Let \(X\) and \(Y\) be independent, each having the standard normal distribution, and let \((R,\Theta )\) be the polar coordinates for \((X,Y)\).
(a) Show that \(X+Y\) and \(X-Y\) are independent, and that \(R^2=[(X+Y)^2+(X-Y)^2]/2\), and conclude that the conditional distribution of \(R^2\) given \(X-Y\) is the chi-squared with one degree of freedom translated by \((X-Y)^2/2\).
(b) Show that the conditional distribution of \(R^2\) given \(\Theta \) is chi-squared with two degrees of freedom.
(c) If \(X-Y=0\), the conditional distribution of \(R^2\) is chi-squared with one degree of freedom. If \(\Theta =\pi /4\) or \(\Theta =5\pi /4\), the conditional distribution of \(R^2\) is chi-squared with two degrees of freedom. But the events \(\{X-Y=0\}\) and \(\{\Theta =\pi /4\}\cup \{\Theta =5\pi /4\}\) are the same. Resolve the apparent contradiction.