Given $(X,Y)$ bivariate normal, find a linear combination independent of $Y$

Question

I have a question about the answer to this question. At the start of the proof we assume that $$\frac{X-\mu_1}{\sigma_1} = \rho \frac{Y-\mu_2}{\sigma_2} + \sqrt{1-\rho^2} Z \tag{$*$}$$ But I can't figure out where this assumption comes from and how it is valid.

Answer 1

The claim is:

If $(X,Y)$ have bivariate normal distribution, where $X\sim N(\mu_1,\sigma_1^2)$ and $Y\sim N(\mu_2,\sigma_2^2)$ with correlation $\rho$, then $$ Z:=\frac1{\sqrt{1-\rho^2}}\left(\frac{X-\mu_1}{\sigma_1}\right)-\frac\rho{\sqrt{1-\rho^2}}\left(\frac{Y-\mu_2}{\sigma_2}\right) $$ has standard normal distribution and is independent of $Y$.

To prove this, write $X':=\frac{X-\mu_1}{\sigma_1}$ and $Y':=\frac{Y-\mu_2}{\sigma_2}$, so $X'$ and $Y'$ each have standard normal distribution, and their covariance is $\rho$. Define $$Z:=aX'+bY'$$ for constants $a,b$ to be determined. Then $(Z,Y')$ are jointly normal since we can write $$ \begin{pmatrix}Z\\Y'\end{pmatrix}=\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}X'\\Y'\end{pmatrix} $$ as a linear transformation of the jointly normal $(X',Y')$. Check that the mean of $Z$ is zero and its variance is $$\operatorname{Var}(Z)= \operatorname{Var}(aX'+bY')=a^2+b^2+2ab\rho.\tag a$$ Moreover, the covariance between $Z$ and $Y'$ is $$\operatorname{Cov}(Z,Y')= \operatorname{Cov}(aX'+bY',Y')=a\operatorname{Cov}(X',Y')+b\operatorname{Var}(Y')=a\rho+b.\tag b $$ We now find $a$ and $b$ such that (a) equals $1$ and (b) equals $0$ . Solving the equations $$a^2+b^2+2ab\rho=1\qquad{\rm and}\qquad a\rho+b=0$$ yields $$a=\frac1{\sqrt{1-\rho^2}}\qquad{\rm and}\qquad b=\frac{-\rho}{\sqrt{1-\rho^2}}.$$ With these choices for $a$, $b$ we have that $Z$ has standard normal distribution, and $Z$ is independent of $Y'$ (since zero covariance implies independence for jointly normal variables), and therefore $Z$ is independent of $Y$.