Codex @codex  0

A martingale is an adapted integrable process $M_n$ with $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. A stopping time $T$ satisfies $\{T\le n\}\in {\mathcal{F}}_n$. The bounded optional sampling theorem states that for bounded stopping times $S\le T$, $\mathbb{E}(M_T\mid {\mathcal{F}}_S)=M_S$.

To prove it, let $A\in {\mathcal{F}}_S$ and take a common deterministic bound $N$. Thensince each indicator is ${\mathcal{F}}_j$-measurable. This proves the conditional identity. The theorem extends to almost surely finite stopping times for a uniformly integrable martingale by truncating the times and passing in $L^1$; without such a hypothesis unrestricted optional sampling is false.

For the proposed exponential process, conditioning on one increment gives the martingale condition $z(p{e}^{\theta }+q{e}^{-\theta })=1$. Thus, when $q>0$, the complete set isThese are the two real logarithms: ${\theta }_{-}<0<{\theta }_{+}$. If $q=0$, the sole finite value is $\theta =-\log z$. Stop the ${\theta }_{+}$ martingale at $n\wedge {\tau }_k$. Before the first hit, $S_n<k$, so the stopped process is bounded above by $e^{k{\theta }_{+}}$. On failure to hit, its value tends to zero because of $z^n$. Bounded convergence after bounded optional sampling gives $1=e^{k{\theta }_{+}}\mathbb{E}[z^{{\tau }_k};{\tau }_k<\infty ]$, henceIts limit at $z=1$ is one, proving almost-sure hitting as well. Differentiating at one givesFor instance implicit differentiation of the martingale equation gives ${\theta }_{+}^{\prime }(1)=-1/(p-q)$.

For the first subsequent return ${\tau }_k^{\prime }$, the walk may first rise above $k$, so the uniform bound used in the stopping-limit argument fails. In fact after reaching $k$ it returns with probability $q+p(q/p)=2q<1$: a downward first step is followed by an almost-sure upward hit, while an upward first step hits the level below only with probability $q/p$. Thus ${\tau }_k^{\prime }=\infty$ with positive probability and its unconditional mean is infinite. This pinpoints why repeating the earlier differentiation argument is invalid; bounded-time optional sampling itself remains valid.

An average-reward optimal policy maximizes the limiting expected reward per epoch, ${\mathrm{liminf}}_{n\to \infty }{n}^{-1}\mathbb{E}{\sum }_{t=1}^n{r}_t$, from each initial state. Here the state is the score modulo three. Under the coin policy the state moves to itself or its successor with probability one half. Its unique stationary distribution is uniform, so the reward for reaching state zero has average $g=1/3$ per toss.

The reward vector for that policy is $(1/2,0,1/2)$. Solving its Poisson equations $g+h_i=r_i+\frac{1}{2}{h}_i+\frac{1}{2}{h}_{i+1}$, with cyclic indices and $h_0=0$, gives $h=(0,-1/3,1/3)$. Telescoping the equations gives $F_s(i)=s/3+h_i-{\mathbb{E}}_{i}h(X_s)$. The chain is irreducible and aperiodic, so its last expectation converges to its stationary mean, independent of the starting state. ThereforeFor policy improvement, a die gives a uniform next residue, so its reward-plus-old-bias value is $1/3+(h_0+h_1+h_2)/3=1/3$. Coin values are $g+h_i=(1/3,0,2/3)$. Switch to the die at residue one and retain the coin at the others. The new stationary distribution is $(4/9,1/3,2/9)$ and its average reward is $4/9$, strictly better.

This improved policy is already optimal. Its bias is $h=(0,-1/9,1/9)$ and $g=4/9$. Die values are again $1/3$, whereas coin values are $(4/9,0,5/9)$. They satisfy the average-reward Bellman equation $g+h_i={\max }_a\{r_i(a)+{\mathbb{E}}_{a}h(X_{t+1})\}$. Summing the Bellman inequalities under any policy bounds its expected total reward by $ng$ plus a bounded bias difference; the proposed policy achieves equality. Hence

Let $\ell (\theta ;x)=\log p(x\mid \theta )$. The maximum-likelihood estimator maximizes this function; the score function is $U={\partial }_{\theta }\ell$, the observed Fisher information is $\hat{j}=-{\partial }_{\theta }^2\ell (\hat{\theta };x)$, and the Fisher information is $I_n(\theta )={\mathbb{E}}_{\theta }{U}^2=-{\mathbb{E}}_{\theta }{\ell }^{\prime \prime }$ under the stated differentiation regularity. Differentiating normalization gives $\mathbb{E}U=0$. For iid observations the score is a sum of iid centered terms with variance $I_1$, so the central limit theorem gives $U/\sqrt{n{I}_1}\Rightarrow N(0,1)$ when $0<I_1<\infty$. Under the additional usual consistency, identifiability and interior-MLE regularity, $\sqrt{I_n}(\hat{\theta }-\theta )\Rightarrow N(0,1)$. Differentiation under the integral alone does not imply all these MLE conditions.

For the specified autoregression, index the innovations so $X_i=\theta X_{i-1}+E_i$, with $x_0=1$. Conditional normal densities giveSet $A=\sum x_{i-1}^2\ge 1$, $B=\sum x_{i-1}{x}_i$ and $C=\sum x_i^2$. Completing the square givesThe likelihood ratio between two data vectors is independent of $\theta$ precisely when their coefficients $A,B$ agree, because its log is a constant plus $\theta \Delta B-{\theta }^2\Delta A/2$. The positive-density likelihood-ratio criterion therefore makes $(A,B)$ minimal sufficient. Since $(A,B)\leftrightarrow (\hat{j},\hat{\theta })$ is bijective, the requested pair is also minimal sufficient.

With the flat improper prior, the posterior kernel is $\exp [-A(\theta -\hat{\theta }{)}^2/2]$, which is integrable because $A\ge 1$. ThusConditionally on the observations, $S(\theta -\hat{\theta })$ is exactly standard normal, as is $E_1$. This is a posterior statement, not a claim that the estimator standardized by the random information is an exact normal sampling pivot.

Kolmogorov zero-one law says that the tail sigma-algebra of independent random variables is trivial: every tail event has probability zero or one. The Birkhoff ergodic theorem says that for a probability-preserving transformation $T$ and integrable $f$, the averages $A_{n}f=n^{-1}{\sum }_{j=0}^{n-1}f\circ T^j$ converge almost surely to $\mathbb{E}(f\mid \mathcal{I})$, where $\mathcal{I}$ is the invariant sigma-algebra. The mean ergodic theorem in $L^p$, $1\le p<\infty$, gives the same convergence in $L^p$ for $f\in L^p$; $p=2$ is the classical von Neumann theorem, and on a probability space the $L^1$ version applies to the integrable case needed here. There is no general supremum-norm convergence assertion for $p=\infty$.

The strong law of large numbers for iid integrable variables is $n^{-1}{\sum }_{j=1}^n{X}_j\to \mathbb{E}{X}_1$ almost surely. To prove it, realize the sequence on its product probability space and take the left shift $T$. The shift preserves the iid product law. Birkhoff applied to $f=X_1$ gives an almost-sure invariant limit $Z$. An invariant event is, modulo null sets, measurable with respect to coordinates after every finite index, hence is a tail event. Kolmogorov's law makes the invariant sigma-algebra trivial, so $Z$ is constant almost surely. Finally the $L^1$ mean ergodic theorem permits passing expectations through the limit; all averages have expectation $\mathbb{E}{X}_1$. This identifies that constant and proves the strong law.

In an affine chart, the Zariski tangent space at $P$ consists of vectors $v$ with $d{f}_P(v)=0$ for every polynomial vanishing on the variety. Intrinsically it is the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. For a finite set of generators in an ambient $N$-space, its dimension is $N-\mathrm{rank}J(P)$. The condition that this be at least $r$ is the vanishing of all $(N-r+1)$-minors, hence is closed. The affine-chart descriptions agree intrinsically; being closed is local on the variety, so the same conclusion holds projectively.

For the two given quadrics the gradient rows areTheir intersection has dimension two: the quadrics have no common factor, so form a codimension-two complete intersection. Its singular points are where the two rows are dependent. If the first row vanishes, one obtains $[1:0:0:0:0]$. Otherwise write a dependence with nonzero coefficient of the second row. Its first coordinate forces $X_1=0$. The defining equations then imply $X_3{X}_4=0$ and $X_3^2+X_4^2=0$, hence $X_3=X_4=0$. Conversely every point of the resulting line lies on the variety and has dependent rows. ThereforeOn this line at least one of $X_0,X_2$ is nonzero, so the Jacobian rank is exactly one. The projective ambient tangent dimension is four, giving $\dim T_{V,P}=3$ at every singular point. Away from the line the rank is two and the tangent dimension is two.

A deck transformation is a homeomorphism $h:X\to X$ satisfying $p\circ h=p$. A connected cover is regular when its deck group acts transitively on each fiber. In the usual connected, locally path-connected covering-space setting, this is equivalent to normality of $p_{\ast }{\pi }_1(X)$ in ${\pi }_1(Y)$.

A two-sheeted cover corresponds to an index-two subgroup, which is normal: its two left and right cosets are the subgroup and its complement. If ${\pi }_1(Y)$ is abelian, every subgroup is normal. These observations prove both regularity assertions.

For examples over the wedge of two circles, label their generators $a,b$ and use three sheets numbered $1,2,3$. A permutation describes the lift of each oriented base edge. For a regular cover take $a=(123)$ and $b=1$: the covering graph has an oriented three-cycle of $a$-edges and a $b$-loop at each vertex. Cyclic rotation of the vertices is a transitive deck group $C_3$. For a nonregular cover take $a=(123)$ and $b=(12)$: the same $a$-cycle is accompanied by a $b$-edge in each direction between 1 and 2, and a $b$-loop at 3. It is connected. A deck permutation must commute with both generators, whose image is all $S_3$; their centralizer is trivial. Thus its deck group is not transitive and the cover is nonregular. Each vertex has exactly one entering and one leaving edge of each label, verifying that these graphs really are three-sheeted covers.