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Let $Q$ denote the given unique equivalent martingale measure. A zero-coupon bond paying one at $T$ has risk-neutral pricing valueWrite $h=T-t$. Split the integrated short rate into its known and future parts:By integrating the future Brownian motion increments in the opposite order, the last integral equals ${\int }_t^T(T-u)d{\hat{W}}_u$. It is independent of ${\mathcal{F}}_t$ and is a Gaussian random variable of mean zero and varianceThe Gaussian exponential moment formula $\mathbb{E}[e^{aZ}]=e^{a^2\mathrm{Var}(Z)/2}$ for centered $Z$ therefore gives the bond price in this Gaussian short-rate model with a deterministic shift:The instantaneous forward rate is the negative maturity derivative of the logarithm of the bond price. ConsequentlyIn particular, $P_t(t)=1$ and $f_t(t)=g(t)+\sigma {\hat{W}}_t=r_t$, as required. The derivative formula is pointwise for continuous $g$ and almost everywhere for locally integrable $g$.

The Gumbel distribution for maxima has distribution function $G(x)=\exp (-e^{-x})$ on the whole real line. For $n\ge 2$ setBoth the centering and scale are elementary, and the scale is positive. For any fixed real $x$, the normalized threshold ${\beta }_n+{\alpha }_{n}x$ is positive for all sufficiently large $n$. Writing $L=\log n$, a Taylor expansion givesConsequently,The logarithm calculation in part (i) now givesThus $F$ belongs to the maximum domain of attraction of the standard Gumbel distribution. The scale also has the Gumbel auxiliary function interpretation: the reciprocal hazard function is $2\sqrt{t}$, whose derivative $1/\sqrt{t}$ tends to zero at the infinite right endpoint. This agrees with the Von Mises conditions for extreme values, and at $t={\beta }_n$ gives exactly ${\alpha }_n=2\log n$.

Let the summands be independent and identically distributed random variables with mean $\mu$, variance ${\sigma }^2>0$, and standardized third cumulant ${\gamma }_1=\mathbb{E}[(Y_1-\mu {)}^3]/{\sigma }^3$. DefineAn Edgeworth expansion refines the central limit theorem using the higher cumulants of one summand. Some smoothness and moment assumptions are essential: finite variance alone does not supply a density expansion. A convenient sufficient set for the expansions here is a smooth density whose derivatives decay faster than every inverse power, together with $\mathbb{E}{e}^{\delta |Y_1|}<\infty$ for some $\delta >0$. These assumptions give all required moments, Fourier smoothing and the nonlattice characteristic-function condition. Less restrictive sufficient conditions are possible, but this set covers the logistic model below.

Write $\varphi$ and $\Phi$ for the standard normal density and standard normal distribution function. The first-order density Edgeworth expansion isThe remainder is uniform on every fixed compact interval under the stated assumptions. To see the origin of the correction, for $Z=(Y_1-\mu )/\sigma$ the cumulant expansion of its characteristic function givesExponentiating and using Fourier inversion yields the displayed correction, since the inverse transform of $(it{)}^3{e}^{-t^2/2}$ is $-{\varphi }^{(3)}(x)=(x^3-3x)\varphi (x)$. The Probabilists' Hermite polynomial convention is being used: $H_2(x)=x^2-1$ and $H_3(x)=x^3-3x$.

The corresponding distribution function expansion isuniformly on fixed compact intervals. The sign follows from $\frac{d}{dx}[H_2(x)\varphi (x)]=-H_3(x)\varphi (x)$. These are expansions of both the density and distribution function; a uniform density error alone would not justify integrating an error bound over the entire real line without the accompanying tail control.

For a fixed $\alpha \in (0,1)$, write $z=z_{\alpha }={\Phi }^{-1}(\alpha )$. At leading order the proposed quantile expansion gives $\Phi (p_0(z))=\Phi (z)$, and strict monotonicity gives $p_0(z)=z$. Substitute $y_{\alpha }=z+p_1(z)n^{-1/2}+O(n^{-1})$ into the distribution function expansion. A Taylor expansion at $z$ givesSince $\varphi (z)>0$, the coefficient of $n^{-1/2}$ must vanish. Thus the first Cornish-Fisher expansion coefficients are

For the logistic distribution, set $Z_i=Y_i-\theta$. Its density $q(z)=e^z/(1+e^z{)}^2$ satisfies $q(-z)=q(z)$ and has exponentially decreasing tails. Symmetry and integrability give $\mathbb{E}{Z}_i=0$, and the supplied integral givesTherefore $\mathbb{E}{Y}_i=\theta$, $\mathrm{Var}(Y_i)={\pi }^2/3$, and the central limit theorem givesWriting $z_{-}=z_{\alpha /2}$ and $z_{+}=z_{1-\alpha /2}$, the condition $z_{-}\le T_n\le z_{+}$ is equivalent toConsequently the confidence intervalhas limiting coverage $\Phi (z_{+})-\Phi (z_{-})=1-\alpha$. Whether the finite endpoints are included makes no probability difference because the sample mean has a density.

The centered logistic distribution is symmetric, so ${\gamma }_1=0$. Its smooth density and exponentially decreasing derivatives satisfy the conditions stated above; in particular all moments exist and Fourier smoothing is available. The first correction in the Edgeworth expansion vanishes, leaving $F_{T_n}(z)=\Phi (z)+O(n^{-1})$ at each of the two fixed quantiles. Subtracting gives the stronger conclusionThe standardized law is independent of $\theta$, so this coverage-error conclusion holds with the same bound for every location parameter.

An exponential dispersion family of order one has density or mass function, relative to a fixed dominating measure, of the formwhere $\theta$ is a scalar natural parameter, $\varphi >0$ is the dispersion parameter, and the support does not depend on $\theta$. For regular families, $b$ is twice differentiable on an open natural-parameter domain and $b^{\prime \prime }>0$. Known observation weights $w$ replace $\varphi$ by $\varphi /w$. “Order one” refers to the scalar canonical statistic $y$, rather than an arbitrary vector of canonical statistics.

Differentiate the normalization $\int p(y;\theta ,\varphi )dy=1$, or its discrete counterpart, assuming differentiation under the integral is valid. The first derivative gives $\mathbb{E}(Y-b^{\prime }(\theta ))=0$. The second derivative givesConsequently,The variance function expresses the second derivative in terms of the mean:

A generalized linear model specifies independent responses from an exponential dispersion family, a linear predictor ${\eta }_i=x_i^T\beta$, and a differentiable monotone link function relating it to the response mean by $g({\mu }_i)={\eta }_i$. The family, variance function, covariate design, link function and dispersion specification together define the model. The canonical link function identifies the linear predictor with the natural parameter:One advantage is that the likelihood equations become particularly simple. With known weights $w_i$ and common dispersion $\varphi$, the score function and negative Hessian matrix for $\beta$ areThe latter does not explicitly depend on the responses, so it equals the expected Fisher information matrix at the same parameter. With full-rank design and positive variances the likelihood is strictly concave in $\beta$, simplifying optimization. These statements concern the coefficient block at a fixed dispersion value.

Three explicit examples, taking unit weights, are as follows.

For a normal distribution with mean $\mu$ and variance $\varphi$,Thus $\theta =\mu$, $b(\theta )={\theta }^2/2$, $c(y,\varphi )=-y^2/(2\varphi )-\frac{1}{2}\log (2\pi \varphi )$, and $V(\mu )=1$. Its canonical link function is $\boxed{g(\mu )=\mu }$, the identity link.

For a Poisson distribution with mean $\mu >0$ and support $\{0,1,\dots \}$,Here $\varphi =1$, $\theta =\log \mu$, $b(\theta )=e^{\theta }$, and $c(y,1)=-\log (y!)$. Therefore $b^{\prime }(\theta )=b^{\prime \prime }(\theta )=e^{\theta }$, $V(\mu )=\mu$, and the Poisson canonical link is $\boxed{g(\mu )=\log \mu }$.

For a Bernoulli distribution with success probability $\mu \in (0,1)$,Taking $\varphi =1$, $\theta =\log (\mu /(1-\mu ))$, $b(\theta )=\log (1+e^{\theta })$, and $c(y,1)=0$ gives the required form. Its mean and variance function are $b^{\prime }(\theta )=e^{\theta }/(1+e^{\theta })=\mu$ and $V(\mu )=\mu (1-\mu )$. Its canonical link function is $\boxed{g(\mu )=\log (\mu /(1-\mu ))}$, the logit link. The fixed dispersion values in the Poisson and Bernoulli cases are part of those models; an arbitrary continuous dispersion is not being asserted for them.

To test $H_0:{\beta }_j=0$, first fit the unrestricted model and then the restricted model with that component fixed at zero, estimating all nuisance parameters, including dispersion when applicable, in each fit. If $\hat{\vartheta }$ and $\tilde{\vartheta }$ denote their full fitted parameter vectors, the likelihood-ratio test usesFor identifiable regular models, an interior null parameter and a fixed-dimensional full-rank design with increasing information, the Wilks theorem gives $D\stackrel{d}{\to }{\chi }_1^2$ under $H_0$. A level-$\alpha$ test rejects for $D$ greater than the $(1-\alpha )$-quantile of the chi-squared distribution with one degree of freedom.

Alternatively the Wald test uses $W={\hat{\beta }}_j^2/\widehat{\mathrm{Var}}({\hat{\beta }}_j)$, with the variance obtained from the appropriate inverse full Fisher information matrix; it has the same limiting ${\chi }_1^2$ null law. The score test fits only the restricted model. If $\lambda$ collects its nuisance parameters, its efficient information is $I_{jj\cdot \lambda }=I_{jj}-I_{j\lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda j}$, and the statistic is $U_j(\tilde{\vartheta }{)}^2/I_{jj\cdot \lambda }(\tilde{\vartheta })$, again asymptotically ${\chi }_1^2$. Accounting for the nuisance block is necessary rather than using the unadjusted diagonal information.

For ordinary homoscedastic fixed-design nonparametric regression, take observations $Y_i=g(x_i)+{ϵ}_i$ at known distinct sites, with independent mean-zero errors of common variance. A degree-$p$ local polynomial estimator at a target $u$ solves the finite-dimensional weighted least squares problemHere $K$ is a nonnegative regression kernel, $h>0$ is the smoothing bandwidth, and sufficient local design rank ensures uniqueness. The estimate of the regression function is $\hat{g}(u)={\hat{\beta }}_0(u)$. Each target has its own locally fitted polynomial; this is not one global polynomial fit. The bandwidth controls the localization and the bias-variance tradeoff.

A natural cubic smoothing spline instead solves the global penalized problemThe Sobolev space here consists of functions with square-integrable weak derivatives through order two. The minimizer is a natural cubic spline with knots at the observation sites and linear tails outside the extreme sites. Equivalently one may perform the minimization on that finite-dimensional spline space: the minimum roughness property of the natural cubic spline interpolant shows that replacing a candidate by the natural spline with the same site values preserves the data loss and decreases the second-derivative penalty. The smoothing parameter $\lambda$ controls curvature; it differs from the local bandwidth $h$.

For the interval-average observation problem, the printed upper limit in the loss is an indexing error: there are only $n-1$ responses and intervals, so a term with $i=n$ would involve the undefined $Y_n$ and $x_{n+1}$. The coherent loss uses the upper limit $n-1$. Put ${\Delta }_i=x_{i+1}-x_i$ and define the linear observation functionalsThe intended criterion on $C^1[a,b]$ is thereforeUnlike the ordinary cubic smoothing spline, this uses interval averages as data and a first-derivative penalty; the resulting spline degree will be two.

Fix any candidate $v\in C^1[a,b]$. By the supplied interpolation property, there is a unique quadratic spline $q\in C^1[a,b]$ with knots $x_1,\dots ,x_n$, constant on both exterior intervals, such that $L_{i}q=L_{i}v$ for every $i$. Put $d=v-q$. ThenOn each interior knot interval $q^{\prime \prime }$ is a constant $c_i$, and on the two exterior intervals $q^{\prime }=0$. Integrate $q^{\prime }{d}^{\prime }$ by parts separately on the intervals. At every knot the boundary contributions cancel because both $q^{\prime }$ and $d$ are continuous; the contributions at $a$ and $b$ vanish because $q^{\prime }$ is zero there. ThusThis orthogonality is the key to the quadratic smoothing spline for interval averages. Expanding the roughness penalty givesThe data-fit terms are equal because the interval averages are equal. Consequently,Equality can occur only if $d^{\prime }=0$ everywhere: the derivative is continuous and its squared integral is zero. Then $d$ is constant, and any of its zero interval averages forces that constant to be zero. Therefore every minimizer must equal its matching quadratic spline $q$.

We conclude that the minimizer is a continuously differentiable quadratic spline with knots at the specified sites and constant tails on both exterior intervals. Existence and uniqueness also follow from the supplied interpolation property. The interval-average map identifies the finite-dimensional space of these quadratic splines with ${\mathbb{R}}^{n-1}$, and the criterion becomes a strictly convex quadratic function of that average vector: its data-fit term is $\Vert Y-y{\Vert }^2$, while its penalty is nonnegative and quadratic. It has a unique minimizer, and the strict equality argument above proves uniqueness in the full function class.

In adaptive experimental quantum control, parametrize a reproducible pulse by a finite real vector $c$, for example $f_c(t)={\sum }_{j=1}^p{c}_j{B}_j(t)$ in a calibrated waveform basis. Prepare the same initial state for each trial, apply that pulse without feedback during the run, and estimate the objective $J(c)$ from repeated measurements. No accurate dynamical model is required if the laboratory can evaluate the objective.

One concrete algorithm is experimental finite-difference gradient ascent. For every parameter, run trials at $c\pm \delta e_j$ and estimateUpdate $c$ to an admissible projection of $c+\eta g$, test the new pulse, and reduce the step size if performance deteriorates. Averaging many shots controls statistical noise; choose $\delta$ large enough for the difference to exceed the measurement uncertainty but small enough to approximate a derivative. Repeat until no statistically significant improvement remains. Multiple starting pulses help explore distinct basins, but this does not guarantee a global optimum. Resource penalties and hardware restrictions belong in the objective or the admissible parameter set.

The feedback is across experimental runs: a measured objective chooses the next pulse. Each individual trial is open-loop control. This distinction matters because the algorithm does not require an instantaneous nondestructive state measurement on the system being controlled.

The feedback law printed in the PDF is not sufficient for the asserted monotonicity. We first derive the exact identity, then give a counterexample and the corrected law. Both density operators evolve unitarily, so $\mathrm{Tr}{\rho }^2$ and $\mathrm{Tr}{\rho }_d^2$ are constant. ThusCyclicity of the trace gives $\mathrm{Tr}([-i{H}_0,{\rho }_d]\rho )=-\mathrm{Tr}({\rho }_d[-i{H}_0,\rho ])$. These drift terms cancel in the derivative of the overlap, leavingHere $C$ is real: $-i[H_1,\rho ]$ is Hermitian, and the trace of the product of two Hermitian matrices is real. This is the Hilbert-Schmidt Lyapunov feedback identity.

For a direct counterexample to the source's overlap law, take $H_0=0$, $H_1={\sigma }_y/2$, ${\rho }_d=(I+{\sigma }_z)/2$ and $\rho =(I+{\sigma }_x)/2$. All are admissible qubit operators and states. The printed overlap is $f=\mathrm{Tr}({\rho }_d\rho )=1/2$, but $-i[H_1,\rho ]=-{\sigma }_z/2$ gives $C=-1/2$. Hence $\dot{V}=1/2>0$: the distance initially increases. This proves that feedback based on state overlap need not be stabilizing.

The intended Lyapunov quantum control replaces the overlap by its control-direction derivative:The conclusion is nonincrease, not strict decrease or automatic convergence. For example, opposite computational-basis pure states, diagonal $H_0$, and $H_1={\sigma }_x$ give $C=0$ and zero control even though $V=2$. Target convergence requires further invariant-set and reachability conditions. A target with a different spectrum is also inaccessible under Quantum Hamiltonian evolution. The density-operator derivatives in the governing equations use the dots visible in the PDF, which the TeX loses.

The stated $M$ is specifically the transfer from the external input to the internal field $b_1$. Combining $b_1=G{b}_0$ with $b_0=\beta b_{min}-\alpha b_1$ givesFor $\gamma >0$ and a nondegenerate beam splitter $|\alpha |<1$, the closed-loop pole is $-{\gamma }_{meff}/2<0$. The return ratio for the Nyquist stability criterion is $L=\alpha G$. Its frequency locus is a circle of radius $|\alpha |<1$, so it cannot reach or encircle $-1$. The open cavity has no unstable poles; hence the closed loop has none. This agrees with the directly derived state damping rate.

The externally accessible output is different. From $b_2=\alpha b_{min}+\beta b_1$,It is again an all-pass cavity response. Confusing $b_1$ and $b_2$ would give the wrong output steady state.

For a constant coherent input mean $B=⟨b_{min}⟩$, the steady mean cavity amplitude follows by setting its derivative to zero:The input-output values also follow from $G(0)=-1$ and $M(0)=-\beta /(1-\alpha )$. For any initial mean amplitude, its difference from the stationary mean decays as $e^{-{\gamma }_{meff}t/2}$.

The stability statement needs endpoint qualifications. At $\alpha =1$, normalization forces $\beta =0$, and the cavity obeys $\dot{a}=0$: it is isolated and undamped, not asymptotically stable. Its external output is nevertheless the decoupled input. At $\alpha =-1$, the algebraic elimination divides by zero; the instantaneous feedback connection is singular and is not a well-posed instance of these reduced equations. Thus strict cavity stability holds for $|\alpha |<1$, not for every real pair satisfying only ${\alpha }^2+{\beta }^2=1$.

Finally, $b_{min}(t)$ is a quantum stochastic field, so arbitrary inputs do not approach a literal time-independent operator amplitude. The stationary expressions above describe a constant coherent drive's means, or the formal zero-frequency response. The full solution contains the filtered input-noise convolutionThe stationary cavity density operator also depends on the input noise statistics. For this ideal passive cavity with a coherent drive and vacuum noise, it is the coherent state with the amplitude just calculated; without such input statistics, the gain function alone does not specify a complete stationary quantum state.

The exponents are real, as required for this family of real continuous functions. We prove the exponential Chebyshev system property by induction on the number of functions. A single nonzero multiple of an exponential function has no zero, so the base case holds.

Suppose the result holds for $n$ distinct real exponents, and consider a nontrivial combination $F(x)={\sum }_{j=0}^n{c}_j{e}^{{\lambda }_{j}x}$. If it had $n+1$ distinct zeros, so wouldsince the prefactor is everywhere positive. Applying Rolle's theorem between consecutive zeros gives at least $n$ distinct zeros ofThe $n$ exponents in this derivative are distinct. By induction, a nontrivial combination of them has at most $n-1$ zeros. Therefore every coefficient $c_j({\lambda }_j-{\lambda }_0)$ must be zero. Distinctness gives $c_j=0$ for $j\ge 1$. Then $g=c_0$ has a zero only if $c_0=0$, contradicting the original nontriviality. Consequently $F$ has at most $n$ zeros, and

Use unit knot spacing and the centered Cardinal cubic B-spline convention $C(t)={\sum }_i{P}_i{M}_3(t-i)$, where $M_3$ is supported on $[-2,2]$. On the span $[i,i+1]$, with $s=t-i$, the active control points are $P_{i-1},P_i,P_{i+1},P_{i+2}$, and their B-spline weights areFor $t=11/4$, take $i=2$, $s=3/4$. The weights and their derivatives areThe PDF has $P_3=(0,0,0)$ and $P_4=(0,1,0)$, so the point and tangent vector areThus, under this centered convention,The original PDF specifies neither the knot origin nor its indexing relative to $P_i$. If instead the convention associates span $[i,i+1]$ with $P_i,P_{i+1},P_{i+2},P_{i+3}$, the identical local weights multiply $P_2,P_3,P_4,P_5$ at the requested parameter, givingThese are parameter translations of the same uniform construction, rather than contradictory evaluations with the same knots. A specified knot convention is necessary to select the numerical pair uniquely; changing the knot spacing would also rescale the derivative.