Two-Dimensional LOTUS for Continuous Variables¶

Theorem¶

For a jointly continuous pair $(X,Y)$ and a Borel-measurable function $g$ with $\iint |g(x,y)|f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y<\infty$ $\infty$ ,

\mathbb{E}[g(X,Y)] =\iint g(x,y)\,f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y.

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

Interpretation¶

Two-dimensional LOTUS computes the mean of a function $g(X,Y)$ of a continuous pair directly from the joint density: weight each value $g(x,y)$ by $f_{X,Y}(x,y)$ and integrate over the plane. The distribution of the new random variable $g(X,Y)$ is never needed, and the pair need not be independent.

The unit square with uniform joint density and level curves of g(x, y) = xy at 0.05, 0.25, 0.5 and 0.75, illustrating how the expectation of XY weights values of g by the density.

On the unit square the density is $1$ , so $\mathbb{E}[g(X,Y)]$ is simply the average of $g$ over the square. Along each blue level curve $xy=c$ the function $g$ has the constant value $c$ , and most of the square lies where $xy$ is small. Here $\mathbb{E}[XY]=\displaystyle\int_0^1\!\!\int_0^1xy\,\mathrm{d}x\,\mathrm{d}y=\tfrac{1}{4}$ $\frac{1}{4}$ , without finding the distribution of $XY$ .

The theorem is the LOTUS of Probability Theory with a double integral instead of a single one, and the continuous counterpart of the discrete two-dimensional version. The integrability condition makes the integral a well-defined finite number; for $g\ge0$ it may be dropped if an infinite answer is allowed. Choosing $g$ gives the standard quantities: $g(x,y)=xy$ gives $\mathbb{E}[XY]$ for the covariance, $g(x,y)=x+y$ gives $\mathbb{E}X+\mathbb{E}Y$ for any dependence provided $\mathbb{E}|X|<\infty$ and $\mathbb{E}|Y|<\infty$ $\infty$ , and an indicator of a region gives its probability.

For the dependent pair with density $2$ on $0<x<y<1$ $1$ , $\mathbb{E}[XY]=\int_0^1\int_0^y2xy\,\mathrm{d}x\,\mathrm{d}y=1/4$ $1/4$ , while $\mathbb{E}X=1/3$ and $\mathbb{E}Y=2/3$ $2/3$ . So $\operatorname{Cov}(X,Y)=1/4-2/9=1/36$ $1/36$ . For independent variables the double integral factors and $\mathbb{E}[XY]=\mathbb{E}X\,\mathbb{E}Y$ $E X E Y$ .

In general $\mathbb{E}[g(X,Y)]\ne g(\mathbb{E}X,\mathbb{E}Y)$ $g (E X, E Y)$ : for the triangle, $\mathbb{E}[XY]=1/4$ but $\mathbb{E}X\,\mathbb{E}Y=2/9$ $2/9$ .

Key Results¶

For $g(x,y)=xy$ $x y$ , the integral gives $\mathbb{E}[XY]$ .
For an indicator of a region $B$ , it gives $\mathbb{P}((X,Y)\in B)$ $B)$ .

Examples¶

For independent uniform coordinates on $(0,1)$ , $\mathbb{E}[XY]=\int_0^1\int_0^1xy\,\mathrm{d}x\,\mathrm{d}y=1/4$ $1/4$ .
For density $2$ on $0<x<y<1$ $1$ , $\mathbb{E}X=\int_0^1\int_0^y2x\,\mathrm{d}x\,\mathrm{d}y=1/3$ $1/3$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 2

Keywords: LOTUS Symbols: element of, infinity, integral

Last updated: 27 October 2026 11:52