Suppose there are two random variables $\theta,Y$, which are dependent. Moreover, I observe the value $f(\theta)=f$ for some known function $f(.)$ (which need not be invertible). I want to write the conditional probability $Pr(\theta=t|f(\theta)=f,Y=y)$ as a function of $Pr(Y=y|\theta=t)$.
Okay stop me when there is a mistake. Using Bayes' rule $$Pr(\theta=t|f(\theta)=f,Y=y)=\frac{Pr(\theta=t)Pr(f(\theta)=f,Y=y|\theta=t)}{Pr(f(\theta)=f,Y=y)}.$$ My thinking is that we cannot simplify to $$Pr(\theta=t|f(\theta)=f,Y=y)=\frac{Pr(\theta=t)Pr(f(\theta)=f|\theta=t)Pr(Y=y|\theta=t)}{Pr(f(\theta)=f)Pr(Y=y)}$$ since $f(\theta)$ and $y$ are not necessarily independent.
Is there another way to rewrite the "conditional joint probability" $Pr(f(\theta)=f,Y=y|\theta=t)$ as a function of $Pr(Y=y|\theta=t)$ using some law of iterated expectations, since I know $f(\theta)$ if given $\theta=t$?
Otherwise I can't straightforwadly apply the definition of joint probability here, since there is still a conditioning set. My attempt is $$Pr(f(\theta)=f,Y=y|\theta=t)=E[Pr(f(\theta)=f,Y=y|\theta=t)|\theta=t]=E[Pr(f(\theta)=f,Y=y)|\theta=t]=E[Pr(Y=y|f(\theta)=f)Pr(f(\theta)=f)|\theta=t]=1\{f(\theta=t)=f\}E[Pr(Y=y|f(\theta)=f)|\theta=t]=1\{f(\theta=t)=f\}Pr(Y=y|\theta=t).$$ Is this correct? If not, is there another way to isolate $Pr(Y=y|\theta=t)$?
Many thanks in advance!