An importance weight has expected value ${\mathbb{E}}_q[w(Z)]=\int p(z)dz=1$. From the ratio in (a),The supplied normal distribution transform applies with quadratic coefficient $-(1-s^{-1})$, giving each factor $(2s-1{)}^{-1/2}$. ThusFor $s>1$, the base $A:=s/\sqrt{2s-1}$ exceeds one because $s^2-(2s-1)=(s-1{)}^2>0$: importance weight variance increases exponentially with dimension. For $s=1$, the importance weight is identically one and its variance is zero. This calculation concerns the importance weight, rather than the weighted integrand.

The moment-generating function of a standard normal distribution gives $\mathbb{E}[e^{t{X}_j}]=e^{t^2/2}$. Using independence of the coordinates,The accepted draws are independent and identically distributed random variables, so their sample mean is an unbiased estimator and has varianceThis counts $n$ accepted draws, requiring $n{\sigma }^d$ proposals in expectation.

Write $s={\sigma }^2\ge 1$. The two multivariate normal distribution probability density functions have ratioHere $\Vert x\Vert$ is the Euclidean norm. Consequently the smallest rejection sampling envelope constant, and its acceptance probability, areThe supremum is attained at $x=0$; hence no smaller constant can dominate the ratio. Draw $Y\sim N(0,s{I}_d)$ and an independent $U$ from the uniform distribution on $(0,1)$. Accept $Y$ whenotherwise repeat with fresh independent random variables. One can generate $Y=\sqrt{s}G$ using a random vector $G$ with independent standard normal distribution coordinates.

For any measurable set $B$, the joint acceptance probability is ${\int }_{B}q(y)p(y)/(M_{\ast }q(y))dy=M_{\ast }^{-1}{\int }_{B}p(y)dy$. Dividing by $1/M_{\ast }$ proves that the returned random vector has probability density function $p$. The number of proposals has a geometric distribution on the positive integers, with expected value $M_{\ast }$. When $s=1$, every proposal is accepted.

With the apsidal origin chosen as specified, $G=0$. For $p_{\varphi }\ne 0$, rearrangement givesLet $u=1/r$. Since $du/d\varphi =-p_r/p_{\varphi }$, the equation is $u^{\prime }=-(u-mk/p_{\varphi }^2)\tan \varphi$. Its solution, continued through points where the intermediate division fails, isAs printed, $e$ has units of inverse length and is not the usual dimensionless eccentricity. Writing $\ell =p_{\varphi }^2/(mk)$ gives the standard form $r=\ell /(1+\epsilon \cos \varphi )$ with $\epsilon =e\ell$. The orbit is an ellipse for $|\epsilon |<1$, parabolic for unit magnitude and hyperbolic for magnitude greater than one on its allowed angular interval. The purely radial case $p_{\varphi }=0$ must be handled separately and is excluded by these divisions.

In the equatorial plane,Differentiating the given combination produces a cosine termand a sine term $p_r{p}_{\varphi }\dot{\varphi }\sin \varphi /m-p_{\varphi }^2\dot{r}\sin \varphi /(m{r}^2)=0$. Thus $\boxed{\dot{G}=0}$. The expression is a Cartesian component of the Laplace-Runge-Lenz vector with a conventional overall normalization, reflecting the additional symmetry of the inverse-square-force problem.

The radial and azimuthal Hamilton equations giveAlso $\dot{\theta }=p_{\theta }/(m{r}^2)$ and ${\dot{p}}_{\theta }=p_{\varphi }^2\cos \theta /(m{r}^2{\sin }^3\theta )$. Differentiating the angular-momentum square,Therefore both $p_{\varphi }$ and $\boxed{J^2=p_{\theta }^2+p_{\varphi }^2/{\sin }^2\theta }$ are constants. Rotational symmetry conserves the full angular-momentum vector; these are its axial component and squared magnitude in spherical canonical variables.

Apply the chain rule to the phase-space function:Thus $\boxed{[F,H]=0⟹F\text{ is conserved}}$. If an explicitly time-dependent function is allowed, the correct equation is $dF/dt={\partial }_{t}F+[F,H]$, so a vanishing bracket alone would not suffice. The source specifies $F(\mathbf{q},\mathbf{p})$, for which that extra term is absent.

For each canonical coordinate and momentum, Hamilton's equations areThey use canonical rather than arbitrary mechanical momenta, a distinction important in a magnetic field. The bracket convention printed in the source is consistent with $\dot{F}=[F,H]$ for a function with no explicit time dependence.