# Binary random variables event-level independence implies random variable independence

Can you help me in proving the following?

For two binary random variables X and Y, the event-level independence ($x^0 \perp y^0$) implies random variable independence $X \perp Y$.

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What does event-level independence mean? –  Dilip Sarwate Mar 19 '12 at 20:42
and does binary mean Bernoulli variables? –  yohBS Mar 19 '12 at 20:45
X and Y are random variables. $X = \{ x^0, x^1 \}$ and $Y=\{ y^0, y^1 \}$. $x^0, x^1, y^0$ and $y^1$ are the events that make up $X$ and $Y$. –  tzsjqnpn Mar 19 '12 at 20:46
Yes, binary random variables are Bernoulli variables. –  tzsjqnpn Mar 19 '12 at 20:46

It is a matter of a system of equations type argument. Using the law of total probability, and the fact that the event $(X=0)$ is the complement of the event $(X=1)$ (same for $Y$), then we know: $$P(X=0) = P(X=0|Y=0)\cdot{}P(Y=0) + P(X=0|Y=1)\cdot{}P(Y=1)$$ $$P(X=0) = P(X=0)\cdot{}P(Y=0) + P(X=0|Y=1)\cdot{}(1-P(Y=0))$$ Note that I used the known independence assumption to simplify the first term on the RHS going from line one to line two.
Now solve this expression for $P(X=0|Y=1)$ $$P(X=0 | Y=1) = \frac{P(X=0) - P(X=0)P(Y=0)}{1-P(Y=0)} = P(X=0)$$
Thus, we now also know that $x^{0}$ is independent of $y^{1}$ too. Now, just repeat the same thing with $P(Y=1)$ with the new knowledge about $y^{1}$ and $x^{0}$ to get the result for $y^{1}$ and $x^{1}$.