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.
 A: In general, if $X, Z$ are random variables defined on the probability space $(\Omega, \mathscr{F}, P)$, the conditional distribution of $X$ given $Z$ is a function $\mu(H, \omega)$ defined on $\mathscr{R}^1 \times \Omega$, with the following two properties (cf. Theorem 33.3 in Probability and Measure by Patrick Billingsley):

*

*For each $\omega \in \Omega$, $\mu(\cdot, \omega)$ is a probability measure on $\mathscr{R}^1$.

*For each $H \in \mathscr{R}^1$, $\mu(H, \cdot)$ is a version of $P[X \in H | \sigma(Z)]$.

By the above definition, it is easy to verify that the conditional distribution of $Y$ given $aY + b$ is
\begin{align}
\mu(H, \omega) = I_{aH + b}(aY(\omega) + b), \tag{1}
\end{align}
where the set $aH + b := \{ax + b: x \in H\}$.
In more elementary notations, suppose $a > 0$, $(1)$ can be rewritten as
\begin{align}
P[Y \leq y | aY + b = z] = 
\begin{cases}
1 & \text{ if } a^{-1}(z - b) \leq y, \\
0 & \text{ otherwise.}
\end{cases}
\end{align}
The intuitive meaning has been noted in the comment:  once you observe the value of $aY + b$, then for any Borel set $H$, whether $Y$ belongs to $H$ is certain, which implies that the conditional probability of $[Y \in H]$ is either $1$ or $0$.
A: @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.
