- #1
Scootertaj
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1. Consider the random variables X,Y where X~B(1,p) and
f(y|x=0) = 1/2 0<y<2
f(y|x=1) = 1 0<y<1
Find cov(x,y)
Cov(x,y) = E(XY) - E(X)E(Y) = E[(x-E(x))(y-E(y))]
E(XY)=E[XE(Y|X)]
E(X) = p (known since it's Bernoulli, can also be proven
E(Y) = [itex]\int Y*1/2 dy[/itex] 0 to 2 + [itex]\int Y*1[/itex] 0 to 1 = 3/2
I'm not sure E(Y) is right.
If this is right, I still don't know how to solve E(XY).
Could we do cov(x,y) =∫ ∫(x-p)(y-3/2)dxdy from 0 to 1, 0 to 2 ?
Thoughts?
f(y|x=0) = 1/2 0<y<2
f(y|x=1) = 1 0<y<1
Find cov(x,y)
Homework Equations
Cov(x,y) = E(XY) - E(X)E(Y) = E[(x-E(x))(y-E(y))]
E(XY)=E[XE(Y|X)]
The Attempt at a Solution
E(X) = p (known since it's Bernoulli, can also be proven
E(Y) = [itex]\int Y*1/2 dy[/itex] 0 to 2 + [itex]\int Y*1[/itex] 0 to 1 = 3/2
I'm not sure E(Y) is right.
If this is right, I still don't know how to solve E(XY).
Could we do cov(x,y) =∫ ∫(x-p)(y-3/2)dxdy from 0 to 1, 0 to 2 ?
Thoughts?