Random variables and expectation

A random variable $X$ is a function from the sample space to the real numbers , it assigns a numerical value to each outcome. Most of the time, instead of thinking about complicated outcome events, we work with the distribution of $X$ : the list of possible values it can take, and their probabilities.

Definition

Formally, $X : \Omega \to \mathbb{R}$ is a real-valued function on the sample space $\Omega$ .

For discrete random variables (the Class XII focus), $X$ takes finitely or countably many values $x_1, x_2, \ldots$ . The probability mass function (PMF) gives the probability of each value: $p_i = P(X = x_i),$ with $\sum_i p_i = 1$ .

Examples of random variables

$X =$ number on a die: values $\{1, 2, 3, 4, 5, 6\}$ , each with probability $1/6$ .
$X =$ number of heads in 3 coin flips: values $\{0, 1, 2, 3\}$ , probabilities $1/8, 3/8, 3/8, 1/8$ .
$X =$ number of red balls drawn in 2 draws (without replacement) from a bag with 4 red, 6 blue.

Probability distribution table

A common way to present a discrete PMF:

$X$	$x_1$	$x_2$	$x_3$	$\cdots$
$P(X)$	$p_1$	$p_2$	$p_3$	$\cdots$

Always check: probabilities are non-negative and sum to 1.

Expectation

The expected value (or mean) of a discrete random variable $X$ is $E[X] = \mu = \sum_i x_i\, p_i.$

It is the weighted average of the values, weighted by their probabilities. Interpretation: if you repeat the experiment many times, $E[X]$ is the long-run average of $X$ .

Properties of expectation:

Linearity: $E[aX + b] = aE[X] + b$ .
For two random variables: $E[X + Y] = E[X] + E[Y]$ (always, even if not independent).
For independent $X, Y$ : $E[XY] = E[X]\,E[Y]$ .

Variance and standard deviation

The variance measures spread: $\mathrm{Var}(X) = \sigma^2 = E[(X - \mu)^2] = \sum_i (x_i - \mu)^2 p_i.$

A useful computational form: $\mathrm{Var}(X) = E[X^2] - (E[X])^2.$

The standard deviation is $\sigma = \sqrt{\mathrm{Var}(X)}$ , in the same units as $X$ .

Properties:

$\mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X)$ . (Shifts don't affect spread; scaling squares.)
For independent $X, Y$ : $\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$ .

Worked examples

Example 1. $X$ = score on a fair die. Find $E[X]$ and $\mathrm{Var}(X)$ .

$E[X] = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 7/2 = 3.5$ .

$E[X^2] = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6$ .

$\mathrm{Var}(X) = 91/6 - (7/2)^2 = 91/6 - 49/4 = 182/12 - 147/12 = 35/12 \approx 2.917$ .

Example 2. Two coins are flipped. $X$ = number of heads. Find the distribution, $E[X]$ , $\mathrm{Var}(X)$ .

$X$	0	1	2
$P$	1/4	1/2	1/4

$E[X] = 0(1/4) + 1(1/2) + 2(1/4) = 1$ .

$E[X^2] = 0 + 1(1/2) + 4(1/4) = 3/2$ .

$\mathrm{Var}(X) = 3/2 - 1 = 1/2$ .

Example 3. A coin is flipped repeatedly until heads appears (or until 4 flips). Let $X$ = number of flips. Find the distribution.

$X = 1$ : heads on first flip, $P = 1/2$ . $X = 2$ : TH, $P = 1/4$ . $X = 3$ : TTH, $P = 1/8$ . $X = 4$ : TTTH or TTTT, $P = 1/16 + 1/16 = 1/8$ (either heads-on-4th or four-tails-and-stop).

Wait, the latter case "TTTT and stop" is also " $X = 4$ " by our definition. So $P(X = 4) = 1/16 + 1/16 = 1/8$ . Check sum: $1/2 + 1/4 + 1/8 + 1/8 = 1$ ✓.

$E[X] = 1(1/2) + 2(1/4) + 3(1/8) + 4(1/8) = 1/2 + 1/2 + 3/8 + 1/2 = 15/8$ .

Example 4. A box has 4 white and 2 black balls. Two balls are drawn without replacement. Let $X$ = number of white balls. Find distribution and mean.

$X = 0$ : both black. $P = \binom{2}{2}/\binom{6}{2} = 1/15$ . $X = 1$ : one of each. $P = \binom{4}{1}\binom{2}{1}/\binom{6}{2} = 8/15$ . $X = 2$ : both white. $P = \binom{4}{2}/\binom{6}{2} = 6/15$ .

Check: $1/15 + 8/15 + 6/15 = 15/15$ ✓.

$E[X] = 0(1/15) + 1(8/15) + 2(6/15) = 0 + 8/15 + 12/15 = 20/15 = 4/3$ .

Example 5. A lottery ticket costs ₹ $10$ . Probability of winning ₹ $1000$ is $0.005$ ; otherwise nothing. Find the expected net gain.

$X$ = net gain. $X = -10$ with probability $0.995$ (lose ₹10), $X = 990$ with probability $0.005$ (win ₹1000 minus ₹10 spent).

$E[X] = -10(0.995) + 990(0.005) = -9.95 + 4.95 = -5$ . Expected loss of ₹5.

Example 6. A random variable $X$ takes values $1, 2, 3$ with probabilities $1/6, 1/3, 1/2$ . Find $E[X], E[X^2], \mathrm{Var}(X)$ .

$E[X] = 1(1/6) + 2(1/3) + 3(1/2) = 1/6 + 4/6 + 9/6 = 14/6 = 7/3$ .

$E[X^2] = 1(1/6) + 4(1/3) + 9(1/2) = 1/6 + 8/6 + 27/6 = 36/6 = 6$ .

$\mathrm{Var}(X) = 6 - 49/9 = 54/9 - 49/9 = 5/9$ .

Try it yourself

$X$ = number of heads in 4 fair coin flips. Find the distribution, $E[X], \mathrm{Var}(X)$ .
A die is rolled; $X$ = score if even, $0$ if odd. Find $E[X]$ .
$X$ = sum of two dice. Find $E[X]$ and $\mathrm{Var}(X)$ .
A box has 5 R, 3 B balls. Two drawn without replacement; $X$ = red balls. Find distribution and $E[X]$ .
Three coins. $X$ = number of tails. Find $E[X], \mathrm{Var}(X)$ .
$X$ takes values $-2, -1, 0, 1, 2$ each with probability $1/5$ . Find $E[X], \mathrm{Var}(X)$ .
A random variable has PMF $P(X = k) = k/10$ for $k = 1, 2, 3, 4$ . Verify and find $E[X]$ .
A gambler bets on red at roulette ( $P = 18/38$ ). $X$ = net gain per ₹ $100$ bet. Find $E[X]$ .
$X$ has $P(X = -1) = 1/2, P(X = 0) = 1/3, P(X = 1) = 1/6$ . Find $E[X], E[X^2]$ .
Two cards from a deck without replacement. $X$ = number of aces. Find distribution, $E[X]$ .
$X$ uniform on $\{1, 2, 3, 4, 5\}$ . Find $E[X], \mathrm{Var}(X)$ .
$Y = 2X + 5$ where $X$ as in previous. Find $E[Y], \mathrm{Var}(Y)$ .
Verify $\mathrm{Var}(X) = E[X^2] - (E[X])^2$ for $X$ in any previous problem.
A random variable is positive with probability $p$ and zero otherwise; positive values are uniform on $\{1, 2, 3\}$ . Find $E[X]$ in terms of $p$ .

Pitfalls and tricks

Always verify $\sum p_i = 1$ . A common error is to compute probabilities that don't sum to 1.
Use linearity of expectation to avoid manipulating distributions. $E[X + Y] = E[X] + E[Y]$ regardless of independence.
The shortcut $\mathrm{Var}(X) = E[X^2] - (E[X])^2$ is easier than computing $E[(X - \mu)^2]$ directly.
Variance is non-negative. If your computed variance is negative, you've made an error.
Standard deviation is in the same units as $X$ ; variance is in squared units.