Consider some discrete real valued random variables $X,Y,Z$.
Is it correct to say that $$ Pr(X=x| Y=y \text{ or } Y=\tilde{y}, Z=z)\geq Pr(X=x| Y=y, Z=z) $$ ?
I though it makes sense because on the lhs we are conditioning on a "larger" world. But then when I try to prove this, it seems to fail: $$ Pr(X=x| Y=y \text{ or } Y=\tilde{y}, Z=z)=\frac{Pr(X=x, Y=y \text{ or } Y=\tilde{y}| Z=z)}{Pr(Y=y \text{ or } Y=\tilde{y}| Z=z)} $$ and $$ Pr(X=x| Y=y, Z=z)=\frac{Pr(X=x, Y=y | Z=z)}{Pr(Y=y | Z=z)} $$ We have that $$ Pr(X=x, Y=y \text{ or } Y=\tilde{y}| Z=z)\geq Pr(X=x, Y=y | Z=z) $$ but also $$ Pr(Y=y \text{ or } Y=\tilde{y}| Z=z)\geq Pr(Y=y | Z=z) $$ Could you help me to understand?
Suppose $X = Y, x = y \neq \tilde{y},$ and $Z$ takes only one value $z$ (so we can ignore it). Also assume none of these have $0$ probability of occurring. Then:
$$Pr(X = x | Y = y, Z = z) = Pr(X = x | X = x) = 1$$
However, if $X = \tilde{y}$ with any probability then:
$$Pr(X = x | Y = y \text{ or } Y = \tilde{y}, Z=z) = Pr(X = x | X = x \text{ or } X = \tilde{y} \neq x) < 1$$
So the inequality doesn't hold in general. Intuitively, it could be that the possibility of $Y$ being $\tilde{y}$ could correlate strongly with $X \neq x,$ so including that possibility drops the probability.