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Whe know that $\mathrm{Cov}(X,Y)=E(X.Y)-E(X).E(Y)$ so if we consider this equation what happened if $X$ & $Y$ being conditional random variable $\mathrm{Cov}((X|Z_i),(Y|Z_j))=?$

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same thing. Just replace $X$ and $Y$ by the conditional ones since this is the definition. –  Patrick Li Oct 31 '12 at 16:48
There is no such thing as a conditional random variable. What you are asking for is the conditional covariance of $X$ and $Y$ given $Z$, that is, $$\text{Cov}(X,Y \mid Z) = E[XY\mid Z] - E[X\mid Z]E[Y\mid Z].$$ You will need the conditional joint distribution of $(X,Y)$ given $Z$ to calculate the first term on the right. –  Dilip Sarwate Oct 31 '12 at 18:39
but there are two different z –  d.soleimani Oct 31 '12 at 19:44
OK, I didn't see that, but your notation makes no sense. There isn't such a thing as a conditional random variable. What you get is $$\text{Cov}(X,Y \mid Z_i, Z_j) = E[XY\mid Z_i, Z_j] - E[X\mid Z_i, Z_j]E[Y\mid Z_i, Z_j].$$ –  Dilip Sarwate Nov 1 '12 at 1:54
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