Here is a problem in my text (Hogg & McKean, Introduction to Mathematical Statistics 7th edition, problem 2.6.6):
Let $X_1, X_2,$ and $X_3$ be three random variables with means, variances, and correlation coefficients, denoted by $\mu_1 , \mu_2 , \mu_3$; $\sigma_1^2, \sigma_2^2, \sigma_3^2$; and $\rho_{12}, \rho_{13}, \rho_{23}$, respectively. For constants $b_2$ and $b_3$, suppose $E(X_1 - \mu_1 | x_2,x_3) = b_2(x_2 - \mu_2) + b_3(x_3 - \mu_3)$. Determine $b_2$ and $b_3$ in terms of the variances and the correlation coefficients.
My attempt is as follows: first, I wrote the conditional pdf of $X_1$ given $X_2, X_3$ as:
$f_{1|2,3}(x_2 , x_3) = \frac{f(x_1, x_2, x_3)}{f_{2,3}(x_2, x_3)}$
where $f$ is the joint pdf of $X_1, X_2, X_3$ and $f_{2,3}$ is the marginal pdf of $X_2, X_3$.
By definition we have:
$E(X_1 - \mu_1 | x_2,x_3) = \int_{-\infty}^{\infty} (x_1 - \mu_1) \frac{f(x_1, x_2, x_3)}{f_{2,3}(x_2,x_3)} dx_1 = \frac{E(X_1)}{f_{2,3}(x_2,x_3)} - E(X_1)\frac{f_{2,3}(x_2,x_3)}{f_{2,3}(x_2,x_3)} = \frac{\mu_1}{f_{2,3}(x_2,x_3)} - \mu_1$
With what we're given we have the equality:
$\frac{\mu_1}{f_{2,3}(x_2,x_3)} - \mu_1 = b_2(x_2 - \mu_2) + b_3(x_3 - \mu_3)$
Multiplying both sides by $f_{2,3}$:
$\mu_1 - f_{2,3}(x_2, x_3)\mu_1 = f_{2,3}(x_2, x_3)[b_2(x_2 - \mu_2) + b_3(x_3 - \mu_3)]$
I've tried several things from here but none of them seem to go anywhere.
What I'm trying to do is multiply and/or integrate both sides by $x_2$ or $x_3$ to kill off the $f_{2,3}$ and get everything in terms of the means, variances, and $\rho$s.
However this doesn't seem possible because of the constant $\mu_1$ term on the left hand side. If I integrate this term over all values of $x_2$ or $x_3$ it simply integrates to infinity.
I've also tried evaluating both sides of the expression at either $\mu_2$ or $\mu_3$ to kill off one of the $b_i$'s and solve for the other, but this also doesn't seem to lead anywhere because I'm left with the term $f_{2,3}(\mu_2, x_3)$ which I can't seem to get rid of.
Does anyone have any ideas/hints on how to proceed from here?
Without loss of generality we may assume $\mu_i=0$ for all $i=1,2,3.$ Denoting $$\Sigma_{ij}=\sigma_i\sigma_j\rho_{ij}$$ with $\rho_{ii}=1$ we compute $b_2,b_3$ defined by $$E(X_1|X_2,X_3)=b_2X_2+b_3X_3\ \ (*)$$ in terms of the covariance matrix $\Sigma=(\Sigma_{ij})$ as follows. Identity (*) implies that for all numbers $x_2, x_3$ we have
$$0=E((X_1-b_2X_2-b_3X_3)(x_2X_2+x_3X_3)=A_2x_2+A_3x_3\ \ (**)$$ where $$ A_2=\Sigma_{12}-b_2\sigma_2^2-b_3\Sigma_{23},\ A_3=\Sigma_{13}-b_2\Sigma_{23}-b_3\sigma_3^2.$$ Since (**) is true for all $x_2,x_3$ this implies that $A_2=A_3=0$ which is a linear system giving the desired values of $b_2$ and $b_3.$