Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why does $ \text{Var}(Y) = E(\text{Var}(Y|X))+ \text{Var}(E(Y|X))$? What is the intuitive explanation for this? In laymen's terms it seems to say that the variance of $Y$ equals the expected value of the conditional variance plus the variance of the conditional expectation.

share|cite|improve this question
up vote 13 down vote accepted

A rigorous proof is here; it relies on the law of total expectation, which says that $E(E(X|Y))=E(X)$. The intuitive explanation of that is that $E(X|Y)$ is the expected value of $X$ given a particular value of $Y$, and that $E(E(X|Y))$ is the expected value of that over all values of $Y$. So $Y$ no longer matters, and we're just looking at $E(X)$.

The variance law is a bit more difficult to parse, but this is what it says to me. "How much does $Y$ vary? We expect it to vary by the average value of the variances we get by fixing $X$. But even when we fix $X$, there is some swing in $Y$, and thus swing in $E(Y|X)$. So we add on the variance of $E(Y|X)$. The first term is the expected variance from the mean of $Y|X$; the second is the variance of that mean."

share|cite|improve this answer

Geometrically it's just the Pythagorean theorem. We may measure the "length" of random variables by standard deviation.

We start with a random variable Y. E(Y|X) is the projection of this Y to the set of random variables wich may be expressed as a deterministic function of X.

We have a hypotenuse Y with squared length Var(Y).

The first leg is E(Y|X) with squared length Var(E(Y|X)).

The second leg is Y-E(Y|X) with squared length Var(Y-E(Y|X))=...=E(Var(Y|X)).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.