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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.2012-10-31
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    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.2012-10-31
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    but there are two different z2012-10-31
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    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 $$\text{Cov}(X,Y \mid Z_i, Z_j) = E[XY\mid Z_i, Z_j] - E[X\mid Z_i, Z_j]E[Y\mid Z_i, Z_j].$$2012-11-01

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