Suppose $X,Y,Z$ are random variables such that $Y,X$ are perfectly correlated. Does it hold that $P(Z|X,Y) = P(Z|Y)$? Suppose $X,Y,Z$ are random variables such that $Y,X$ are perfectly correlated. I am wondering if it holds that $P(Z|X,Y) = P(Z|Y)$? It would seem that $X$ would then be a substitute of $Y$. Is there a way to formally prove this?
 A: In short the answer is that yes this is true, but proving it requires some manipulation of the measure theoretic definitions of probability: largely because the definition of conditional independence is heavily steeped in measure theory.
Note that perfect correlation implies that almost surely $X = aY + b$ for some constants $a,b$.
Now we turn to the measure theory, recall that by Bayes formula
$$\mathbf P[ Z = z \, | X = ay + b , \, Y =y ] = \frac{ \mathbf P[Z = z, \, X = ay + b,\, Y = y ] }{\mathbf P[ X = ay + b,\, Y = y ]}.$$
Moreover, measure theoretically if the random variables $X,Y,Z \colon \Omega \rightarrow \mathbf R$ then the event $\{X = x\}$ etc. are defined as
$$ \{X = x\} = \{ \omega \in \Omega \, \colon \, X(\omega) = x\},$$
further since $X$ is almost surely equal to $aY + b$ we have that up to a sets of measure $0$
$$\{X = ay + b\} = \{Y = y\},$$
and hence their intersection is, up to sets of measure $0$
$$\{X = ay + b\} \cap \{Y = y\} = \{Y = y\}.$$
This can now be substituted into Bayes formula to derive the desired result
\begin{align*}\mathbf P[ Z = z \, | X = ay + b , \, Y =y ] &= \frac{ \mathbf P[Z = z, \, X = ay + b,\, Y = y ] }{\mathbf P[ X = ay + b,\, Y = y ]}\\
&=\frac{ \mathbf P[Z = z, \,  Y = y ] }{\mathbf P[ Y = y ]}\\
&=\mathbf P[ Z = z \, | \, Y =y ] .\end{align*}
Note that there is some level of detail omitted from this (largely around the sets of measure $0$ argument), but the essence is there. 
