 Solution

The assertion is false for the algorithm as printed. Its second output depends only on $V$. For $0<V\le 1/2$ its sine factor is nonnegative. For $1/2<V<1$, its radius is at most $\sqrt{2\log 2}$, so in every case $Y\ge -\sqrt{2\log 2}$. Therefore $\mathrm{Pr}(Y<-2)=0$, whereas a standard normal variable has positive probability below $-2$. The printed output cannot be a pair of independent standard normals.

The minimal correction is to use $U$ in both logarithms. Let $R=\sqrt{-2\log U}$, $\Theta =2\pi V$, and set $X=R\cos \Theta$, $Y=R\sin \Theta$. Independence of $U,V$ gives independent $R,\Theta$ with densities $r{e}^{-r^2/2}$ for $r>0$ and $1/(2\pi )$ for the angle. The polar-coordinate Jacobian is $r$, so the joint Cartesian density isThis proves independent standard normal outputs for the corrected Box-Muller transform. The private transcription retains the source's original $\log V$ and records this correction separately.