The law of iterated expectations states the following:
$$\text{E}\left[ \text{E} \left[ \mathbb{X} | \mathbb{Y} \right] \right] = \text{E}\left[ \mathbb{X} \right]$$
Given a bivariate Gaussian (normal) joint pdf with means $\mu_1$ and $\mu_2$, variances $\sigma_1^2$ and $\sigma_2^2$ and correlation coefficient $\rho$
$$\text{E}[\mathbb{X}|\mathbb{Y} = y] = \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(y – \mu_2)$$
Does this concur with:
$$\text{E}\left[ \text{E} \left[ \mathbb{X} | \mathbb{Y} \right] \right] = \text{E}\left[ \mathbb{X} \right]$$
Attempted Solution:
Since $\mu_1$, $\mu_2$, $\sigma_1$, $\sigma_2$, and $\rho$ are all numbers and $\text{E}[\bullet]$ is a linear operator, I did the following:
$$\text{E}\left[ \text{E} \left[ \mathbb{X} | \mathbb{Y} \right] \right] = \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(\text{E}[y] – \mu_2)$$
and
$$\text{E}[\mathbb{Y}] = \mu_2$$
so the above reduces to:
$$\text{E}\left[ \text{E} \left[ \mathbb{X} | \mathbb{Y} \right] \right] = \mu_1 $$
Question:
Am I allowed to assume that $\text{E}[\mathbb{Y}] = \mu_2$ by definition of a bivariate normal? Or do I have to prove it?
Thanks in advance!
Best Answer
$E(X|Y=y)=g(y)$ is a number but $E(X|Y)=g(Y)$ is a random variable.
by definition you get: $$g(y)=E(X|Y=y)=\mu_1 + \rho \frac{\sigma_1}{\sigma_2}(y - \mu_2)$$
so
$$g(Y)=E(X|Y)=\mu_1 + \rho \frac{\sigma_1}{\sigma_2}(Y - \mu_2)$$