Condition Expectation of normal variables

Question

Let X,Y be jointly normal. Then I know that $E(X|Y)=E(X)+\frac{Cov(X,Y)}{V(Y)}(Y-E(Y))$.

Do I need joint normality for this result? Does it also hold, if just X is normal and Y is normal?

Answer 1

Counter example: Let $Y\sim\mathcal{N}(0,1)$. Define the random variable $X$ as follows: \begin{equation} X = \begin{cases} Y, & \text{if } |Y| \geq 1\\ -Y, & \text{if } |Y| < 1. \end{cases}. \end{equation} Note that by symmetry, $-Y$ is also $\mathcal{N}(0,1)$. Then, we compute the cdf of $X$ as follows: \begin{align} \mathbb{P}\{X\leq x\} &= \mathbb{P}\{X\leq x, |Y|\geq 1\} + \mathbb{P}\{X\leq x, |Y|< 1\} \\ &= \mathbb{P}\{Y\leq x, |Y|\geq 1\} + \mathbb{P}\{-Y\leq x, |Y|< 1\} \\ &= \mathbb{P}\{Y\leq x, |Y|\geq 1\} + \mathbb{P}\{Y\leq x, |Y|< 1\} \\ &= \mathbb{P}\{Y\leq x\}. \end{align} Thus, $X\sim\mathcal{N}(0,1)$. Next, computing the conditional expectation, we get \begin{align} \mathbb{E}\{X|Y\} &= Y\delta_{ |Y|\geq 1} - Y\delta_{ |Y| < 1}, \end{align} where $\delta$ denote the indicator function. Further, the RHS gives \begin{equation} \mathbb{E}\left\{X\right\}+\frac{\text{Cov}\{X,Y\} }{\text{Var}\{Y\} } \left(Y-\mathbb{E}\{Y\}\right) = \text{Cov}\{X,Y\} Y. \end{equation} Since $\text{Cov}\{X,Y\} $ is a real number independent of $X$ and $Y$, we get that LHS$\neq$RHS.