A problem on correlation and independence Here is a little challenge/problem for you guys.
Let $(X,Y)$ be jointly discrete RVs such that each have at most two mass points (not necessarily $1$ and $0$, i.e. not necessarily indicator variables). Suppose $X$ and $Y$ are uncorrelated. Are they independent?
 A: Since you're dealing with correlation, you can restrict consideration to 0-1 variables without loss of generality. Any linear rescaling of the 0-1 variables to $a$ and $b$ leaves the correlation unaltered. So if it's true of 0-1 variables, it's true of distributions with mass points at some arbitrary $a$ and $b$, with $a\neq b$.
Since the Pearson correlation for binary variables is the phi coefficient, and $n \phi^2$ is the chi-square, the immediate implication for binary variables is that zero correlation implies independence.
A: Here is an answer with no jargon about $\chi^2$ random variables statistics.
If $X$ and $Y$ are Bernoulli random variables with parameters $p$ and $q$
respectively, then 
$E[X] = p = P\{X=1\}$ and $E[Y] = q = P\{Y=1\}$. Also, 
$$E[XY] = 1\cdot 1\cdot P\{X=1,Y=1\} = P\{X=1,Y=1\}.$$
Now, if the random variables $X$ and $Y$ are 
uncorrelated, then $$\operatorname{cov}(X,Y) = E[XY]-E[X]E[Y] = 0
~\Rightarrow ~E[XY] = pq$$
which shows that $P\{X=1, Y = 1\} = P\{X=1\}P\{Y=1\}$, that is, the
events $\{X=1\}$ and $\{Y=1\}$ are independent events.  It follows 
from standard properties of two independent events that
the events $\{X=1\}$ and $\{Y=0\}$ also are independent events, as are
$\{X=0\}$ and $\{Y=1\}$, as well as $\{X=0\}$ and $\{Y=0\}$. In other
words, $$P\{X=i, Y=j\} = P\{X=i\}P\{Y=j\}~ \text{for all}~ i,j \in \{0,1\}$$
showing that $X$ and $Y$ are independent random variables. 
Now, if $X$ and $Y$ are any independent random variables,
then so are $g(X)$ and $h(Y)$ independent random variables (for
arbitrary measurable functions $g$ and $h$), so that
if $X$ and $Y$ are uncorrelated Bernoulli random variables, and thus
independent random variables, $aX+b$ and $cY+d$ are also
independent random variables and hence uncorrelated random variables.
The latter can also be inferred using Glen_b's remark that
linear transformations don't affect the covariance and
the (Pearson) correlation coefficient, that is,
$$\begin{align}\operatorname{cov}(aX+b, cX+d) &= \operatorname{cov}(X,Y),\\
\rho_{aX+b,cY+d} &= \rho_{X,Y}\end{align}$$
