Let X and Y be discrete random variables with joint probability function p(x,y) given by the following table:

X\Y | 0 _ 1___ p(x)

0 __| 0 _0.2__ 0.2

1 __| 0.4 0.2 __0.6

2 __| 0.2 0 ___0.2

p(y)| 0.6 0.4 ___1

Find the Variance of Y - X.

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Hi ! I've been staring at this for a over an hour. It shouldn't be hard but I don't see my mistake...These are my calculations:

E(X) = 1(0.6) + 2(0.4) = 1

E(X^2) = 1.4

Var(X) = 0.2

E[Y] = 0.4

E[Y^2] = 0.4

Var[Y] = 0.24

E[XY] = 1(1)(0.2) = 0.2

Cov(X,Y) = 0.152

So Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y) = 0.136

Where did I go wrong? Thanks !!!

(ps: sry for the poor table formatting)

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Answer is 1.04