Conditional expectation of a bivariate normal distribution

Question

Can you help me?

Find the conditional expectation $\mathbb{E}[X|Y]$ if $(X,Y)$ possesses a bivariate normal distribution.

Is $\mathbb{E}[X|Y=y]=\mu_X+\sigma_X\rho(\frac{\displaystyle y-\mu_Y}{\displaystyle \sigma_Y})$ the solution?

My question: Is the same $\mathbb{E}[X|Y=y]$ and $\mathbb{E}[X|Y]$?

Answer 1

1. Sort ... of

$\mathbb{E}[X|Y=y]=\mu_X+\sigma_X\rho(\frac{\displaystyle y-\mu_Y}{\displaystyle \sigma_Y})$

but

$\mathbb{E}[X|Y]=\mu_X+\sigma_X\rho(\frac{\displaystyle \color{red}{Y}-\mu_Y}{\displaystyle \sigma_Y})$

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2. No.

$E[X|Y=y]$ is a number since $(Y=y)$ is an event. If $\mathbb P(Y=y) > 0$, it is defined as

$$E[X|Y=y] = \frac{1}{\mathbb P(Y=y)}\int_{Y=y} X d \mathbb P$$

O/w, if Y is a continuous random variable,

$$E[X|Y=y] = \int_{\mathbb R} x f_{X|Y}(x|y) dx$$

where

$$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$

assuming $f_{X|Y}(x|y), f_{X,Y}(x,y)$ and $f_Y(y)$ exist.

For a continuous random variable that is not absolutely continuous (pdf doesn't exist), I have no idea.

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$\mathbb{E}[X|Y]$ aka $\mathbb{E}[X|\sigma(Y)]$ is a random variable that depends on the value of $Y$ . It is defined as the random variable $Z$ s.t.

1. $Z$ is $Y$-measurable or $\sigma(Z) := \sigma(\mathbb{E}[X|Y]) \subseteq \sigma(Y)$

2. $\forall G \in \sigma(Y)$,

$$\int_G Z d \mathbb P := \int_G \mathbb{E}[X|Y] d \mathbb P = \int_G X d \mathbb P$$

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Say $Y$ takes values in $\mathbb N$.

Then $\sigma(Y) = \sigma((Y=1),(Y=2),(Y=3), ...)$ and

$$\mathbb{E}[X|Y] = \mathbb{E}[X|Y=1]1_{Y=1} + \mathbb{E}[X|Y=2]1_{Y=2} + \cdots = \sum_{i=1}^{\infty} \mathbb{E}[X|Y=i]1_{Y=i}$$

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2 implies that:

$\forall y \in \mathbb{N}$,

$$\int_{Y=y} Z d \mathbb P := \int_{Y=y} \mathbb{E}[X|Y] d \mathbb P = \int_{Y=y} X d \mathbb P$$

Hopefully, it is clear that in the above case and in general, if $Y=y$, then $E[X|Y=y] = \mathbb{E}[X|Y]$.