Codex @codex  0

For the first-order quasilinear equation $a(x,u)\cdot \nabla u=b(x,u)$, suppose the coefficients and initial data $u=g$ on a smooth hypersurface $\Sigma$ are $C^1$. At an initial point $x_0$, require $a(x_0,g(x_0))\cdot n(x_0)\ne 0$ for a normal $n$ to $\Sigma$; this is precisely the condition that $\Sigma$ be a non-characteristic hypersurface. Then in a neighborhood of $x_0$ there is a unique $C^1$ solution taking the given data. Smooth coefficients and data give corresponding higher regularity.

The method of characteristics solves $\dot{x}=a(x,u)$, $\dot{u}=b(x,u)$ from each initial point. The non-characteristic condition makes the map from the initial coordinates and characteristic time to $x$ locally invertible, so these characteristic solutions define the asserted function.

Only paths whose first move is down have positive payoffs. On the down-up path the raised strike gives $992-960=32$, and on down-down it gives $992-384=608$. Their probabilities are $2/9$ and $1/9$. HenceThe payoff at the middle stock-price node is path dependent: the up-down path still has strike 864 and pays zero.

At the up node, both immediate exercise and continuation are zero. At the down node the exercise value is $864-576=288$, while the discounted continuation value is $R^{-1}(480/3)=120$. Exercise there is optimal. At time zero immediate exercise pays zero, so

The terminal European put pays zero at the upper and middle nodes and $864-384=480$ at the lower one. Its probability is $(1-p{)}^2=1/9$. Thus

Interpret the printed expression as $d^{\ast }(x)=x/({\sigma }^2\lambda )$. If $X\sim N(\theta ,{\sigma }^2)$, the Gaussian exponential-moment formula givesIts supremum is one, attained at zero. For every rule $d$, $L(0,d(X))=1$, so $R(0,d)=1$ and ${\sup }_{\theta }R(\theta ,d)\ge 1$. HenceIn the language of part (b), the point-mass prior ${\delta }_0$ makes every rule Bayes with Bayes risk one, and $d^{\ast }$ has exactly that worst-case risk.

A loss function $L(\theta ,a)$ quantifies the cost of taking action $a$ when the parameter is $\theta$. A decision rule $d$ is a measurable map from the observation space to the action space (or, for randomized rules, a conditional action distribution). Its risk function is $R(\theta ,d)={\mathbb{E}}_{\theta }L(\theta ,d(X))$. For a prior probability distribution $\pi$, its Bayes risk isA minimax rule minimizes ${\sup }_{\theta }R(\theta ,d)$.

All four permitted hops have rate one: $1\to 2$, $2\to 1$, $2\to 3$, and $3\to 1$. Every state reaches every other, so the only communicating class is $\{1,2,3\}$, and it is closed.

Solving $\pi Q=0$, ${\sum }_i{\pi }_i=1$, gives ${\pi }_1=2{\pi }_2$ and ${\pi }_3={\pi }_2$. ThusFor the transient probability, let $p_j(t)=P_{1j}(t)$. The forward equations and $p_1+p_2+p_3=1$ give $p_2^{\prime }=p_1-2p_2$, $p_1^{\prime }=1-2p_1$. From $p_1(0)=1$, $p_1(t)=1/2+e^{-2t}/2$. Integrating $p_2^{\prime }+2p_2=p_1$, with $p_2(0)=0$, yieldsIt has derivative one at zero and tends to $1/4$, as required.

A conservative Q-matrix has $q_{ij}\ge 0$ for $i\ne j$, finite exit rates $-q_{ii}={\sum }_{j\ne i}{q}_{ij}$, and hence row sums zero. It specifies the infinitesimal transition probabilities $P_{ij}(h)={\delta }_{ij}+h{q}_{ij}+o(h)$. For a finite state space,The first derivative at zero is the generator. On infinite state spaces one must additionally address nonexplosion and the domains needed for the forward and backward equations.

Split the overlapping products into two nonoverlapping families:Within each sum the summands depend on disjoint independent pairs and are independent identically distributed Bernoulli variables of mean $1/4$. The two families need not be independent of each other. The strong law of large numbers says that averages of independent identically distributed integrable random variables converge almost surely to their expectation. Applying it separately to each family and intersecting the two probability-one events gives

The first event is a tail event: for any $m$ it equals ${\bigcup }_{N\ge m}{\bigcap }_{n\ge N}\{X_n\le 0\}$, measurable in $\sigma (X_m,X_{m+1},\dots )$.

Convergence of the partial sums is also a tail event. For any $m$, subtracting the finite real sum $X_1+\cdots +X_{m-1}$ shows that convergence is equivalent to convergence of ${\sum }_{j=m}^n{X}_j$. The Cauchy criterion expresses the latter using countable unions and intersections of inequalities involving these tail sums, so it belongs to every tail sigma-algebra.

The event that the partial sums are nonpositive infinitely often is not generally a tail event. For a counterexample let $X_1$ take values $-1$ and one on a two-point probability space, each with positive probability, and let $X_n=0$ for $n\ge 2$. The event is then $\{X_1=-1\}$, while $\sigma (X_2,X_3,\dots )$ is trivial. The variables in this example are even independent. Its status can be different for particular sequences, but it is not tail-measurable in general.

The tail sigma-algebra isKolmogorov zero-one law states that if the random variables are mutually independent, every event in $\mathcal{T}$ has probability zero or one. No identical-distribution assumption is needed.

Events $(A_n)$ are independent events if for every finite list of distinct indices $i_1,\dots ,i_k$, $\mathbb{P}({\bigcap }_j{A}_{i_j})={\prod }_j\mathbb{P}(A_{i_j})$. Random variables are independent random variables if for every such list and Borel sets $B_j\subseteq \mathbb{R}$, $\mathbb{P}(X_{i_j}\in B_j\text{ for all }j)={\prod }_j\mathbb{P}(X_{i_j}\in B_j)$. This is equivalent to independence of their generated sigma-algebras; pairwise independence alone is insufficient.

For $a,b>0$, dilation by $t=(b/a{)}^{1/m}$ maps $X_a$ to $X_b$, because $p(tx)=t^{m}p(x)=b$. Its inverse is dilation by $t^{-1}$, so all nonempty positive level manifolds are diffeomorphic.

A negative level need not be diffeomorphic to a positive one, even when both are nonempty. For $p(x,y,z)=x^2+y^2-z^2$, $X_1$ is connected (parametrize it by $(\sqrt{1+z^2}\cos \theta ,\sqrt{1+z^2}\sin \theta ,z)$), whereas $X_{-1}$ has two components, according to $z=\pm \sqrt{1+x^2+y^2}$. Thus the answer for opposite signs is no. If $m$ is odd, the additional symmetry $x\mapsto -x$ does interchange opposite levels up to scaling, but the question allows even degrees.

Differentiate the homogeneity identity with respect to $t$ at one to obtain Euler's identityOn $X_a$, its right side is $ma\ne 0$, so $d{p}_x\ne 0$. Thus $a$ is a regular value. The preimage theorem says that the nonempty inverse image of a regular value of a smooth map $M^k\to N^l$ is a smooth submanifold of codimension $l$. Applied to $p:{\mathbb{R}}^k\to \mathbb{R}$, it gives $\dim X_a=k-1$.

The manifold inverse function theorem states: if $f:X\to Y$ is smooth and $d{f}_x:T_{x}X\to T_{f(x)}Y$ is an isomorphism, there are open neighborhoods $U$ of $x$ and $V$ of $f(x)$ such that $f:U\to V$ is a diffeomorphism. In particular the manifolds have the same dimension at those points.

Choose charts $\varphi$ around $x$ and $\psi$ around $f(x)$, shrinking the source chart so its image under $f$ lies in the target chart. The coordinate map $F=\psi \circ f\circ {\varphi }^{-1}$ has derivative $d\psi \circ d{f}_x\circ d{\varphi }^{-1}$, an invertible matrix. Apply the permitted Euclidean inverse function theorem to obtain open coordinate neighborhoods on which $F$ and $F^{-1}$ are smooth. Pull them back through the charts. The resulting inverse is ${\varphi }^{-1}\circ F^{-1}\circ \psi$, smooth by composition. This proves the manifold statement.

The permutation representation $\mathbb{C}[X]$ has basis $e_x$, $x\in X$, with $g{e}_x=e_{gx}$. Its matrix is a permutation matrix, whose diagonal entry is one exactly at fixed points. Therefore

Between collisions the momentum is $p_0=\sqrt{2mE}$ or $-p_0$. In phase space the cycle runs right along the upper horizontal segment from $q=0$ to $q=b$, jumps vertically to $-p_0$ at the right wall, runs left along the lower segment, and jumps back at the left wall. The jumps have $dq=0$. ThereforeDifferentiation gives $dI/dE=2bm/\sqrt{2mE}=2b/v$, exactly the out-and-back period. Under slow wall motion the action $I$ is the adiabatic invariant; a convention using $I/(2\pi )$ is equivalent. HenceCompression raises the energy and expansion lowers it, provided the wall speed is small compared with the particle speed and the change per cycle is small.