The distribution of $Y|aY+b$ What is the distribution of $Y|aY+b$?  I'm assuming that $a\neq 0$.  A related question here discusses the distribution of $X|X$, which is degenerate.  My guess is that the distribution of $Y|aY+b$ will also be degenerate but I'm not entirely sure how to work that out rigorously.  My thought was to try working with the definition
$$
f_{Y|aY+b}=\frac{f_{Y,aY+b}}{f_{aY+b}},
$$
but the joint distribution in the numerator doesn't make sense (I'm guessing it's degenerate).
Context:
I am working a problem where I have two random variables $Y$ and $X=a Y+b+\epsilon$, where $\epsilon\sim\mathcal N(0,\sigma^2)$.  I would like to compute the Fisher information matrix for $X$, which is too complicated; however, the fisher information matrix for $(X,Y)$ is easy to compute.  Furthermore, we have the relation
$$
I_X(\theta)=I_{X,Y}(\theta)-I_{Y|X}(\theta).
$$
If $\sigma$ is small it seems that (through Monte Carlo simulation)
$$
I_X(\theta)\sim I_{X,Y}(\theta).
$$
If I could show that $I_{Y|X}(\theta)\to\mathbf 0$ as $\sigma\to 0$ then this would explain by the approximation makes sense. My thought was to show that $I_{Y|aY+b}(\theta)=\mathbf 0$ but this requires (I think) the distribution of $Y|aY+b$, which I am not sure how to derive.

@Zhanxiong's answer got me thinking about this more, which led to a less rigorous but intuitive solution to the problem.  What caused the issue was the fact that in this question $Y$ is conditioned on a function of itself. However, note that the random variable $Y|aY+b=z$ can, after defining $Z=aY+b$, be written as $Y|aY+b=z\overset{d}{=}(Z-b)/a|Z=z$.  In this form it becomes obvious that this random variable is degenerate since
$$
\mathsf E(Y|aY+b=z)=\mathsf E((Z-b)/a|Z=z)=(z-b)/a
$$
and
$$
\mathsf{Var}(Y|aY+b=z)=\mathsf{Var}((Z-b)/a|Z=z)=0.
$$
So,
$$
Y|aY+b=z\sim\delta(y-(z-b)/a),
$$
which agrees with @Zhanxiong's answer.
