Examples of pairewise independent but not independent continuous random variables By considering the set $\{1,2,3,4\}$, one can easily come up with an example (attributed to S. Bernstein) of pairwise independent but not independent random variables. 
Counld anybody give an example with continuous random variables? 
 A: 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.
A: 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 
$x, y, z \in (-\infty,\infty)$
A: 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).
A: Here's a potentially very simple construction for $k$-wise independent random variables, uniformly distributed on $[0, 1]$ (though admittedly I didn't work it out very carefully so hopefully the condition checks out)? There is a standard way of constructing discrete $k$-wise independent random variables: let $X_1,\ldots, X_{k}$ be uniform independent rvs drawn from $F$, $F$ a finite field of order $q$ ($q$ sufficiently larger than $k$, of course). Then, for $u\in F$, let $Y_u = X_1 + uX_2+\cdots + u^{k-1}X_k$. Then, the random variables $\{Y_u : u\in F\}$ are $k$-wise independent.
Now, just divide $[0, 1]$ into $q$ evenly spaced subintervals. Let $Z_u$ be uniform from the $Y_u$th subinterval. The rvs $\{Z_u : u\in F\}$ are $k$-wise independent (their joint CDF would decompose as the product of the individual CDFs, since the $\{Y_u\}$ are $k$-wise independent), and are uniform over $[0, 1]$ since each $Y_u$ is uniform over $F$.
