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Lax's equation in the convention used here isLet $\mathcal{L}\psi =\lambda \psi$ with $(\psi ,\psi )=1$. Since $\mathcal{L}$ is self-adjoint and $\mathcal{A}$ is anti-self-adjoint, differentiating $\lambda =(\psi ,\mathcal{L}\psi )$ givesThe first and third terms equal $\lambda ,d(\psi ,\psi )/dt=0$. For the middle term,Thus $\dot{\lambda }=0$, proving the Isospectral Lax equation property: every eigenvalue is independent of $t$.

Let $V(p)$ be the maximal expected future reward minus search costs at posterior $p=(p_1,\dots ,p_n)$, with stopping value zero. Searching $i$ costs $c_i$, succeeds with probability ${\alpha }_i{p}_i$ and earns $R_i$, and otherwise produces the posterior $p^{(i)}$ displayed in part (a). The rewarded Bayesian box search Bellman equation is

Suppose ${\sum }_i{c}_i/({\alpha }_i{R}_i)<1$. At every posterior $p$, some index must satisfyotherwise summing the reverse inequalities would give $1={\sum }_i{p}_i<1$. For this box,even if one stops immediately after a failed search. Hence $V(p)>0$ at every posterior, and stopping before discovery is never optimal.

Once search necessarily continues until the ball is found, the expected reward ${\sum }_i{p}_i{R}_i$ is independent of the order of searches. Maximizing expected net reward therefore amounts to minimizing expected search cost. Part (a) applies, so the policy maximizing ${\alpha }_i{P}_i^t/c_i$ remains optimal.

At a given time write $p_i=P_i^t$ and $q_i={\alpha }_i{p}_i$, the conditional probability that the next search succeeds if box $i$ is chosen. Compare two consecutive searches $i,j$, keeping the continuation after two failures fixed. Their expected costs up to that common continuation areThe probability of reaching the common continuation is the same in either order. Moreover,Repeated adjacent interchanges move a box with maximal index to the front without increasing expected cost; applying the same argument after every failure proves the optimal index for Bayesian box search policyFor completeness, after an unsuccessful search of box $i$, Bayes' theorem gives the posterior in the Bayesian box search problem:

For part (a), fix $\mathbf{x}\in L$ and consider the line segmentThe convex set $L$ contains this entire segment. Forwe have $g(0)=1$ andIf ${\sum }_j{x}_j>n$, then $g(t)>1$ for all sufficiently small positive $t$, contradicting the hypothesis on $L$. Thus

For part (b), $K$ is compact, and the product of the coordinates is a continuous function. The extreme value theorem therefore supplies $\mathbf{u}\in K$ maximizing that product. Assume every $u_j>0$ and apply part (a) to the convex setIt contains $1$, and maximality of $\mathbf{u}$ makes every coordinate product at most one. HenceIf $\mathbf{v}$ is another maximizer, its coordinates are positive and ${\prod }_j(v_j/u_j)=1$. The arithmetic-geometric mean inequality giveswhile the displayed inequality gives the reverse bound. Equality in the arithmetic-geometric mean inequality forces every $v_j/u_j=1$. Therefore the maximizer $\mathbf{u}$ is unique.

Let $\tau =T-t$. Conditional on ${\mathcal{F}}_t$ under the risk-neutral measure,Using the moment-generating function of a normal distribution in the risk-neutral expectation gives the Power payoff in the Black-Scholes model valueIn particular,The delta hedge holdsunits of the risky asset.

The price process is a geometric Brownian motion withPutand define the equivalent measure $Q$ on ${\mathcal{F}}_T$ by the exponential Brownian martingaleThe Cameron-Martin-Girsanov theorem says that $W_t^Q=B_t+\vartheta t$ is Brownian motion under $Q$. Thereforeso $e^{-rt}{S}_t$ is a $Q$-martingale. This is the unique equivalent martingale measure, since cancellation of the single Brownian risk forces the displayed drift shift.

By risk-neutral valuation,Writing $W_T^Q=\sqrt{T}Y$ with $Y\sim N(0,1)$ gives

Since the MLE has constant risk $p$, its worst-case risk is $p$, so the minimax risk is at most $p$. On the other hand, the worst-case risk of every estimator is at least its average risk under any prior, and therefore at least the optimal Bayes risk for that prior. Using ${\pi }_c=N_p(0,c^2{I}_p)$ givesLetting $c\to \infty$ yields a lower bound $p$. The bounds agree, proving the minimaxity of the usual multivariate normal mean estimator:

Combining the Gaussian likelihood with the prior $\theta \sim N_p(0,c^2{I}_p)$ gives the posterior precision and meanThus the posterior in the Gaussian normal-mean shrinkage Bayes estimator isThe Bayes estimator under squared error loss is the posterior mean, soIts posterior expected loss is the trace of the posterior covariance and does not depend on $X$. Therefore its Bayes risk is

An admissible estimator is one for which there is no estimator $\delta$ satisfyingfor every $\theta$, with strict inequality for at least one $\theta$. For $p\ge 3$, the James–Stein estimatorhas riskIt strictly dominates $X$, and hence

The risk function of an estimator $\hat{\theta }$ under the stated quadratic loss isThe multivariate normal density is proportional, as a function of $\theta$, toso maximum likelihood estimation gives ${\hat{\theta }}_{\mathrm{MLE}}=X$. Writing $X=\theta +Z$ with $Z\sim N_p(0,I_p)$,for every $\theta \in {\mathbb{R}}^p$.

SetBy variance additivity for independent random variables,Thus $Y_N$ differs in $L^2$ by $o(1)$ from the deterministic scalar $m_N$. The L2 convergence of averages of independent variables with bounded second moments gives the necessary and sufficient condition

From convergence in probability, choose a subsequence $X_{n_k}\to X$ almost surely. The Fatou lemma then givesso $X\in L^2$.

The bounded second moments make $(X_n)$ uniformly integrable, and $X$ is integrable. Convergence in probability together with uniform integrability therefore givesConvergence in $L^2$ need not follow. On $[0,1]$ with Lebesgue measure, letThen $X_n\to 0$ in probability and $\mathbb{E}|X_n{|}^2=1$, but $\Vert X_n{\Vert }_2=1$ for every $n$. This is the bounded second moments do not upgrade convergence in probability to convergence in L2 example.

The hypothesis says exactly that $X_n\rightharpoonup 0$ in the Hilbert space $L^2$. We give the constructive weak Banach–Saks theorem in a Hilbert space argument. Choose $n_1<n_2<\cdots$ inductively so that, for $Y_k=X_{n_k}$,This is possible because each earlier $Y_j$ is an admissible test variable in the weak-convergence assumption. Since $\Vert Y_k{\Vert }_2\le 1$,Therefore

The Cauchy-Schwarz inequality givesThus $Z{X}_n$ is integrable for every $n$.

Let $M_t(x)=R(t)x+T(t)$ be the given family. Composing throughout with $M_0^{-1}$ lets us assume $M_0$ is the identity; the family remains regular and preserves $\gamma$. Differentiating at $t=0$ produces a nonzero infinitesimal proper Euclidean motionFor every $x\in \gamma$, the path $t\mapsto M_t(x)$ lies in $\gamma$, so $V(x)$ is tangent to $\gamma$. The flow of this complete vector field consequently preserves $\gamma$ for all times.

The orbit of a one-parameter proper Euclidean motion is a line if $a=0$, and otherwise a helix about an axis parallel to $a$. Since $\gamma$ is compact, it cannot contain an entire line or a helix with nonzero axial pitch. Thus the flow is a pure rotation about an axis. The curve cannot lie entirely in the fixed axis, because no subset of a line is a simple closed curve. Choose a point off the axis. Its orbit is a circle contained in $\gamma$; as a one-dimensional orbit it is open in the connected one-dimensional manifold $\gamma$, and it is closed by compactness. It must therefore equal the whole curve. Hence the continuous rigid-motion symmetry of a simple closed space curve gives

Choose unequal positive numbers $a,b$. The planar ellipseis a simple closed curve and is not a circle. The nonidentity proper rigid motionis rotation through $\pi$ about the $z$-axis, and $R\gamma (t)=\gamma (t+\pi )$. It therefore preserves the curve as a set.

For a unit-speed curve with nonzero curvature of a space curve $\kappa$, define the Frenet frameThe torsion of a space curve iswith this choice of convention.

For the requested example, take the unit-speed circleand the unit-speed circular helixBoth have the constant nonzero curvature $\kappa =1$, but their torsions are respectively $0$ and $1$. A proper Euclidean motion of Euclidean three-space preserves torsion, so no such rigid motion relates the two curves.

Fix $t_0\in [a,b]$ and defineBecause $\gamma$ is a regular curve, $s^{\prime }(t)=\Vert {\gamma }^{\prime }(t)\Vert >0$. The inverse function theorem gives a smooth inverse $t=t(s)$ onto the corresponding interval. For $\tilde{\gamma }(s)=\gamma (t(s))$,Thus the arc-length parametrization exists.