Eve’s Law: Total Variance¶

Theorem¶

For random variables $X$ and $Y$ with $\mathbb{E}[Y^2]<\infty$ $\infty$ ,

\operatorname{Var}(Y)=\mathbb{E}[\operatorname{Var}(Y\mid X)] +\operatorname{Var}(\mathbb{E}[Y\mid X]).

Consequence: the residual $Y-\mathbb{E}[Y\mid X]$ is uncorrelated with every function $h(X)$ that has $\mathbb{E}[h(X)^2]<\infty$ (or satisfies another assumption that makes the products integrable):

\begin{aligned} \operatorname{Cov}(Y-\mathbb{E}[Y\mid X],\,h(X))=0. \end{aligned}

The first term averages the Conditional Variance Given a Random Variable; the second is the ordinary Variance and Standard Deviation of the conditional mean. To derive the identity, apply Adam’s Law: Total Expectation to $Y^2$ and expand both variances as second moment minus squared mean.

Interpretation¶

Total spread consists of average spread within conditional groups plus spread between their conditional means. Knowing $X$ splits the variance of $Y$ into a part that $X$ explains, how much the group centres move, and a part that remains, how much $Y$ varies inside a group.

Points of Y in three groups for X = 1, 2 and 3, each with a short line at its group mean, and a long line at the overall mean, showing the spread within groups and between group means.

The figure groups values of $Y$ by $X=1,2,3$ $1, 2, 3$ . The short red lines are the group means $\mathbb{E}[Y\mid X=x]$ and the dashed line is the overall mean $\mathbb{E}Y$ . The first term, $\mathbb{E}[\operatorname{Var}(Y\mid X)]$ $X)]$ , averages the spread of the points around their own red line (the arrow "within"). The second term, $\operatorname{Var}(\mathbb{E}[Y\mid X])$ $X])$ , measures how far the red lines lie from the dashed line (the arrow "between").

Two extreme cases show what each term means. If $Y$ is a function of $X$ , every group is a single point, the first term is $0$ and all variance lies between groups. If $Y$ is independent of $X$ , every group mean equals $\mathbb{E}Y$ by conditional expectation and independence, so the second term is $0$ and $X$ explains nothing. Both terms are nonnegative, which gives the Key Result $\operatorname{Var}(\mathbb{E}[Y\mid X])\le\operatorname{Var}(Y)$ $Var (Y)$ : replacing $Y$ by its conditional mean can only reduce variance.

The practical form of the identity is the conditioning trick for variance. A frequent slip is to drop one of the two terms, or to write the first term as the random variable $\operatorname{Var}(Y\mid X)$ instead of its average $\mathbb{E}[\operatorname{Var}(Y\mid X)]$ $X)]$ .

Key Results¶

$\operatorname{Var}(\mathbb{E}[Y\mid X])\le\operatorname{Var}(Y)$ $Var (Y)$ .
The residual $R=Y-\mathbb{E}[Y\mid X]$ has mean zero and $\operatorname{Cov}(R,h(X))=0$ for every square-integrable $h(X)$ .

Examples¶

If $X\sim\operatorname{Exp}(2)$ and $Y\mid X\sim\operatorname{Pois}(X)$ $Pois (X)$ , then $\operatorname{Var}(Y)=\mathbb{E}X+\operatorname{Var}(X)=1/2+1/4=3/4$ $3/4$ .
If $X\sim\operatorname{Unif}(0,1)$ and $Y\mid X\sim\operatorname{Bin}(n,X)$ $Bin (n, X)$ , then $\operatorname{Var}(Y)=n/6+n^2/12=n(n+2)/12$ $2) /12$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 6

Aliases: law of total variance, variance decomposition formula Symbols: infinity

Last updated: 22 October 2026 19:04