Is there a *simple* example showing that uncorrelated random variables need not be independent? 
Is there a simple example showing that given $X,Y$ uncorrelated
  (covariance is zero), $X,Y$ are not independent?

I have looked up two references, however, I am dissatisfied with both.


*

*In Reference $1$, $X,Y$ are assumed to be independent uniform RVs
from $(0,1)$, construct $Z = X+Y, W = X - Y$, then the claim is that
$Z,W$ is uncorrelated but not independent. Unfortunately, finding the
PDF of $Z,W$ is not trivial.

*In Reference $2$, $\phi$ is assumed to be uniform RV from $(0,
   2\pi)$, and construct $X = \cos(\phi)$, $Y = \sin(\phi)$. Then the claim is that $X,Y$ are uncorrelated but not independent. Unfortunately, the PDFs of $X,Y$ takes on the form of rarely mentioned arcsine distribution.
I just wish to have an example at hand where I can whip out to show that uncorrelated does not necessarily implies independent. Is this do-able? 
 A: Really boring example:
$$ \begin{array}{c|ccc}
Y \backslash X & -1 & 0 & 1 \\ \hline
-1 & p & p & p \\
0 & p & 1-8p & p \\
1 & p & p & p
\end{array}. $$
Then the marginal distributions are both
$$  \begin{array}{ccc}
-1 & 0 & 1 \\ \hline
3p & 1-6p & 3p
\end{array}, $$
so $E[X]=E[Y]=0$. A similar calculation shows $E[XY]=0$, but an independent joint distribution would have be the product of the marginals,
$$ \begin{array}{c|ccc}
 & -1 & 0 & 1 \\ \hline
-1 & 9p^2 & 3p(1-6p) & 9p^2 \\
0 & 3p(1-6p) & (1-6p)^2 & 3p(1-6p) \\
1 & 9p^2 & 3p(1-6p) & 9p^2
\end{array}. $$
It is easy to see this corresponds to the original table precisely when $p=1/9$ or $0$: $p$ anything between $0$ and $1/9$ exclusive gives a counterexample.
A: $X$ is uniform over $[-1,1]$. 
$Y = |X|$ 
$E[XY]=0$ since it is symmetric about $0$
$Y$ is uniform over $[0,1]$.
$Y$ given $X$ is deterministic.
So they are not independent. 
A: Let $X$ be any symmetric, square integrable random variable, and let $Y$ be independent of $X$, with $P(Y=-1)=P(Y=1)=\frac{1}{2}$.
Then $X$ and $XY$ are trivially uncorrelated, but certainly not independent, as $\lvert X\rvert=\lvert XY\rvert$.
A: Here's a (perhaps) simpler example. Let $X$ be $N(0,1)$ and $Y = X^2.$ Then $$ E(XY) = E(X^3) = 0 =E(X)E(Y),$$ so $X$ and $Y$ are uncorrelated, but clearly they aren't independent (if you know $X$, then you know $Y).$
A: A very simple example: Let (X,Y) take values from {(0,0), (1,1), (2,0)}, equiprobably. Y is functionally dependent on X, yet the covariance is 0
A: Two fair coins are tossed independently; the first has sides labelled $0$ and $1,$ the second has sides labelled $1$ and $-1.$ Let $X$ be the number that comes up on the first coin, and let $Y$ be the product of the two numbers that come up.
The variables $X$ and $Y$ are uncorrelated: since $XY=Y,$
$$E(XY)=E(Y)=0=\frac12\cdot0=E(X)E(Y).$$
The variables $X$ and $Y$ are not independent:
$$P(X=0,Y=0)=P(X=0)=\frac12\ne\frac12\cdot\frac12=P(X=0)P(Y=0).$$
A: Another easy to explain on the run:
for $\theta$ Uniform [0 , 2$\pi$] set X = sin($\theta$), Y = cos($\theta$).
Then $X^2 + Y^2 = 1$ - clearly X and Y are not independent.
However as correlation is a linear relationship it is unable to produce anything but zero for the correlation between X and Y
A: How about $(X,Y)$ taking values $(1,0)$, $(-1,0)$, $(0,1)$ and $(0,-1)$
each with probability $1/4$? Then $E(X)=E(Y)=0$ and $XY=0$, so the covariance is zero, but $X$ and $Y$ are not independent.
