Uncorrelated implies independence?

192 Views Asked by At

Let $X\sim b(1,p)$ and $Y\sim b(1,q)$, in other words, $X$ and $Y$ are in Bernoulli distribution.
Question: If $X$ and $Y$ are uncorrelated, does it imply $X$ and $Y$ are independent?

These are my steps:
$E(X)=p$ and $E(Y)=q$

Since $X$ and $Y$ are uncorrelated
$\begin{align} Cov(X,Y)&=0\\ E(XY)-E(X)E(Y)&=0\\ E(XY)&=pq\\ \sum_{x} \sum_{y}\ xy \cdot p(x,y) &= pq\ \\ p(1,1) &= pq \end{align}$

Since $P_{X}(1)=p$, then
$\begin{align} P(1,0)+P(1,1) &= p\\ P(1,0) &= p-qp \\ &= p(1-q) \end{align}$

For the same reason, $P(0,1)=q(1-p)$
Then $P(0,0)=1-pq-p(1-q)-q(1-p)=(1-p)(1-q)$

Since $p(1,1)$ is known, the remaining 3 probabilities $p(x,y)$ is also determined and equal to $p_{X}(x) \cdot p_{Y}(y)$
Therefore, in this case, $X$ and $Y$ are uncorrelated and independent.

Am I right for this proof?