## dependent probability

if X and Y are independent then E(X)E(Y)=E(XY)

I have found a lot of examples for this for example if X-values gives tail or head and Y is the sides of a square

but i cant find an example for a dependent function where E(X)E(Y)=E(XY) does not apply and I want to have an example tht shows mathematically that E(X)E(Y)=E(XY) does not apply- Does anyone have such an illustrating example that describers this mathematically?
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 I think an example would be the case where X gives the number of hearts in a poker hand and Y gives the number of red cards in a poker hand. Here knowing X affects the expected value of Y. Isn't this really a prob/stats question rather than a number theory one, though?
 Blog Entries: 1 Recognitions: Homework Help The best example is the drastic example. Flip a coin - X=1/2 if the coin is heads, X=-1/2 if the coin is tails. Y=1/2 if the same coin is heads, Y=-1/2 if the coin is tails. E(X)=E(Y)=0 but XY=1/4 regardless of whether the coin is heads or tails, so E(XY)=1/4

## dependent probability

 Quote by Office_Shredder The best example is the drastic example. Flip a coin - X=1/2 if the coin is heads, X=-1/2 if the coin is tails. Y=1/2 if the same coin is heads, Y=-1/2 if the coin is tails. E(X)=E(Y)=0 but XY=1/4 regardless of whether the coin is heads or tails, so E(XY)=1/4
thanks! That was a neat efficient proof:)

Thanks both!

EDIT: I did unfortunately run into an issue for my self here. How can I show that the quoted example above is not independent?
 Office_Shredder, correct me if I'm wrong, but I think in this example you'll want to go about it like this: In general, the way you show independence is that P[X=x and Y=y] = P[X=x]*P[Y=y]. If you do P[X=1/2 and Y= -1/2], that probability is 0, because the same coin cannot be heads-up and tails-up at once, but P[X=1/2]*P[Y= -1/2] = 1/2 probability of heads*1/2 probability of tails = 1/4. Hence they aren't independent, because 0 != 1/4.