I tried solving exercise 12.6 in the second edition of the book probability essentials by Jacod & Protter. The exercise is as follows
$12.6$ Let $X, Y$ have finite variances and let $$ Z=\left(\frac{1}{\sigma_{Y}}\right) Y-\left(\frac{\rho_{X, Y}}{\sigma_{X}}\right) X $$ Show that $\sigma_{Z}^{2}=1-\rho_{X, Y}^{2}$, and deduce that if $\rho_{X, Y}=\pm 1$, then $Y$ is a non-constant affine function of $X$.
In my opinion the variance won't work out as $\sigma_{Z}^{2}=1-\rho_{X, Y}^{2}$ but instead will be $\sigma_{Z}^{2}=1-3\rho_{X, Y}^{2}$. I found a solution on the internet which in my opinion also shows that $\sigma_{Z}^{2}=1-3\rho_{X, Y}^{2}$ is the true value, but in this solution they also claim that $\sigma_{Z}^{2}=1-\rho_{X, Y}^{2}$. Here is the part of the solution which is wrong in my opinion.
$$\text { Let } Z=\left(\frac{1}{\sigma_{Y}}\right) Y-\left(\frac{\rho_{X Y}}{\sigma_{X}}\right) X$$ $$\Rightarrow \sigma_{Z}^{2}=\left(\frac{1}{\sigma_{Y}^{2}}\right) \sigma_{Y}^{2}-\left(\frac{\rho_{X Y}^{2}}{\sigma_{X}^{2}}\right) \sigma_{X}^{2}-2\left(\frac{\rho_{X Y}}{\sigma_{X} \sigma_{Y}}\right) \operatorname{Cov}(X, Y). $$
But $\rho_{XY}$=$\dfrac{\text{Cov(X,Y)}}{\sigma{(X)}\sigma{(Y)}}$ and therefore $$2\left(\frac{\rho_{X Y}}{\sigma_{X} \sigma_{Y}}\right) \operatorname{Cov}(X, Y) = \rho_{XY}^2$$
Now, am I not getting something really basic or what's going on ?
Hint: you have a small mistake in how you compute the variance of $Z$. It should be:
$$ \sigma_{Z}^{2}=\left(\frac{1}{\sigma_{Y}^{2}}\right) \sigma_{Y}^{2}+\left(\frac{\rho_{X Y}^{2}}{\sigma_{X}^{2}}\right) \sigma_{X}^{2}-2\left(\frac{\rho_{X Y}}{\sigma_{X} \sigma_{Y}}\right) \operatorname{Cov}(X, Y).$$