I am trying to establish a relationship between the following conditional expectation and random variable based on the a given identity:
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Let $X,Y \in \{0,\dots, n\}, n \in \mathbb{N}$ and $Z \in [0,1]$ be random variables on said space. Suppose it holds that \begin{align} \tag{1} \label{1} P(X=Y \mid Z = z) = z \quad \forall z \in [0,1] \end{align} What is the relationship between $\mathbb{E}[1_{X=Y} \mid Z]$ and $Z$?
I can show this implies equality in distribution. Let $W = \mathbb{E}[1_{X=Y} \mid Z]$, then for $w \in [0,1]$ \begin{align} P(W \leq w) = \int_0^w \mathbb{E}[1_{X=Y} \mid Z = z] \,f(z) \,dz \overset{\ref{1}}{=} \int_0^w z \,f(z) \,dz = P(Z \leq z) \end{align} My questions are:
- Is the other direction also true, i.e. \ref{1} $\Leftrightarrow \mathbb{E}[1_{X=Y} \mid Z] \overset{P}{=} Z$?
- Does \ref{1} $\Leftrightarrow \mathbb{E}[1_{X=Y} \mid Z] \overset{a.s.}{=} Z$ hold?
The answer on 2. is: "yes" (hence also the answer of 1. is: "yes").
$$\mathbb E\mathsf1_{X=Y}=P(X=Y)\tag2$$ and according to the same principle: $$\mathbb E[\mathsf1_{X=Y}\mid Z=z]=P(X=Y\mid Z=z)\text{ for all }z\in[0,1]\tag3$$
So $(1)$ in your question can be translated into: $$\mathbb E[\mathsf1_{X=Y}\mid Z=z]=z\text{ for all }z\in[0,1]\tag4$$ which on its turn comes to the same as:$$\mathbb E[\mathsf1_{X=Y}\mid Z)=Z\tag5$$