Probability Theory 16 | Variance [dark version]

Variance turns “how much a random variable fluctuates around its mean” into a precise number. After defining expectation as the average value a random variable centers around, variance measures the spread by looking at deviations from that expectation, squaring them to remove sign, and then averaging the squared deviations. The key requirement is that the expectation involved in the variance calculation must exist; otherwise the spread measure isn’t well-defined.

Formally, variance is built from the random variable’s deviation from its mean: start with X − E[X], note that this new quantity has expectation 0, then square it and take its expectation. That yields Var(X) = E[(X − E[X])²]. Expanding the square gives a more computationally convenient form: Var(X) = E[X²] − (E[X])². This identity follows from linearity of expectation and the fact that E[c] = c for constants (including E[1] = 1 under any probability measure). In practice, this means variance can be computed by finding the expected value of X² and subtracting the square of the expected value of X.

To compute variance without abstract measure-theory language, the transcript connects the definition to the two standard cases. For discrete random variables, expectations become sums over outcomes weighted by a probability mass function. For continuous random variables, expectations become integrals weighted by a probability density function. In both settings, the only change from the mean calculation is that X is replaced by X² inside the expectation.

A first worked example uses a discrete uniform distribution over n outcomes {X1, …, Xn}, where each outcome has probability 1/n. The expectation becomes the arithmetic mean X̄ = (1/n)∑_{j=1}^n Xj. Using the variance definition, the spread becomes Var(X) = (1/n)∑_{j=1}^n (Xj − X̄)²—an average of squared deviations from the mean, scaled by 1/n. This generalizes the familiar “fair die” intuition to any finite set of equally likely outcomes.

The second example switches to a continuous exponential distribution with rate parameter λ. From the earlier expectation result, E[X] = 1/λ. To get variance, the calculation needs E[X²], which becomes the integral ∫_0^∞ x² · (λ e^{-λx}) dx. Solving via integration by parts twice gives E[X²] = 2/λ². Plugging into Var(X) = E[X²] − (E[X])² yields Var(X) = 1/λ², the standard variance formula for the exponential distribution.

Overall, variance is presented as a mean-squared deviation: compute E[X²], subtract (E[X])², and interpret the result as the width of the distribution around its expectation. The next step promised is to study additional properties that make variance especially useful in later probability work.