Is it possible to compute $P(X\mid Y,Z)$ by calculating $P(X\mid Y)$ given the probability $P(\cdot\mid Z)$? Similarly, is it possible to get at the density $f_{X\mid Y,Z}$ by calculating the desity $f_{X\mid Y}$ given $P(\cdot\mid Z)$?
More precisely, let $(\Omega,\mathcal{A},P)$ be a probability space and let $X,Y,Z$ be random variables. Consider the conditional probability induced on $\mathcal{A}$ by conditioning on $Z$: $P(\cdot\mid Z=z)$. Suppose for each $z$ we calculate the conditional distribution $P(X\mid Y=y)$ in the modifed probability space $(\Omega,\mathcal{A},P(\cdot\mid Z))$. Is the resulting function of $(y,z)$ equal to the conditional distribution $P(X\mid Y=y, Z=z)$?
Suppose for each $z$ the conditional density $f_{X\mid Y}$ exists given the modified probability space described above. Is the resulting function of $(y,z)$ equal to the conditional density $f_{X\mid Y,Z}$?
Best Answer
For every $z$, let $Q_z=P(\ \mid Z=z)$, your question is whether for every $(x,y)$, $$ Q_z(X=x\mid Y=y)=P(X=x\mid Y=y,Z=z). $$ The answer is "yes", since, by definition, $$ Q_z(X=x\mid Y=y)=\frac{Q_z(X=x,Y=y)}{Q_z(Y=y)}=\frac{P(X=x,Y=y\mid Z=z)}{P(Y=y\mid Z=z)}, $$ that is, $$ Q_z(X=x\mid Y=y)=\frac{P(X=x,Y=y, Z=z)}{P(Y=y, Z=z)}=P(X=x\mid Y=y,Z=z). $$ In particular, if, for every $z$, $f_{X\mid Y}^{(z)}$ is the density of $X$ conditionally on $Y$ with respect to $Q_z$, then $f_{X\mid Y}^{(Z)}$ is the density of $X$ conditionally on $(Y,Z)$.