Random Variables and Expected Value

A random variable assigns a numerical value to each outcome of a random process. It is the bridge between probability (which deals with events) and statistics (which deals with numbers). Instead of saying “the event is heads,” a random variable lets you say “the value is 1.” This shift from events to numbers opens the door to averages, spread, and all the quantitative tools of statistics.

What Is a Random Variable?

A random variable is a variable whose value is determined by the outcome of a random process. Random variables are typically denoted by capital letters such as $X$, $Y$, or $Z$.

Example: Flip a fair coin 3 times and let $X$ = the number of heads. The possible values of $X$ are 0, 1, 2, and 3. Before you flip, you do not know what $X$ will be — but you can describe the probability of each possible value.

There are two types of random variables:

• Discrete random variables take on a countable number of values (0, 1, 2, 3, …). Examples include the number of heads in 3 flips, the number of defective items in a batch, or the number of customers per hour.
• Continuous random variables can take any value within a range (like height, temperature, or time). These require different tools (probability density functions) and are covered in a later topic.

This page focuses entirely on discrete random variables.

Probability Distribution Tables

A probability distribution lists every possible value of a random variable alongside its probability. It provides a complete picture of the random variable’s behavior.

Every valid probability distribution must satisfy two rules:

1. Every probability is non-negative: $P(X = x) \geq 0$ for all values $x$
2. All probabilities sum to exactly 1: $\sum P(X = x) = 1$

Example 1: Number of Heads in 3 Coin Flips

When you flip a fair coin 3 times, the sample space has $2^3 = 8$ equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Counting the number of heads in each outcome gives us this distribution:

Verification: $\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{8}{8} = 1$ $\checkmark$

The distribution tells us that getting exactly 1 or exactly 2 heads is much more likely (probability $\frac{3}{8}$ each) than getting 0 or 3 heads (probability $\frac{1}{8}$ each).

Expected Value (Mean of a Random Variable)

The expected value (or mean) of a discrete random variable is the long-run average value you would observe if you repeated the random process many times. It is calculated by multiplying each value by its probability and summing:

$E(X) = \mu = \sum x_i \cdot P(x_i)$

The expected value does not have to be a value the random variable can actually take. It represents the theoretical average.

Example 2: Expected Number of Heads

Using the distribution from Example 1:

$E(X) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8}$

$= 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = 1.5$

Interpretation: On average, you would expect 1.5 heads per 3 flips. Of course you cannot flip exactly 1.5 heads in a single trial, but over 1,000 sets of 3 flips, the average number of heads would be very close to 1.5.

Example 3: Insurance Payout

An insurance policy pays out as follows:

What is the expected payout per policy?

$E(\text{payout}) = 0(0.90) + 2000(0.08) + 10000(0.02)$

$= 0 + 160 + 200 = 360$

The expected payout is $360 per policy. If the insurance company charges a premium of $500, it expects to profit $140 per policy on average. This is exactly how insurance companies set their prices — by calculating expected value and adding a margin.

Verification: $2000 \times 0.08 = 160$. $10000 \times 0.02 = 200$. $0 + 160 + 200 = 360$ $\checkmark$

Variance and Standard Deviation of a Random Variable

Expected value tells you the center of a distribution, but it says nothing about how spread out the values are. Variance measures the average squared distance from the mean:

$\text{Var}(X) = \sigma^2 = \sum (x_i - \mu)^2 \cdot P(x_i)$

There is an equivalent shortcut formula that is often easier to compute:

$\text{Var}(X) = E(X^2) - [E(X)]^2$

The standard deviation is the square root of the variance, bringing the measure of spread back to the original units:

$\text{SD}(X) = \sigma = \sqrt{\text{Var}(X)}$

Example 4: Variance of Coin Flips

Using the distribution from Example 1 with $\mu = 1.5$:

$\text{Var}(X) = (0 - 1.5)^2 \cdot \frac{1}{8} + (1 - 1.5)^2 \cdot \frac{3}{8} + (2 - 1.5)^2 \cdot \frac{3}{8} + (3 - 1.5)^2 \cdot \frac{1}{8}$

$= 2.25 \cdot \frac{1}{8} + 0.25 \cdot \frac{3}{8} + 0.25 \cdot \frac{3}{8} + 2.25 \cdot \frac{1}{8}$

$= 0.28125 + 0.09375 + 0.09375 + 0.28125 = 0.75$

$\text{SD}(X) = \sqrt{0.75} \approx 0.866$

Verification: $\frac{2.25}{8} = 0.28125$. $\frac{0.25 \times 3}{8} = 0.09375$. Sum: $0.28125 + 0.09375 + 0.09375 + 0.28125 = 0.75$ $\checkmark$

Interpretation: The number of heads in 3 coin flips has a mean of 1.5 and a standard deviation of about 0.87. Most outcomes will fall within one standard deviation of the mean (roughly between 0.63 and 2.37), which aligns with what we see — 1 and 2 heads are the most common results.

Linear Transformations

If you transform a random variable using a linear function $Y = aX + b$, the expected value and variance follow predictable rules:

$E(Y) = E(aX + b) = a \cdot E(X) + b$

$\text{Var}(Y) = \text{Var}(aX + b) = a^2 \cdot \text{Var}(X)$

Adding a constant ($b$) shifts the mean but does not change the spread. Multiplying by a constant ($a$) scales both the mean and the standard deviation, and scales the variance by $a^2$.

Example: A store marks up each item by doubling the wholesale price and adding $5. If the wholesale price $X$ has $E(X) = 20$ and $\text{Var}(X) = 9$, then the retail price $Y = 2X + 5$ has:

$E(Y) = 2(20) + 5 = 45$

$\text{Var}(Y) = 2^2(9) = 36 \quad \Rightarrow \quad \text{SD}(Y) = 6$

Real-World Application: Retail — Expected Daily Revenue

A small bookstore tracks the number of books sold per day and finds the following distribution:

Each book sells for $25. What is the expected daily revenue?

Step 1: Find $E(X)$, the expected number of books sold.

$E(X) = 0(0.05) + 1(0.15) + 2(0.30) + 3(0.25) + 4(0.15) + 5(0.10)$

$= 0 + 0.15 + 0.60 + 0.75 + 0.60 + 0.50 = 2.60$

Verification: $0.05 + 0.15 + 0.30 + 0.25 + 0.15 + 0.10 = 1.00$ $\checkmark$

Step 2: Since revenue $R = 25X$, use the linear transformation rule:

$E(R) = 25 \cdot E(X) = 25 \times 2.60 = 65$

Answer: The bookstore expects to earn $65 in daily book revenue on average.

Step 3 (bonus): Find the standard deviation of daily revenue.

$E(X^2) = 0^2(0.05) + 1^2(0.15) + 4(0.30) + 9(0.25) + 16(0.15) + 25(0.10)$

$= 0 + 0.15 + 1.20 + 2.25 + 2.40 + 2.50 = 8.50$

$\text{Var}(X) = 8.50 - (2.60)^2 = 8.50 - 6.76 = 1.74$

$\text{Var}(R) = 25^2 \times 1.74 = 625 \times 1.74 = 1087.50$

$\text{SD}(R) = \sqrt{1087.50} \approx 32.98$

The expected daily revenue is $65, with a standard deviation of about $33 — there is meaningful day-to-day variation.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• A random variable assigns a number to each outcome of a random process. Discrete random variables take countable values; continuous ones take any value in a range.
• A probability distribution lists every value with its probability. All probabilities must be non-negative and sum to 1.
• Expected value $E(X) = \sum x_i \cdot P(x_i)$ gives the long-run average — the theoretical center of the distribution.
• Variance $\text{Var}(X) = \sum (x_i - \mu)^2 \cdot P(x_i)$ measures spread. Standard deviation is $\sigma = \sqrt{\text{Var}(X)}$.
• Linear transformations: if $Y = aX + b$, then $E(Y) = aE(X) + b$ and $\text{Var}(Y) = a^2 \text{Var}(X)$.
• Expected value is used everywhere — insurance pricing, inventory planning, game theory, and business revenue forecasting.

Return to Statistics for more topics in this section.