I'm reading through Ross's A First Course in Probability , and am having some difficulty understanding a step within the derivation of the conditional bivariate normal.

$$f(x, y) = \frac{1}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}\exp{[-\frac{1}{2(1-\rho^2)}[(\frac{x-\mu_x}{\sigma_x})^2+(\frac{y-\mu_y}{\sigma_y})^2 -2\rho\frac{(x-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y}]] }$$

Their approach to finding $f(x|y)$ is to collect any terms in the conditional distribution that are not a function of $x$, leading to,

$$f(x|y)=\frac{f(x,y)}{f(y)}$$ $$=C_1 f(x,y)$$ $$=C_2 \exp{[-\frac{1}{2(1-\rho^2)}[(\frac{x-\mu_x}{\sigma_x})^2-2\rho\frac{x(y-\mu_y)}{\sigma_x\sigma_y}]] }$$

On the rightmost term $2\rho\frac{x(y-\mu_y)}{\sigma_x\sigma_y}$, I don't get how the $(x-\mu_x)$ term of the original distribution simply become $x$. As best I can tell there's no assumption that $\mu_x=0$. Can anyone help me understand this?

Full image for reference: Screenshot


Presumably you would have been happy with

$$=C_2 \exp{\left[-\frac{1}{2(1-\rho^2)}\left[\left(\frac{x-\mu_x}{\sigma_x}\right)^2-2\rho\frac{(x-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y}\right]\right] }$$

which can be written as

$$=C_2 \exp{\left[-\frac{1}{2(1-\rho^2)}\left[\left(\frac{x-\mu_x}{\sigma_x}\right)^2-2\rho\frac{x(y-\mu_y)}{\sigma_x\sigma_y}+2\rho\frac{\mu_x(y-\mu_y)}{\sigma_x\sigma_y}\right]\right] }$$

and here the $2\rho\frac{\mu_x(y-\mu_y)}{\sigma_x\sigma_y}$ term can be subsumed into the original multiplicative constant


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