Derive linear regression model from the conditional distribution of Y|X

Question

Suppose that $Y|X=x \sim N(\mu_Y + \frac{\sigma_Y(x-\mu_X)\rho}{\sigma_X}, \sigma_Y^2(1-\rho^2))$. The question asks to specify a simple regression model under this conditional distribution.

A simple regression model is that

$$y = b_0 + b_1 x + \epsilon.$$

I first thought that I have to specify $b_0$ and $b_1$. We have that $\mu_Y+\frac{\sigma_Y(x-\mu_X)\rho}{\sigma_X} = b_0 + b_1x +\epsilon$. However, we have one question but two unknowns, so I guess that this approach is wrong. I haven't come across this type of question. What does it mean exactly "specify a simple regression model"?

Answer 1

Assume you set the model $y=b_0+b_1x+\epsilon$ and it satisfies OLS assumptions. Then, $$E[Y]=b_0+b_1E[X]+E[\epsilon]=b_0+b_1\mu_X$$ $$E[Y|X=x]=b_0+b_1x+E[\epsilon|x]=b_0+b_1x$$ Substitute the first into the second and you'll have $b_1=\rho{\sigma_Y\over\sigma_X}, b_0=\mu_Y-b_1\mu_X$.

And, we verify the variance as follows: $$\operatorname{var}(Y|X=x)=\operatorname{var}(b_0+b_1X+\epsilon|X=x)=\operatorname{var}(\epsilon|X=x)=\sigma_Y^2 (1-\rho^2)$$ $$\operatorname{var}(Y)=b_1^2\operatorname{var}(X)+\operatorname{var}(\epsilon)=b_1^2\sigma_X^2+\sigma_Y^2(1-\rho^2)=\sigma_Y^2$$ which checks out. We also assumed no correlation among error and regressor terms, which is implied by Strict exogeneity assumption.