Conditional Expectation and Independence¶

Proposition¶

Let $\mathbb{E}|Y|<\infty$ $\infty$ . If $X$ and $Y$ are independent ( $X ⊥$ $X\perp Y$ $Y$ ), then $\mathbb{E}[Y\mid X]=\mathbb{E}[Y]$ almost surely.

Interpretation¶

If the response $Y$ is independent of the observed variable $X$ , observing $X$ cannot change the expected value of $Y$ : the best prediction of $Y$ given $X$ is the constant $\mathbb{E}Y$ .

Two panels. Left: a 6 by 6 grid of dice outcomes with a flat line at the conditional mean 3.5. Right: the parabola x = y squared, where the conditional mean of Y is 0 for every x although X and Y are dependent.

The red lines show $\mathbb{E}[Y\mid X=x]$ $x]$ . For independent fair dice this equals $3.5$ in every column. A constant conditional mean does not imply independence: if $Y\sim\operatorname{Unif}(-1,1)$ and $X=Y^2$ $Y^{2}$ , then given $X=x$ the values $\pm\sqrt{x}$ are equally likely, so $\mathbb{E}[Y\mid X]=0$ almost surely, yet $X$ and $Y$ are dependent.

Under independence the conditional distribution of $Y$ given $X=x$ is the unconditional distribution of $Y$ , for every $x$ , so its mean is $\mathbb{E}Y$ . This gives more than the mean: $\mathbb{E}[h(Y)\mid X]=\mathbb{E}[h(Y)]$ for every integrable $h(Y)$ . The words almost surely are needed because $\mathbb{E}[Y\mid X]$ is only determined up to events of probability zero.

The proposition is the first step of a chain: independence implies a constant conditional mean, which implies zero covariance when the variances are finite. The second step uses taking out what is known and Adam’s law: $\mathbb{E}[XY]=\mathbb{E}[X\,\mathbb{E}[Y\mid X]]=\mathbb{E}X\,\mathbb{E}Y$ $E X E Y$ . Neither step can be reversed. The figure shows the first; for the second, $X\sim\operatorname{Unif}(-1,1)$ and $Y=X^2$ are uncorrelated, but $\mathbb{E}[Y\mid X]=X^2$ is not constant.

Note the direction: the proposition is about $\mathbb{E}[Y\mid X]$ $X]$ . In the right panel $\mathbb{E}[Y\mid X]=0$ is constant, but $\mathbb{E}[X\mid Y]=Y^2$ is not, which already shows the dependence.

Key Results¶

Independence gives $\mathbb{E}[h(Y)\mid X]=\mathbb{E}[h(Y)]$ for every integrable $h(Y)$ .
The converse fails: a constant conditional mean $\mathbb{E}[Y\mid X]=\mathbb{E}Y$ does not by itself imply independence.

Examples¶

For independent fair dice $X,Y$ , $\mathbb{E}[Y\mid X]=3.5$ $3.5$ .
For $Y\sim\operatorname{Unif}(-1,1)$ and $X=Y^2$ $Y^{2}$ , symmetry gives $\mathbb{E}[Y\mid X]=0=\mathbb{E}Y$ $E Y$ , yet $X$ and $Y$ are dependent.

Prerequisites¶

Correlation

Sources¶

Nicky van Foreest · Probability Distributions · Week 4

Symbols: infinity, square root

Last updated: 20 October 2026 03:02