Mean of a Random Variable vs Expected Value

The mean of a random variable $X$ is $\mu_X = \sum x_i P(x_i)$ (mu-X, the sum of each value times its probability), taken across every value of a distribution whose probabilities total 1. Each value enters in proportion to how likely it is, which is what separates it from an ordinary average of the values. It is a parameter: one fixed number belonging to the distribution, settled before any data exist.

Let $X$ be the number of defective items in a box, with $P(0) = 0.5$, $P(1) = 0.3$ and $P(2) = 0.2$. Then $\mu_X = 0(0.5) + 1(0.3) + 2(0.2) = 0.7$ defects. The physical name for that number is the balance point. Put weights of 0.5, 0.3 and 0.2 at positions 0, 1 and 2 along a beam and it balances at 0.7, because the pull on the left is $0.5(0.7) = 0.35$ and the pull on the right is $0.3(0.3) + 0.2(1.3) = 0.09 + 0.26 = 0.35$.

"The mean is 0.7, so about 70 percent of boxes have a defect" is the misreading to kill. A mean is not a proportion, and it is not capped at 1; it only looks like a percentage here because the counts happen to be small. The share of boxes with at least one defect is $0.3 + 0.2 = 0.50$. Change the values to 0, 1 and 10 with those same probabilities and $\mu_X$ jumps to $0(0.5) + 1(0.3) + 10(0.2) = 2.3$ while that share stays at 0.50.

The distinction that matters downstream is $\mu_X$ against $\bar{x}$ (x-bar). Inspect 20 real boxes and their average will land somewhere near 0.7 without matching it; inspect 20,000 and it will sit closer. The mean of the distribution is the fixed number a sample mean converges toward, never a forecast for the next box.

This is topic 2.9 of Unit 2, Parameters of Random Variables.

The expected value of a random variable $X$ is written $E(X)$ (E of X) or $\mu_X$ (mu sub X), and it is the probability-weighted average of the possible values:

Each value is multiplied by its own probability rather than counted once, and that weighting is the whole difference between an expected value and an ordinary average.

Let $X$ be the number of cars a household owns, with $P(0) = 0.2$, $P(1) = 0.4$, $P(2) = 0.3$ and $P(3) = 0.1$. Then $E(X) = 0(0.2) + 1(0.4) + 2(0.3) + 3(0.1) = 1.3$ cars.

That 1.3 carries two lessons. First, an expected value need not be a value the variable can actually take: no household owns 1.3 cars. It describes the average over many households, not a prediction about the next one. Second, it is not the most likely outcome either. Owning 1 car is the most probable single result here, and most likely and average are separate questions. The mode answers the first, the expected value answers the second.

The other frequent slip is dropping the weights entirely. The plain average of 0, 1, 2 and 3 is 1.5, and that would be correct only if all four outcomes were equally likely. They are not. Weight only moves the average through the value it sits on, so the 0.2 parked on 0 contributes nothing at all: $0(0.2) = 0$. The pull comes from the top of the table, where 3 carries only 0.1 of the weight instead of the 0.25 a flat average assumes, and that is what brings the expected value down to 1.3.

Some distributions come with a shortcut. A binomial variable with $n$ trials and success probability $p$ has $\mu_X = np$ directly, so 10 free throws shot at $p = 0.3$ give an expected 3 makes without summing eleven separate terms. The law of large numbers is what makes any of this useful in practice: as the number of repetitions grows, the observed sample mean converges on $E(X)$. That also fixes the relationship worth keeping straight. $E(X)$ is a parameter of the probability distribution, a fixed number you calculate; a sample mean is a statistic that varies from one batch of data to the next and settles toward it.

The sum here is the discrete version. For a continuous variable the same weighted-balance idea becomes an integral of $x$ against the density instead of a sum, which is the calculus notion of the average value of a function.

The same weighted sum, with dollar payoffs, is how expected value is calculated in economics: how to calculate expected value.