ThmDex – An index of mathematical definitions, results, and conjectures.
Conditional expectation need not preserve equality in distribution
Formulation 0
Let $X, Y \in \text{Bernoulli}(1/2)$ each be a D3999: Standard Bernoulli random boolean number such that
(i) $X, Y$ is an D2713: Independent random collection
(iii) \begin{equation} Z : = X \end{equation}
Then \begin{equation} \mathbb{E}(X \mid Z) \overset{d}{\neq} \mathbb{E}(Y \mid Z) \end{equation}
Proofs
Proof 0
Let $X, Y \in \text{Bernoulli}(1/2)$ each be a D3999: Standard Bernoulli random boolean number such that
(i) $X, Y$ is an D2713: Independent random collection
(iii) \begin{equation} Z : = X \end{equation}
Applying result R5324: Conditional expectation of a random real number conditioned on itself, we have \begin{equation} \mathbb{E}(X \mid Z) = \mathbb{E}(X \mid X) \overset{a.s.}{=} X \end{equation} while from the results
(i) R5325: Conditional expectation of random real number conditioned on an independent random real number
(ii) R5283: Expectation of a standard bernoulli random boolean number

we know that \begin{equation} \mathbb{E}(Y \mid Z) = \mathbb{E}(Y \mid X) \overset{a.s.}{=} \mathbb{E} Y = \frac{1}{2} \end{equation} Clearly, $X \overset{d}{\neq} 1 / 2$. $\square$