= Solution
{c}
For a <joint probability density function>, <independence> is equivalent to $f(x,y)=f_X(x)f_Y(y)$ almost everywhere. The displayed density is valid for $|\rho|<1$. Complete the square:
$$
\frac{x^2-2\rho xy+y^2}{1-\rho^2}
=x^2+\frac{(y-\rho x)^2}{1-\rho^2}.
$$
Integrating the second normal factor over $y$ gives
$$
\boxed{f_X(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}.}
$$
Dividing the <joint probability density function> by this marginal yields the conditional law
$$
\boxed{Y\mid X=x\sim N(\rho x,1-\rho^2).}
$$
For $\rho=0$ the <joint probability density function> factors and the variables are independent. If $\rho\neq0$, the <conditional mean> depends on $x$, so <independence> fails; equivalently their <covariance> is $\rho$.
Define $Z=(Y-\rho X)/\sqrt{1-\rho^2}$. In the change of variables $y=\rho x+\sqrt{1-\rho^2}z$, the Jacobian $\sqrt{1-\rho^2}$ cancels the remaining density factor, leaving the product of two <standard normal> densities. Thus $Z$ is <standard normal> and independent of $X$. One choice is
$$
\boxed{a=-\frac{\rho}{\sqrt{1-\rho^2}},\qquad
b=\frac1{\sqrt{1-\rho^2}}.}
$$
Changing both signs gives the other standardized choice. The density transformation proves <independence> directly rather than inferring it from zero <covariance> without a joint-normal justification.
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