Examples of pairewise independent but not independent continuous random variables

Solution 1:

Let $x,y,z'$ be normally distributed, with $0$ mean. Define $$z=\begin{cases} z' & xyz'\ge 0\\ -z' & xyz'<0\end{cases}$$ The resulting $x,y,z$ will always satisfy $xyz\ge 0$, but be pairwise independent.

Solution 2:

An answer of mine on stats.SE gives essentially the same answer as the one given by vadim123.

Consider three standard normal random variables $X,Y,Z$ whose joint probability density function $f_{X,Y,Z}(x,y,z)$ is not $\phi(x)\phi(y)\phi(z)$ where $\phi(\cdot)$ is the standard normal density, but rather

$$f_{X,Y,Z}(x,y,z) = \begin{cases} 2\phi(x)\phi(y)\phi(z) & ~~~~\text{if}~ x \geq 0, y\geq 0, z \geq 0,\\ & \text{or if}~ x < 0, y < 0, z \geq 0,\\ & \text{or if}~ x < 0, y\geq 0, z < 0,\\ & \text{or if}~ x \geq 0, y< 0, z < 0,\\ 0 & \text{otherwise.} \end{cases}\tag{1}$$

We can calculate the joint density of any pair of the random variables, (say $X$ and $Z$) by integrating out the joint density with respect to the unwanted variable, that is, $$f_{X,Z}(x,z) = \int_{-\infty}^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dy. \tag{2}$$

• If $x \geq 0, z \geq 0$ or if $x < 0, z < 0$, then $f_{X,Y,Z}(x,y,z) = \begin{cases} 2\phi(x)\phi(y)\phi(z), & y \geq 0,\\ 0, & y < 0,\end{cases}$ and so $(2)$ reduces to $$f_{X,Z}(x,z) = \phi(x)\phi(z)\int_{0}^\infty 2\phi(y)\,\mathrm dy = \phi(x)\phi(z). \tag{3}$$

• If $x \geq 0, z < 0$ or if $x < 0, z \geq 0$, then $f_{X,Y,Z}(x,y,z) = \begin{cases} 2\phi(x)\phi(y)\phi(z), & y < 0,\\ 0, & y \geq 0,\end{cases}$ and so $(2)$ reduces to $$f_{X,Z}(x,z) = \phi(x)\phi(z)\int_{-\infty}^0 2\phi(y)\,\mathrm dy = \phi(x)\phi(z). \tag{4}$$

In short, $(3)$ and $(4)$ show that $f_{X,Z}(x,z) = \phi(x)\phi(z)$ for all $x, z \in (-\infty,\infty)$ and so $X$ and $Z$ are (pairwise) independent standard normal random variables. Similar calculations (left as an exercise for the bemused reader) show that $X$ and $Y$ are (pairwise) independent standard normal random variables, and $Y$ and $Z$ also are (pairwise) independent standard normal random variables. But $X,Y,Z$ are not mutually independent normal random variables. Indeed, their joint density $f_{X,Y,Z}(x,y,z)$ does not equal the product $\phi(x)\phi(y)\phi(z)$ of their marginal densities for any choice of $x, y, z \in (-\infty,\infty)$

Solution 3:

The continuous analog of the Bernstein example: Divide up the unit cube into eight congruent subcubes of side length $1/2$. Select four of these cubes: Subcube #1 has one vertex at $(x,y,z)=(1,0,0)$, subcube #2 has one vertex at $(0,1,0)$, subcube #3 has one vertex at $(0,0,1)$, and subcube #4 has one vertex at $(1,1,1)$. (To visualize this, you have two layers of cubes: the bottom layer has two cubes in a diagonal formation, and the top layer has two cubes in the opposite diagonal formation.)

Now let $(X,Y,Z)$ be uniform over these four cubes. Clearly $X, Y, Z$ are not mutually independent, but every pair of variables $(X,Y)$, $(X,Z)$, $(Y,Z)$ is uniform over the unit square (and hence independent).