Conditional expectation of a Gaussian given a r.v.?

Question

Let $X$ be a standard Gaussian random variable and $B$ a standard Bernoulli random variable. Let $Y = B \cdot X$.

Computing $E(Y\mid X)$ is rather straightforward. I simply substitute $Y = B \cdot X$ and apply independence of $X$ and $B$, obtaining $\frac{X}{2}$ .

But how to compute $E(X\mid Y)$?

Answer 1

The answer is $E(X\mid Y)=Y$. Here are two solutions, depending on which level of probability theory you are studying.

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1^st Way. We know that $E(X\mid Y)=u(Y)$, where $u(y)=E(X\mid Y=y)$. So we only need to determine the function $u(y)$. But

1. If $y = 0$, then $$E(X\mid Y=0)=E(X\mid B=0\text{ or } X=0).$$ Since $P(B=0)=\frac{1}{2}$ and $P(X=0)=0$, we may neglect the contribution from the condition $X=0$ and write $$=E(X\mid B=0)=E(X)=0$$ where the second step follows from the independence.

2. If $y \neq 0$, then $$E(X\mid Y=y)=E(X\mid B=1\text{ and }X=y)=E(X\mid X=y)=y,$$ where the second step follows from the independence again.

Altogether, we get $u(y)=y$ and hence

$$E(X\mid Y)=Y.$$

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2^nd Way. We know that $E(X\mid Y)$ is the $P$-a.s. unique random variable $Z$ which is $\sigma(Y)$-measurable and satisfies

$$ E(X\mathbf{1}_A) = E(Z\mathbf{1}_A), \qquad \forall A \in \sigma(Y). \tag{*}$$

Now we claim that

$$E(X\mid Y)=Y, \qquad P\text{-a.s.}$$

This amounts to showing that $\text{(*)}$ holds with the choice $Z = Y$. Since the events of the form $\{Y\leq y\}$ generates $\sigma(Y)$, it suffices to check $\text{(*)}$ only for this form of events by the Monotone Class Theorem. Now,

1. If $y < 0$, then $\{Y \leq y\} = \{B=1, X\leq y\}$ and hence $$ X\mathbf{1}_{\{Y\leq y\}} = BX\mathbf{1}_{\{Y\leq y\}} = Y\mathbf{1}_{\{Y\leq y\}} .$$ This then shows that $$ E[X\mathbf{1}_{\{Y\leq y\}}] = E[Y\mathbf{1}_{\{Y\leq y\}}]. $$

2. If $y \geq 0$, then $\{Y \leq y\} = \{B=1, X\leq y\} \cup \{B = 0\}$, and so, $$ X\mathbf{1}_{\{Y\leq y\}} = X\mathbf{1}_{\{B=1, X\leq y\}} + X\mathbf{1}_{\{B = 0\}} .$$ Similarly as before, the first term can be treated as \begin{align*} X\mathbf{1}_{\{B=1, X\leq y\}} &= BX\mathbf{1}_{\{B=1, X\leq y\}} \\ &= Y\mathbf{1}_{\{B=1, X\leq y\}} \\ &= Y\mathbf{1}_{\{B=1, X\leq y\}} + Y\mathbf{1}_{\{B=0\}} \\ &= Y\mathbf{1}_{\{Y \leq y\}}, \end{align*} where the third step follows from $ Y\mathbf{1}_{\{B=0\}} = 0 $. In light of this, we get \begin{align*} E[X\mathbf{1}_{\{Y\leq y\}}] &= E[Y\mathbf{1}_{\{Y \leq y\}}] + E[X\mathbf{1}_{\{B=0\}}] \\ &= E[Y\mathbf{1}_{\{Y \leq y\}}] + E[X]P(B = 0) \\ &= E[Y\mathbf{1}_{\{Y \leq y\}}]. \end{align*}

Therefore the desired conclusion follows.