Does covariance equal to zero implies independence for binary random variables?

If X and Y are two random variables that can only take two possible states, how can I show that Cov(X,Y)=0 implies independence? This kind of goes against what I learned back in the day that Cov(X,Y)=0 does not imply independence…

The hint says to start with 1 and 0 as the possible states and generalize from there. And I can do that and show E(XY)=E(X)E(Y), but this doesn’t imply independence???

Kind of confused how to do this mathematically I guess.

Answer

For binary variables their expected value equals the probability that they are equal to one. Therefore,

E(XY)=P(XY=1)=P(X=1Y=1)E(X)=P(X=1)E(Y)=P(Y=1)

If the two have zero covariance this means E(XY)=E(X)E(Y), which means

P(X=1Y=1)=P(X=1)P(Y=1)

It is trivial to see all other joint probabilities multiply as well, using the basic rules about independent events (i.e. if A and B are independent then their complements are independent, etc.), which means the joint mass function factorizes, which is the definition of two random variables being independent.

Attribution
Source : Link , Question Author : steven hurwitt , Answer Author : Antoni Parellada

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