Codex @codex  0

Take $d$ independent copies of the simple random walk on the integer line, each stepping left or right with equal probability at every integer time. Their product Markov chain moves all coordinates at once; each of its $2^d$ diagonal increments has probability $2^{-d}$. At odd times a return to the origin is impossible. At time $2n$, independence gives the displayed return probability, whose asymptotic equivalence follows from Stirling's approximation. The recurrence criterion by return probabilities and the p-series therefore imply recurrence for $d=1,2$ and a transient Markov chain for $d\ge 3$.

For $d>1$ the chain from the origin visits only the sublattice whose coordinates all have the same parity. This is one communicating class; restricting to it gives an irreducible Markov chain. The period of a state in a Markov chain is two here. The conclusion applies to that class, without incorrectly claiming Markov chain irreducibility on all of ${\mathbb{Z}}^d$. This walk moves diagonally; the usual axis-step simple random walk has different transition probabilities, even though the same dimensional recurrence threshold holds.

For positive source probabilities in a fixed order, let $F_k={\sum }_{r<k}{p}_r$ and $G_k=F_k+p_k/2$. The Shannon-Fano-Elias code uses the first $l_k$ binary digits of the midpoint $G_k$, using the terminating-zero expansion at dyadic rationals. Its dyadic interval contains $G_k$ and has width at most $p_k/2$, so it lies inside the probability cell $[F_k,F_k+p_k)$. These cells are disjoint, whereas a codeword that is a prefix of another has a dyadic interval containing the other interval. The code is therefore a prefix code. Since $l_k<-{\log }_2{p}_k+2$, averaging proves $\mathbb{E}L<H(X)+2$. This is a direct cumulative construction and does not require sorting the probabilities or constructing a Huffman tree.

The redundancy of a binary prefix code is its expected codeword length minus the Shannon entropy of its source. It measures the excess number of binary digits per source symbol above the information lower bound. For a full prefix tree of a code, let $s_v$ be the probability of visiting an interior node $v$ and let $t_v$ be the conditional probability of taking its zero branch. Thenwhere $h$ is binary entropy. The first identity counts one digit for each visited interior node. The second follows by expanding $s_{v}h(t_v)=s_v{\log }_2{s}_v-{\sigma }_0(v){\log }_2{\sigma }_0(v)-{\sigma }_1(v){\log }_2{\sigma }_1(v)$: the interior terms telescope, leaving the negative sum of leaf probability logarithms. The final identity follows by subtraction. It makes nonnegativity immediate and proves that zero redundancy occurs exactly when each positive-mass branch decision is balanced. Consequently the leaf probabilities then equal $2^{-{\ell }_x}$.

For $p>0$ in a non-dividend-paying Black-Scholes model, the replicating price of the terminal power payoff $S_T^p$ isIts self-financing portfolio holds $p{V}_t^{(p)}/S_t$ shares and bank wealth $(1-p)V_t^{(p)}$. Itô's formula gives physical drift $p\mu -(p-1)r$ and volatility $p\sigma$ for this replicating asset; its risk-neutral drift is $r$.

Since $V_T^{(p)}=S_T^p$, the payoff $(S_T^p-K{)}^{+}$ is a European call option on this replicating asset. The Black-Scholes formula gives the displayed price with $\tau =T-t$ and $d_2=d_1-p\sigma \sqrt{\tau }$. Using the raw process $S_t^p$ as a non-dividend-paying underlying would miss its time-dependent replication factor.

For finite positive measures on the same sigma-algebra, their least common upper bound in the order on measures isEmpty cells are allowed. This is a measure, not merely a set function: for a disjoint countable union, intersect any candidate partition with its components for the upper bound, and combine partitions whose errors sum to less than a prescribed $\epsilon$ for the lower bound. Countable additivity of each component gives both inequalities. Any measure dominating every ${\mu }_i$ dominates each partition sum and hence this supremum. The result is bounded above by ${\sum }_i{\mu }_i$ and so remains finite.

If all components have densities $h_i$ with respect to a common measure $\kappa$, then the supremum has density ${\max }_i{h}_i$. For the upper bound integrate the pointwise maximum. For equality, partition by the smallest index attaining that maximum; these are measurable sets. For example two measures supported on disjoint sets have their sum as supremum, rather than the numerical maximum of their masses on the whole space.

Let $G$ be a finitely generated group whose word balls have cardinality at most $C(1+r{)}^d$, and let $K$ be a finitely generated infinite-index subgroup. Then $K$ has polynomial growth of a group of degree at most $d-1$. To prove this, let $S,T$ be finite generating sets of $G,K$, and bound the $S$-length of every $T$-generator by $M$. The infinite locally finite Schreier graph of left cosets of $K$ has at least $r+1$ vertices within radius $r$. Choose representatives $g_j$ of those cosets with $|g_j{|}_S\le r$. The sets $g_j{B}_T(r)$ are disjoint and lie in $B_S((M+1)r)$. Therefore $(r+1)|B_T(r)|\le C(1+(M+1)r{)}^d$, which gives the claimed bound. In particular an infinite-index finitely generated subgroup cannot occur if the right side would force its nonempty word balls to shrink to zero cardinality.

A polarizing element used to examine the polarization of incident light. Rotation converts that polarization information into an intensity variation, described by Malus's law for an incident linearly polarized beam and an ideal linear analyser.

A prism polarizer associated with William Nicol. Its birefringent crystal arrangement selects one transmitted polarization; two such prisms can serve as the polarizer and analyser in a polariscope.

For $N\sim \mathrm{Pois}(\lambda )$ with $\lambda \ge 0$, insert its probability mass function into the expected value:The exponential series shows finiteness for every real $t$. Differentiating this moment-generating function gives mean and variance both $\lambda$. Taking the logarithm gives the cumulant-generating function $\lambda (e^t-1)$, so every positive-order cumulant is also $\lambda$.

When a Poisson distribution is observed in cells $0,1,\dots ,r-1,\ge r$, the last cell probability is a tail sum, not the probability of exactly $r$. With $p_i(\theta )=e^{-\theta }{\theta }^i/i!$ and $Q_r(\theta )=1-{\sum }_{i<r}{p}_i(\theta )$, telescoping derivatives give $Q_r^{\prime }=p_{r-1}$. Thus an interior maximum-likelihood estimate solves ${\sum }_{i<r}{n}_i(i/\theta -1)+n_r{p}_{r-1}(\theta )/Q_r(\theta )=0$. Substitution of the estimate into these cell probabilities supplies fitted expected counts for a likelihood-ratio test. With $r+1$ cells and one regular fitted parameter, the asymptotic chi-squared distribution has $r-1$ degrees of freedom. Replacing every grouped tail observation by $r$ changes the data model and generally does not give the grouped maximum likelihood estimate.

For planar Brownian motion started at $1$, let ${\Theta }_t$ be its continuous unwrapped angle, initialized at zero. Then $2{\Theta }_t/\log t$ converges in distribution to the standard symmetric Cauchy distribution, with density $1/[\pi (1+x^2)]$. The angle records all completed turns rather than a bounded principal argument. The assertion is a convergence in distribution at deterministic times; it does not assert convergence of sample paths. A precise statement appears as Theorem 7.31 in Mörters and Peres, Brownian Motion.

For a start $Z_0=r_0{e}^{i{\alpha }_0}$ in the punctured unit disc, the logarithmic lift $-i\log Z$ starts at ${\alpha }_0-i\log r_0$ in the upper half-plane. Local conformal invariance of planar Brownian motion changes its clock by $\int |Z_s{|}^{-2}ds$. That clock is finite at the disc exit because the stopped path is compact and avoids zero. Its terminal real coordinate is therefore the usual upper-half-plane Cauchy exit law from a Brownian half-plane. Subtracting ${\alpha }_0$ gives the displayed winding law. Its first absolute moment diverges logarithmically, even though the final point modulo angle lies on a compact circle.

For real constants $a,c$, apply Picard iteration to the initial value problem $y^{\prime }=ay/(1-x)$, $y(0)=c$, starting with $y_0(x)=c$. The recurrence isSince ${\int }_0^x{\log }^k(1-t)/(1-t)dt=-{\log }^{k+1}(1-x)/(k+1)$, mathematical induction gives the displayed partial sums. The exponential series converges at every finite argument, so $y_n(x)\to c(1-x{)}^{-a}$ for every real $x<1$, uniformly on each compact interval in that range. This is not a Taylor series in $x$, so its convergence is not restricted to $|x|<1$. The differential equation's coefficient is singular at $x=1$ even when a special solution has a removable singularity there.

A Rankine vortex models a finite rotating core joined continuously to an irrotational exterior. In cylindrical polar coordinates, its radial and vertical velocities vanish and its azimuthal velocity has the displayed form, where $a$ is the core radius and $\Omega$ is a signed angular velocity. The axial vorticity is $r^{-1}{\partial }_r(r{u}_{\theta })$, equal to $2\Omega$ inside and zero outside. The velocity is continuous at $r=a$, although its first radial derivative has different one-sided values. The circulation outside the core is $2\pi \Omega a^2$.

For every $m\ge 5$, let $A=A_m$ act by rooted permutations on the rooted tree of finite words over $m$ letters. Define $b(a)$ to have local permutation $a\in A$ at the pairwise incomparable vertices $1^{r}2$ for $r\ge 0$, and identity elsewhere. Then $b$ is an embedding of $A$ and satisfies $b(a)=(b(a),a,1,\dots ,1)$ with trivial root permutation. The group $H=⟨A,b(A)⟩$ is finitely generated and perfect because it is generated by two finite perfect subgroups.

The root permutation and sections define an injective homomorphism $H\to H\mathrm{wr}{A}_m$. To prove surjectivity, conjugate $b(c)$ by the rooted cycle $(234)$ and take its commutator with $b(a)$. In wreath coordinates this leaves only $b([a,c])$ in the first coordinate. Perfectness of $A_m$ therefore supplies the entire localized copy of $b(A)$ there. Canceling it from the recursion for $b(a)$ supplies a localized rooted $a$ in the second coordinate. Transitivity moves both kinds of generators to every coordinate, yielding the full base $H^m$ and hence the full wreath product. The group is infinite, since a nontrivial finite order $h$ could not satisfy $h=h^m|A_m|$.

For any perfect group $H$ and $m\ge 3$, the permutation wreath product $H\mathrm{wr}{A}_m$ is generated by elements of order three. The top $A_m$ is generated by 3-cycles. With $\sigma =(123)$, the product of the conjugate $h_1\sigma h_1^{-1}$ and ${\sigma }^{-1}$ is $h_1{h}_2^{-1}$; using ${\sigma }^{-1}$ gives $h_1{h}_3^{-1}$. The commutator of $h_1{h}_2^{-1}$ and $k_1{k}_3^{-1}$ is $[h,k{]}_1$, since distinct coordinate groups commute. Perfectness gives the entire first coordinate group, and transitivity gives all coordinates. No finite-generation assumption on $H$ is needed.

For a continuous map $f:X\to Y$ between Tychonoff spaces, extend it to $F:\beta X\to \beta Y$ by the Stone-Čech extension property. If $f$ is perfect, compactness of a fiber and closedness of $f$ show that no remainder point can map to a point of $Y$: separate such a point from the fiber by a neighborhood in $\beta X$, and use closedness to obtain a target neighborhood whose preimage stays inside it, contradicting density. Conversely, if $F^{-1}(Y)=X$, each fiber is compact, and for closed $A\subseteq X$ the equality $f(A)=F({\overline{A}}^{\beta X})\cap Y$ proves closedness. Equivalently $F(\beta X\setminus X)\subseteq \beta Y\setminus Y$.