 \qquad \text{Var}(F) = 7.2^2 = 51.84

(Adding $32$ changes the mean but not the spread.)

8. (Challenge) $E(X) = 2$ and $\text{Var}(X) = 3$ .

(a) Find $E(X^2)$ and $E\big((2X + 1)^2\big)$ .
(b) Find constants $a \gt 0$ and $b$ so that $Y = aX + b$ has $E(Y) = 0$ and $\text{Var}(Y) = 1$ .

Solution

(a) $\text{Var}(X) = E(X^2) - [E(X)]^2$ , so $E(X^2) = 3 + 4 = 7$ .

E\big((2X + 1)^2\big) = E(4X^2 + 4X + 1) = 4(7) + 4(2) + 1 = 37

(b) $\text{Var}(Y) = a^2(3) = 1$ , so $a = \dfrac{1}{\sqrt{3}}$ (taking $a \gt 0$ ). $E(Y) = \dfrac{2}{\sqrt{3}} + b = 0$ , so $b = -\dfrac{2}{\sqrt{3}}$ .

So $Y = \dfrac{X - 2}{\sqrt{3}}$ : subtract the mean and divide by the standard deviation. That’s standardizing, just like a z-score.

9. (Challenge) $X$ takes only the values $0$ and $a$ , where $a \gt 0$ , with $P(X = a) = p$ . Given that $E(X) = 2$ and $\text{Var}(X) = 6$ , find $a$ and $p$ .

Solution

$E(X) = ap = 2$ and $E(X^2) = a^2p$ , so

\text{Var}(X) = a^2p - (ap)^2 = a^2p(1 - p) = 6

Write $a^2p(1 - p) = a(ap)(1 - p) = 2a(1 - p) = 6$ , so $a(1 - p) = 3$ , which means $a - ap = 3$ . Since $ap = 2$ , $a = 5$ , and then $p = \dfrac{2}{5} = 0.4$ .

Check: $E(X) = 5(0.4) = 2$ and $\text{Var}(X) = 25(0.4)(0.6) = 6$ . ✓