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Let $p=\Phi (1)-\Phi (0)$, and write $\phi$ for the standard normal distribution density. For importance sampling, choose the proposal density $q(x)=1_{[0,1]}(x)$, namely the uniform distribution on the target interval. Its importance sampling weight isFor iid uniform draws $U_1,\dots ,U_N$,is an unbiased estimator of the target mean, since its expectation integrates $xf(x)$. If one chooses not to evaluate $p$, the self-normalized importance sampling alternative $\sum U_i\phi (U_i)/\sum \phi (U_i)$ has statistical consistency, but generally is not an unbiased estimator.

Alternatively, draw independent random variables $V_1,\dots ,V_N$ with the standard normal distribution, let $A_i=1_{\{0\le V_i\le 1\}}$ and $K=\sum A_i$, and useConditionally on $K=k>0$, the retained draws are iid from the truncated normal distribution, so their average is an unbiased estimator with variance ${\tau }^2/k$, where ${\tau }^2$ is its variance. If $K=0$ the ratio is undefined; one can continue drawing until a prescribed number of retained observations is obtained.

Here importance sampling uses every draw inside the interval, whereas only $p\approx 0.3413$ of the normal proposals are retained. The advantage can also be quantified. Integration using ${\phi }^{\prime }(x)=-x\phi (x)$ givesFor uniform importance sampling,For the retained-draw ratio with a fixed total proposal budget, its asymptotic variance coefficient is ${\tau }^2/p\approx 0.233347$. Thus this importance choice has substantially smaller error for the same proposal count; it also avoids wasting about two thirds of the standard normal distribution draws. This comparison is specific to this well-matched importance sampling proposal; arbitrary proposal choices do not share this guarantee.

Use the spectral theorem for real symmetric matrices to write $\Sigma =Q\Lambda Q^T$, with $Q$ orthogonal and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)$, ${\lambda }_i\ge 0$. Set $L=Q{\Lambda }^{1/2}$ and generateThen $L{L}^T=\Sigma$. The characteristic function iswhich identifies the multivariate normal distribution $N_d(\mu ,\Sigma )$. Equivalently, every linear combination is normal, with the required mean and variance. A Cholesky decomposition can be used for a positive-definite matrix. The spectral construction also works for singular $\Sigma$, producing a degenerate Gaussian distribution on an affine subspace; in that case there need not be a density on all of ${\mathbb{R}}^d$.

Use $v={\sigma }^2$ as the variance parameter, with prior density relative to $d\mu dv$. Let $\bar{x}=n^{-1}\sum x_i$ and $S=\sum (x_i-\bar{x}{)}^2$. Multiplying the prior by the likelihood function gives the unnormalized Bayesian posteriorThe printed request for an “improper posterior” should not be read as a claim that this posterior is always improper. Although the prior is improper, integrating over $\mu$ gives a marginal kernel $v^{-(n+1)/2}{e}^{-S/(2v)}$, which is integrable exactly when $n>1$ and $S>0$. Under these usual nondegenerate-data conditions it is a proper posterior. If those conditions fail, the displayed kernel is improper and there is no posterior probability distribution for a sampler to target.

Completing the square and using the inverse-gamma distribution give the full conditionalsThe inverse-gamma distribution convention is density proportional to $v^{-\alpha -1}{e}^{-\beta /v}$. Equivalently the precision satisfiesusing shape and rate, as in the question.

A Gibbs sampler starts from $v^{(0)}>0$ and alternatesThis is the normal mean-variance posterior with a log-uniform variance prior. For a marginal check, $v\mid x\sim \mathrm{IG}((n-1)/2,S/2)$: integrating out $\mu$ reduces the shape by $1/2$. This marginal shape is different from the full-conditional shape $n/2$.

The support condition for importance sampling is missing from the printed assertion. Assume $M>1$, $E_f|h|<\infty$, and $Mg(z)>f(z)$ for $f$-almost every $z$. For a proposal $Z\sim g$, rejection has conditional probability $1-f(Z)/(Mg(Z))$. Its unconditional probability is $(M-1)/M$, so the conditional density of a rejected draw isThus the rejected-draw weight is precisely $f(z)/r(z)$, andConditionally on any positive rejection count $n$ among $N$ independent proposals, the rejected locations have this iid conditional density. Henceis an unbiased estimator conditionally on $n$, and therefore also conditionally on the event $0<n<N$. This is importance sampling from the rejected-proposal distribution.

The nonstrict assumption $f\le Mg$ alone does not suffice: the rejected distribution may completely miss part of the target support. For example, let $g$ be uniform on $[0,1]$, $f(z)=21_{[0,1/2]}(z)$, $M=2$, and $h=1$. All rejected proposals lie in $(1/2,1]$, where $f=0$, so the displayed estimator is always $0$, whereas $E_{f}h=1$. This remains a counterexample after conditioning on $0<n<N$. A strict envelope, such as choosing $M$ larger than the essential supremum of $f/g$, repairs the missing support condition. Near-equality can still give high-variance weights, even when it remains an unbiased estimator.

The printed identity needs an expectation on its right-hand side. For fixed $r\ge 1$, condition on $X_2,\dots ,X_r$ and use independence of $X_1$:The expression inside this expectation is generally random and cannot equal the unconditional probability by itself.

For each independent Monte Carlo method replicate, draw $R_j$ from its given distribution. If $R_j=r\ge 1$, draw $r-1$ independent $F$-variables, form $W_j={\sum }_{i=2}^r{X}_{j,i}$ and set $H_j=F(x-W_j)$. For $r=1$ the empty sum is $0$. If $R$ may equal zero, use $H_j=1_{\{x\ge 0\}}$ when $R_j=0$, since $S_0=0$. The law of total expectation givesSince $0\le H_j\le 1$, the estimator has statistical consistency by the strong law of large numbers. It is the conditional Monte Carlo estimator obtained by Rao-Blackwellization of the direct indicator $1_{\{S_R\le x\}}$, and the law of total variance gives $\mathrm{Var}(H_j)\le l(1-l)$.

This version assumes that the cumulative distribution function $F$ can be evaluated. An easy sampler alone does not automatically supply an easy cumulative distribution function evaluation. If only sampling is available, use the direct indicator estimator, or replace each conditional cumulative distribution function by the average of several independent indicators $1_{\{X_{j,1,k}\le x-W_j\}}$; the nested version remains an unbiased estimator. With $L$ such inner draws its variance per outer replicate is $\mathrm{Var}(H_j)+L^{-1}E[H_j(1-H_j)]$.

Use the ergodic theorem for a positive Harris recurrent Markov chain: for an invariant distribution $\pi$ and an integrable function $h$, the time average converges almost surely to $\int hd\pi$ under the usual ergodicity conditions. This is the relevant theorem for correlated Metropolis–Hastings algorithm output, rather than the strong law of large numbers. Here $h(x)=x_1$ is integrable. Indeed, puttingwe have $V(x)\ge (x_1-4{)}^2+(x_2-4{)}^2-32$, which bounds the unnormalized density by a constant times an integrable Gaussian density and also gives finite first moments.

For a proposal scale $s>0$, initialize $X_0$ anywhere in ${\mathbb{R}}^2$. At each step draw independent random variables $Z_1,Z_2$ with the standard normal distribution and an independent uniform $U$, propose $Y=X_t+s(Z_1,Z_2)$, and accept whenOtherwise set $X_{t+1}=X_t$. The normal proposal is symmetric, so its density cancels; the target normalizing constant $c$ also cancels. Include every state in the average, including repeated states after rejection. The ergodic theorem for a positive Harris recurrent Markov chain then givesA fixed discarded initial segment does not change this limit.

Small proposal variance gives high acceptance but tiny moves and strong serial dependence. Large variance gives more ambitious moves, but many proposals enter very low-density regions and are rejected, creating long runs at one state. An intermediate scale should be assessed by exploration and effective sample size of a Markov chain, not acceptance rate alone.

This density has two modes, at $(2+\sqrt{3},2-\sqrt{3})$ and their interchange. To see this, stationary points satisfy $x_1(1+x_2^2)=x_2(1+x_1^2)=4$, implying either $x_1=x_2$ or $x_1{x}_2=1$. The unequal solutions have $x_1+x_2=4$ and are local minima of $V$; the equal solution is a saddle. Mode switching is therefore a material part of proposal-scale selection: a chain confined to one mode can have high acceptance and a misleading finite-run estimate.

Write $h=h_n$ and $I=[t_0,t_0+h]$, restricting to meshes small enough that $t_0+h\le 1$. This discontinuous indicator function does not satisfy the hypothesis of part (b). Instead apply part (a) with $\Vert g_n{\Vert }_{\infty }=1/h$ and handle its two boundary cells directly.

Let $q_n(s)=g_n(t_{i-1})$ on each observation cell $[t_{i-1},t_i)$. Away from cells containing the two ends of $I$, $q_n=g_n$ almost everywhere. Even if an endpoint is itself an observation time, at most two cells contribute, and their total length is at most $2\Delta$. Thus the quadrature error obeysThe fluctuation has variance at most $2S\Delta /h^2$. Its zero expected value givesusing $\Delta \le 1$.

The remaining bias is the difference between an interval average and the spot variance. By the Hölder continuity of ${\sigma }^2$,Retaining the centered fluctuation and bounding $(B_n+A_h{)}^2\le 2B_n^2+2A_h^2$ yieldsFor $h={\Delta }^{1/(2+2\beta )}$, both $\Delta /h^2$ and $h^{2\beta }$ equal ${\Delta }^{2\beta /(2\beta +2)}$. Therefore $D_1=10$ and $D_2=2C^2$ satisfy both requested bounds:The constants are independent of the observation partition, $\sigma$, $t_0$ and $\beta$. The exponent follows from a bias-variance tradeoff for a shrinking window.

For real $v$, $\overline{\phi (v)}=\phi (-v)$. Expand the complex covariance of the sample averages. Terms indexed by distinct observations vanish because the observations are independent; each diagonal term equals $\mathbb{E}{e}^{i(u-v)X_1}-\phi (u)\overline{\phi (v)}$. HenceTaking $v=u$ and using $\phi (0)=1$ gives the exact variance:The upper bound follows from $|\phi (u)|\le \mathbb{E}|e^{iu{X}_1}|=1$, and holds without any moment assumptions on $X_1$.

Use the Radon-Nikodym derivative at $T_1$ and the discounted bond price martingale from part (b):For the variance, put $b_t^i={\int }_t^{T_i}\sigma (t,u)du$ and $d_t=b_t^2-b_t^1={\int }_{T_1}^{T_2}\sigma (t,u)du$, for $t\le T_1$. The ratio $R_t=P(t,T_2)/P(t,T_1)$ has, by the Itô formula,Under the T-forward measure ${\mathbb{Q}}_{T_1}$, $d{W}_t=d{W}_t^{T_1}-b_t^{1}dt$, so its drift becomes $-d_t^2/2$. Since $R_{T_1}=P(T_1,T_2)$,The first two terms are deterministic; the Itô isometry supplies the variance of the last. The two requested answers areIn fact the terminal zero-coupon bond price has a log-normal distribution under this forward measure, with the deterministic negative half-variance correction ensuring the displayed mean.

An effective bandwidth measures how much capacity a random traffic source requires when unusually large demands matter. A mean-rate description loses burstiness; a peak-rate description can waste capacity when peaks are rare. The parameter $s>0$ in an effective bandwidth controls the emphasis placed on such bursts through the moment-generating function:Here $X$ is the traffic in a chosen time unit. If $X(t)$ instead measures traffic over length $t$, the corresponding rate is $\alpha (s,t)=\log \mathbb{E}[e^{sX(t)}]/(st)$. With independent stationary increments, this is independent of $t$ whenever the moment-generating function exists. More generally both the time scale and dependence structure matter.

The Chernoff bound follows directly from the Markov inequality, since $e^{sX}$ is nonnegative and $X\ge b$ implies $e^{sX}\ge e^{sb}$:Optimize over the admissible positive values of $s$ to obtainFor independent sources $X_1,\dots ,X_N$, moment-generating functions multiply and their effective bandwidths add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb{P}({\sum }_i{X}_i\ge c)\le \exp \{s({\sum }_i{\alpha }_i(s)-c)\}$. In particular, for independent and identically distributed random variables,This is the positive-parameter part of the Legendre transform of a cumulant-generating function. The size of its exponent describes how a service margin reduces the chance of a large aggregate demand. A buffer-overflow event involves traffic over many intervals; a one-time Chernoff bound alone is not automatically a bound on that event. A time-dependent model and, for example, bounds on the union of relevant interval events are then needed.

Assume a finite moment-generating function in a neighborhood of zero. Differentiating the cumulant-generating function gives $K(0)=0$, $K^{\prime }(0)=\mathbb{E}[X]$ and $K^{\prime \prime }(0)=\mathrm{Var}(X)$. Therefore the small-parameter limit is the mean traffic rate:The variance term shows how an effective bandwidth charges extra capacity for burstiness. Also $K$ is a convex function with $K(0)=0$, so $K(s)/s$ is nondecreasing for $s>0$ in its domain.

For the other limit, let $M=\mathrm{esssup}X$, the essential supremum. If $M<\infty$, $X\le M$ almost surely gives $\alpha (s)\le M$. For every $a<M$, $p_a=\mathbb{P}(X\ge a)>0$, andTaking a lower limit and then $a\uparrow M$ proves the large-parameter limit is peak traffic:If $X$ has no finite upper bound and all positive exponential moments exist, the same lower bound for arbitrarily large $a$ makes the limit infinite. If positive exponential moments cease to exist at a finite parameter, the effective bandwidth is infinite beyond that domain, and the infinite limit is understood in this extended sense. For example, a normal distribution with mean $m$ and variance $v$ has $\alpha (s)=m+sv/2$: its unbounded upper tail has no finite peak. Thus effective bandwidth interpolates between mean and peak demand while permitting independent sources to be aggregated by addition.

Put $\beta =k\xi \mu$. The random phase screen producesInterpret the normality assumption as a jointly Gaussian process, not merely Gaussian one-point marginal distributions. Write its normalized autocorrelation as $\rho (\zeta )=⟨W(z+\zeta )W(z)⟩$. The difference $W(z+\zeta )-W(z)$ is a zero-mean Gaussian random variable with variance $2(1-\rho (\zeta ))$. Its characteristic function therefore gives the Gaussian phase-screen correlationWithout joint Gaussianity the autocorrelation alone does not determine this expectation.

For an acoustic pressure amplitude $\psi$, with background mass density ${\rho }_0$, the time-averaged acoustic energy flux isAn acoustic potential convention changes the dimensional prefactor. The physical field includes the carrier $e^{ikx}$: a transverse Fourier transform mode has axial wavenumber ${\kappa }_{\nu }=k-{\nu }^2/(2k)$ in the parabolic wave equation. Thus its spectral acoustic flux is proportional to ${\kappa }_{\nu }|\hat{E}(x,\nu ){|}^2$, or to $k|\hat{E}(x,\nu ){|}^2$ at leading paraxial order. Differentiating the reduced field alone would omit the carrier contribution.

The infinite stationary illumination has a spectral measure of a stationary random field, rather than a finite ordinary value of $⟨|\hat{E}{|}^2⟩$ at each $\nu$. To specify the normalization, truncate the screen to an aperture of length $L$, propagate this truncated initial field, and setThe propagation multiplier has modulus one. Using the Gaussian phase-screen correlation and the overlap of the two aperture intervals givesMultiplication by ${\kappa }_{\nu }/(2{\rho }_0\omega )$ gives the ensemble-averaged pressure-normalized modal acoustic energy flux for this finite-aperture convention.

The infinite-aperture spectral normalization consistent with the specified Fourier transform isThe integral and limit are understood as spectral measures of a stationary random field when necessary. By Parseval's identity, this is power per unit transverse length: ${\int }_{\mathbb{R}}{S}_E(d\nu )=C_E(0)=1$. The ensemble-averaged axial spectral acoustic flux is independent of $x$:Here the notation includes Dirac delta distributions in the spectral measure of a stationary random field; the leading-order total flux is $I_{\mathrm{inc}}$. The ${\kappa }_{\nu }$ correction is meaningful within $|\nu |\ll k$, the range of the paraxial approximation.

If $\rho (\zeta )\to 0$ and $e^{{\beta }^2\rho (\zeta )}-1$ is integrable, the stationary phase-screen power spectrum separates into coherent and diffuse wave fields:The coherent fraction is $e^{-{\beta }^2}$ and the diffuse fraction is $1-e^{-{\beta }^2}$. If additionally $|\beta |\ll 1$, the stationary phase-screen power spectrum is $(1-{\beta }^2)\delta +{\beta }^2\hat{\rho }+O({\beta }^4)$ with the same Fourier transform convention. Small $\mu$ alone does not imply small accumulated phase $k\xi \mu$.

The binomial theorem makes the sum of the proposed probability mass function equal to $(p+1-p{)}^N=1$; every term is nonnegative. Realize the associated binomial distribution as $K=B_1+\cdots +B_N$, where the $B_j$ are independent variables with the Bernoulli distribution of parameter $p$. Each has expected value $p$ and variance $p(1-p)$. Linearity of expectation and variance additivity for independent random variables yieldAt $p=0$ or $p=1$ this is a constant random variable; the same formulas apply, with the endpoint mass understood by continuity.

For the constant monomial $f_0=1$, the binomial theorem gives $B_n(f_0)(x)=(x+1-x{)}^n=1$. For $f_1$, the binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ givesThus $\boxed{B_n(f_0)=1,B_n(f_1)=x}$, including $x=0,1$ by evaluation of the polynomials.

For $m\ge 1$ and $n\ge 2$, substitute the preceding coefficient identity into the Bernstein polynomial sum and write $j=k-1$. This yields the Bernstein monomial recurrenceThe coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n{)}^{m-1}\to 1$ and the others tend to zero. The weights in any Bernstein polynomial sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le 1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x{x}^{m-1}=x^m$: the $\ell =0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the supremum norm proves uniform convergence on $[0,1]$.

As an independent quantitative bound, take a binomial distribution variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb{E}[(K/n{)}^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the Cauchy-Schwarz inequality and the variance of $K/n$ giveFor $m=0$ the error is exactly zero. Finally the requested sum equals $2^m{B}_{2n}(f_m)(1/2)$, so

For a target $a(\theta )$, the bias of an integrable estimator $T$ is ${\mathbb{E}}_{\theta }T-a(\theta )$; it is an unbiased estimator if this is zero at every allowed parameter value. Here $Y$ has a zero-truncated Poisson distribution, $\theta >0$, and $p=1-e^{-\theta }\in (0,1)$. Write an estimator based only on $Y$ as $t(Y)$, assuming a finite expectation for every $\theta >0$. Unbiasedness requiresAbsolute integrability at every positive parameter ensures that the power series on the left converges absolutely on every complex disc. Uniqueness of coefficients in a power series therefore gives $t(y)=0$ for odd $y$ and $t(y)=2$ for positive even $y$. Conversely these values give the displayed identity, sois the unique unbiased estimator of this form. The uniqueness claim concerns nonrandomized functions of $Y$; allowing external randomness would permit addition of independent mean-zero noise.

For this parity estimator for a zero-truncated Poisson count, usefulness depends on the loss, but under squared-error loss it performs poorly despite being an unbiased estimator. Since $\mathbb{P}(Y\text{ even})=p/2$, $\mathbb{E}{T}^2=2p$ andIt takes the inadmissible value $2$ with positive probability, and its variance tends to one rather than zero as $\theta \to \infty$. Clipping to the parameter interval gives $T_c=\min (T,1)=1_{\{Y\text{ even}\}}$. This has bias $-p/2$, butThus a simple biased estimator strictly improves its mean squared error at every parameter value; uniqueness among unbiased estimators does not make it optimal for this loss.

Since $\mathbb{P}(X_1=0)=e^{-\lambda }$, the indicator $Y$ is an unbiased estimator of $\theta$. The joint probability mass function isThe Fisher-Neyman factorization theorem therefore makes $T={\sum }_i{X}_i$ a sufficient statistic. More explicitly, conditioning on $T=t$ gives a multinomial distribution with $t$ trials and cell probabilities $1/n$, so $X_1\mid T=t$ has the binomial distribution $\mathrm{Bin}(t,1/n)$. Consequently conditioning by the Rao-Blackwell theorem givesFor $n=1$, interpret the value at $T=0$ as $1$ and at $T>0$ as $0$; thus the same expression uses $0^0=1$ in this probability convention. For $\lambda =0$ all observations and $T$ are zero almost surely. For $n>1$ and $\lambda >0$, the reduction in variance is strict: given $T=1$, the event $X_1=0$ still has a nondegenerate conditional probability. Directly, using the probability generating function of $T\sim \mathrm{Poisson}(n\lambda )$, the new variance is $e^{-2\lambda }(e^{\lambda /n}-1)$.