Two-Dimensional LOTUS for Discrete Variables¶

Theorem¶

For a discrete pair $(X,Y)$ and a function $g$ with $\displaystyle \sum_{x,y}|g(x,y)|p_{X,Y}(x,y)<\infty$ $\infty$ ,

\begin{aligned} \mathbb{E}[g(X,Y)]=\sum_x\sum_y g(x,y)\,p_{X,Y}(x,y). \end{aligned}

For $g\ge0$ the formula also holds without this assumption, as an extended (possibly infinite) expectation.

Interpretation¶

A function of two variables can be averaged directly using their joint PMF, even when the variables are dependent: each pair $(x,y)$ contributes the value $g(x,y)$ times its mass. There is no need to work out the distribution of $g(X,Y)$ first.

A 6 by 6 grid of two-dice outcomes with the anti-diagonal pairs that sum to 7 highlighted; averaging the indicator over the grid gives 1/6.

For two fair dice, take $g(x,y)=1$ if $x+y=7$ and $g(x,y)=0$ otherwise, the indicator of the event $\{X+Y=7\}$ $7}$ . It equals $1$ on the six red pairs of the anti-diagonal and $0$ on the other $30$ pairs. Each pair has mass $\tfrac{1}{36}$ , so $\mathbb{E}[g(X,Y)]=6\cdot\tfrac{1}{36}=\tfrac{1}{6}$ $\frac{1}{6}$ , which is $\mathbb{P}(X+Y=7)$ $7)$ : the expectation of an indicator is the probability of its event.

This is LOTUS with the pair $(X,Y)$ in place of a single variable, so the weights are joint masses, not the two marginal PMFs. The difference matters: for $X=Y\sim\operatorname{Bern}(\tfrac{1}{2})$ $Bern (\frac{1}{2})$ , $\mathbb{E}[XY]=\mathbb{E}[X^2]=\tfrac{1}{2}$ $\frac{1}{2}$ , while for independent fair Bernoulli variables, with the same marginals, $\mathbb{E}[XY]=\tfrac{1}{4}$ $\frac{1}{4}$ . The summability condition guarantees that the double sum has the same finite value in any order of summation.

Choosing $g$ gives familiar rules. With $g(x,y)=x+y$ the double sum splits into $\displaystyle \sum_xx\,p_X(x)+\sum_yy\,p_Y(y)$ $y \sum y p_{Y} (y)$ , which is linearity of expectation, $\mathbb{E}[X+Y]=\mathbb{E}X+\mathbb{E}Y$ $E Y$ , without any independence assumption, provided both means exist. For independent $X$ and $Y$ the joint PMF factorizes, and $g(x,y)=xy$ gives $\mathbb{E}[XY]=\mathbb{E}X\,\mathbb{E}Y$ $E X E Y$ . The continuous version replaces the sums by an integral against a joint density.

Key Results¶

Choosing $g(x,y)=xy$ gives $\mathbb{E}[XY]$ .
Choosing $g(x,y)=\indicator_{\{x+y\le c\}}$ gives $\mathbb{P}(X+Y\le c)$ $c)$ .

Examples¶

For independent fair Bernoulli variables, $\mathbb{E}[XY]=1(1/4)=1/4$ $1/4$ .
For independent fair dice, $\mathbb{E}[\indicator_{\{X+Y=7\}}]=6(1/36)=1/6$ $1/6$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 1

Keywords: LOTUS Symbols: infinity, sum

Last updated: 20 October 2026 03:02