# 2D LOTUS: joint PDF on unit square $\{ (x, y) : x, y \in [0, 1] \}$

My textbook, Introduction to Probability by Blitzstein and Hwang, presents the following example:

Example 7.2.2 (Expected distance between two Uniforms). Let $$X$$ and $$Y$$ be i.i.d. Unif$$(0, 1)$$ r.v.s. Find $$E(|X - Y|)$$.

Solution:

Since the joint PDF is 1 on the unit square $$\{ (x, y) : x, y \in [0, 1] \}$$, 2D LOTUS gives

\begin{align} E(|X - Y|) &= \int_0^1 \int_0^1 |x - y| \ dx dy \\ &= \int_0^1 \int_y^1 (x - y) \ dxdy + \int_0^1 \int_0^y (y - x) \ dxdy \\ &= 2 \int_0^1 \int_y^1 (x - y) \ dxdy = 1/3 \end{align}

First we broke up the integral into two parts so we could eliminate the absolute value; then we used symmetry.

The textbook defines the 2D LOTUS as follows:

Theorem 7.2.1 (2D LOTUS). Let $$g$$ be a function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$. If $$X$$ and $$Y$$ are discrete, then

$$E(g(X, Y)) = \sum_x \sum_y g(x, y) P(X = x, Y = y).$$

If $$X$$ and $$Y$$ are continuous with joint PDF $$f_{X, Y}$$, then

$$E(g(X, Y)) = \int_{-\infty}^\infty \int_{-\infty}^\infty g(x, y) f_{X, Y}(x, y) \ dx dy.$$

LOTUS means Law of the Unconscious Statistician.

My multiple integral knowledge is rusty, so I would appreciate it if people could please take the time to explain what's going on for each step here:

\begin{align} E(|X - Y|) &= \int_0^1 \int_0^1 |x - y| \ dx dy \\ &= \int_0^1 \int_y^1 (x - y) \ dxdy + \int_0^1 \int_0^y (y - x) \ dxdy \\ &= 2 \int_0^1 \int_y^1 (x - y) \ dxdy = 1/3 \end{align}

Thank you.

EDIT:

The PDF of a continuous uniform random variable is $$f(x) = {\begin{cases}{\frac {1}{b-a}}&\mathrm {for} \ a\leq x\leq b,\\[8pt]0&\mathrm {for} \ xb\end{cases}}$$

Since $$X$$ and $$Y$$ are independent, my understanding is that the product of their individual PDFs is equal to the joint PDF:

$$p_{X, Y}(x, y) = p_X(x) p_Y(y) = (1)(1) = 1$$

By the LOTUS, we have $$E(\vert X - Y \vert) = \int_{0}^{1}\int_{0}^{1} \vert x - y \vert \, \mathrm{d}x \,\mathrm{d}y.$$ The $$\vert x - y\vert$$ term inside the integral takes one of the two forms $$|x-y| = x-y$$ for $$x - y \geq 0$$ and $$|x -y| = y - x$$ for $$x - y < 0.$$ So we split the region of interest into two regions (see attached image). 1
In the region defined by $$x - y \geq 0$$ (i.e. the region of the square below the line $$y = x$$), $$y$$ can take values between $$0$$ and $$1$$, and for each fixed $$y$$, $$x$$ runs from $$y$$ to $$1$$. Hence the integral of the function over this region can be written as $$\int_{0}^{1}\int_{y}^{1} (x-y)\,\mathrm{d}x\,\mathrm{d}y.$$ Similarly, for the region defined by $$x - y < 0$$, for each fixed $$y$$ between $$0$$ and $$1$$, $$x$$ runs from $$0$$ to $$y$$. So the integral of the function over this region is $$\int_{0}^{1}\int_{0}^{y} (y-x)\,\mathrm{d}x\,\mathrm{d}y.$$ This explains the step going from the first line to the second: $$\int_{0}^{1}\int_{0}^{1} \vert x - y \vert \, \mathrm{d}x \,\mathrm{d}y = \int_{0}^{1}\int_{y}^{1} (x-y)\,\mathrm{d}x\,\mathrm{d}y + \int_{0}^{1}\int_{0}^{y} (y-x)\,\mathrm{d}x\,\mathrm{d}y.$$ In fact, the region in the second integral can be specified in a different manner by noting that $$x$$ runs from $$0$$ to $$1$$ and for each fixed $$x$$, $$y$$ runs from $$x$$ to $$1$$. This allows us to reverse the order of integration to obtain $$\int_{0}^{1}\int_{0}^{y} (y-x)\,\mathrm{d}x\,\mathrm{d}y = \int_{0}^{1}\int_{x}^{1}(y-x)\,\mathrm{d}y\,\mathrm{d}x.$$ The value of the definite integral is independent of the dummy variables $$x$$ and $$y$$, so we may replace them as we please. In particular, we can switch their places to obtain $$\int_{0}^{1}\int_{0}^{y} (y-x)\,\mathrm{d}x\,\mathrm{d}y = \int_{0}^{1}\int_{y}^{1}(x-y)\,\mathrm{d}x\,\mathrm{d}y.$$ Note that this is an identical copy of the other integral in the second line, so this explains how the $$2$$ is obtained in the step from the second line to the third, and it remains to evaluate the result.
• Thanks for the answer. Out of curiosity, couldn't we do the multiple integral the other way -- that is, in the region defined by $x - y \geq 0$, we could have $x$ take values from $0$ to $1$, and for each fixed $x$, $y$ runs from $x$ to $1$? – The Pointer Oct 14 '19 at 11:18
• Yes that's correct, we could also choose to do the integral that way by integrating with respect to $y$ first. Though in the region $x - y \geq 0$, if $x$ takes values from $0$ to $1$ then for each $x$, $y$ will run from $0$ to $x$ rather than from $x$ to $1$. A helpful way to check this (or to determine the bounds in the first place) is by sketching the region in the $xy$-plane – Zac Oct 14 '19 at 12:18
\begin{align} & E(|X - Y|) = \\ &= \int_0^1 \int_0^1 |x - y| \ dx dy \\ &= \int_0^1 \int_y^1 (x - y) \ dxdy + \int_0^1 \int_0^y (y - x) \ dxdy \text{\qquad Seperate for y