Law of Total Variance

Var(X) splits into an average within-group variance plus the variance of the group means: within-group + between-group.

1. The idea

The law of total variance decomposes $Var (X)$ into two interpretable pieces: $Var (X) = E [Var (X ∣ Y)] + Var (E [X ∣ Y])$ .

The first term, $E [Var (X ∣ Y)]$ , is the average 'within-group' variability — how spread out $X$ is inside each scenario of $Y$ . The second term, $Var (E [X ∣ Y])$ , is the 'between-group' variability — how much the conditional means themselves differ across scenarios.

This decomposition explains why mixing regimes with different means increases overall variance even if each regime individually has low variance: the between-group term adds extra spread.

2. Why it matters

This identity underlies why regime-switching or mixture models can look far riskier in aggregate than any single regime, a frequent subtlety in risk and volatility interview questions.

3. Core mathematics

Law of total variance

Var (X) = E [Var (X ∣ Y)] + Var (E [X ∣ Y])

4. How to recognize it

Think of this when…

X's distribution differs across regimes/groups defined by Y
You are asked for overall variance given per-group means and variances
The question tests whether mixing populations can increase variance beyond any individual group's

5. Worked example

Worked example

A trading strategy is in regime A (prob 0.5, mean return 2, variance 1) or regime B (prob 0.5, mean return 6, variance 1). Find the unconditional variance of return.

1
Within-group term: both regimes have variance 1, so the average is 1.
$E [Var (X ∣ Y)] = 0.5 (1) + 0.5 (1) = 1$
2
Between-group term: the overall mean is $0.5 (2) + 0.5 (6) = 4$ . Compute variance of the conditional means around 4.
$Var (E [X ∣ Y]) = 0.5 (2 - 4)^{2} + 0.5 (6 - 4)^{2} = 0.5 (4) + 0.5 (4) = 4$
3
Add the two components.
$Var (X) = 1 + 4 = 5$
Answer: $Var (X) = 5$ , far above either regime's own variance of 1

6. Common mistakes

Reporting only the within-group variance and ignoring the between-group term
Computing the between-group term as the variance of Y instead of the variance of the conditional means E[X|Y]
Assuming variance of a mixture equals the weighted average of the component variances

7. Quick check

If all regimes have the same conditional mean but different variances, the between-group term equals:

8. Practice

2 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Conditional Expectation Jensen's Inequality

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