Check my proof on showing two Bernoulli RV's are independent

[Runty_Grunty]

I've got a pretty good answer to this one already, yet I'd like to see how solid it is. I'll list the question first in quotes.

Show that two Bernoulli random variables $$X$$ and $$Y$$ are independent if and only if $$P(X=1,Y=1)=P(X=1)P(Y=1)$$.

Here's my work below. I credit http://arxiv.org/PS_cache/arxiv/pdf/0909/0909.1685v4.pdf" for the answer.

$$X$$ and $$Y$$ are independent if and only if $$P(X=i,Y=j)=P(X=i)P(Y=j)$$ where $$i,j=0,1$$.
Two Bernoulli random variables are independent if and only if they are uncorrelated, and thus have a covariance of zero.
$$Corr(X,Y)=0\Leftrightarrow Cov(X,Y)=0$$

Let $$p(x)$$ be the pmf of $$X$$, and let $$p(y)$$ be the pmf of $$Y$$.
If $$X$$ and $$Y$$ are independent then by definition
$$Cov(X,Y)=p(xy)-p(x)p(y)=P(X=i,Y=j)-P(X=i)P(Y=j)=0$$,
as $$P(X=i,Y=j)=P(X=i)P(Y=j)$$ for $$i,j=0,1$$.
If on the other hand we have that $$Cov(X,Y)=0$$, then
$$p(xy)-p(x)p(y)=0\Rightarrow p(xy)=p(x)p(y)$$.
Therefore, $$X$$ and $$Y$$ are independent.

This should properly answer the question, though I've been told by another source that summing up the marginal pmfs is also necessary to show independence. I don't know whether or not that's really necessary, though, and could use a second opinion.

Is there anything about my proof that could use improvement?

 

[statdad]

Note:

$$\begin{align*}<br /> cov(X,Y) & = E(XY) - E(X)E(Y) \\<br /> &= \left(0 \cdot 0 \cdot p(0,0) + 0 \cdot 1 \cdot p(0,1) + 1 \cdot 0 \cdot p(1,0) + 1 \cdot 1 \cdot p(1,1)\right) - \left(0 p_x(0) + 1 p_x(1)\right) \left(0 \cdot p_y(0) + 1 \cdot p_y(1)\right) \\<br /> &= p(1,1) - p_x(1)p_y(1) \tag{A}<br /> \end{align*}$$

I think this relates to the "summing" comment. Also, nothing in the above calculation is based on an assumption of independence or dependence. From (A) you can say:

If X, Y are independent, then ...

and then

If cov(X,Y) = 0 it must be true that ... so ...

 

[Runty_Grunty]

Note:

$$\begin{align*}<br /> cov(X,Y) & = E(XY) - E(X)E(Y) \\<br /> &= \left(0 \cdot 0 \cdot p(0,0) + 0 \cdot 1 \cdot p(0,1) + 1 \cdot 0 \cdot p(1,0) + 1 \cdot 1 \cdot p(1,1)\right) - \left(0 p_x(0) + 1 p_x(1)\right) \left(0 \cdot p_y(0) + 1 \cdot p_y(1)\right) \\<br /> &= p(1,1) - p_x(1)p_y(1) \tag{A}<br /> \end{align*}$$

I think this relates to the "summing" comment. Also, nothing in the above calculation is based on an assumption of independence or dependence. From (A) you can say:

If X, Y are independent, then ...

and then

If cov(X,Y) = 0 it must be true that ... so ...

Thanks, I was wondering what was meant by that "summing" part.