2
$\begingroup$

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$.

  • 1
    What does event-level independence mean?2012-03-19
  • 0
    and does binary mean [Bernoulli variables](http://en.wikipedia.org/wiki/Bernoulli_distribution)?2012-03-19
  • 0
    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 $Y$.2012-03-19
  • 0
    Yes, binary random variables are Bernoulli variables.2012-03-19

1 Answers 1

2

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 $x^{1}$.