Consider 3 random variables $X$, $Y$ and $Z$. Under which conditions would we have $P(X\mid Y,Z) = P(X\mid Z)$?

  • $\begingroup$ Your title asks about "dropping" the condition, regardless of the rest of the question, you can drop it by $P(X|Z) = \sum_y P(X|Y = y,Z) P(Y=y)$ by the law of total probability. $\endgroup$
    – Tim
    Commented Oct 6, 2023 at 6:37

1 Answer 1


This is equivalent to $X$ and $Y$ being conditionally independent given $Z=z$. We have $$ f_{X|Y,Z}(x|y,z) = \frac{f_{X,Y,Z}(x,y,z)}{f_{Y,Z}(y,z)} =\frac{f_{X,Y|Z}(x,y|z)f_Z(z)}{f_{Y|Z}(y|z)f_Z(z)} =\frac{f_{X,Y|Z}(x,y|z)}{f_{Y|Z}(y|z)} $$

We want this to be equal to $f_{X|Z}(x|z)$, which is equivalent to $$ f_{X,Y|Z}(x,y|z) =f_{X|Z}(x|z) f_{Y|Z}(y|z) \,, $$ i.e. conditional independence.

  • $\begingroup$ Thanks. Yes, that is clear. Is there some condition on pairwise relations between $X$, $Y$, $Z$ implying conditional independence of $X$ and $Y$ given $Z=z$? $\endgroup$
    – Nidjsi
    Commented Oct 6, 2023 at 7:29
  • $\begingroup$ You can often spot this from how the random variables are defined. For example, you could have $X,Y | Z=z \sim \mathrm{Poisson}(z)$, where $Z \sim \mathrm{Gamma}(\alpha, \beta)$. $X$ and $Y$ are not independent, because their distributions depend on the common (stochastic) parameter $Z$, but they are conditionally independent for a fixed $Z=z$. $\endgroup$ Commented Oct 6, 2023 at 10:52

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