Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / S / S2

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.

For the independent standard normals stipulated in this part, simulate $n$ values and average the exceedance indicators:Each indicator is Bernoulli with probability $z_c$, so expectation and independence giveThus the Monte Carlo estimator is unbiased.

Let $\varphi$ denote the standard normal density. Under the shifted sampling density $q(y)=\varphi (y-c)$, the likelihood ratio is $\varphi (y)/q(y)=e^{-cy+c^2/2}$. Simulate independent $Y_k\sim N(c,1)$ and useMultiplying the weight by $q$ restores the original density, so each summand $H$ has expectation ${\int }_c^{\infty }\varphi (y)dy=z_c$. The estimator is unbiased. For the second moment,because $y\ge c$. Since $n{v}_c={\mathbb{E}}_q{H}^2-z_c^2$,For $c>0$, $e^{-c^2/2}<1$ and $z_c>0$, so $n{v}_c<z_c-z_c^2=n{v}_0$, proving $v_c<v_0$. Nonnegativity of variance also implies $z_c^2\le e^{-c^2/2}{z}_c$; division by the positive $z_c$ gives