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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.

The diagram has common multipliers $u=5/3,d=2/3$ on every branch, with gross riskless return $R=4/3$. The martingale condition $pu+(1-p)d=R$ givesSince $d<R<u$, these probabilities lie strictly between zero and one at each node, and the market is arbitrage-free. All valuations below use discounted conditional expectations under this measure.

The probability generating function encodes the population distribution in one analytic function. Its coefficients and derivative recoverprovided the mean is finite. Higher derivatives at one give factorial moments, and normalization says $\varphi (1,t)=1$.

Interpret the supplied birth and death expressions as transition rates: over a small time $dt$ their probabilities are $(\alpha +\beta n)dt+o(dt)$ and $\gamma ndt+o(dt)$. Incoming probability from $n-1$ and $n+1$, minus both outgoing rates, giveswith $P(-1,t)=0$. Multiplying by $s^n$ and summing yieldsAt stationarity, ${\varphi }_s/\varphi =\alpha /(\gamma -\beta s)$, so integration and normalization giveAlternatively differentiating the evolution equation at one gives $m^{\prime }=\alpha -(\gamma -\beta )m$, so $m(t)=\alpha /(\gamma -\beta )+[m(0)-\alpha /(\gamma -\beta )]e^{-(\gamma -\beta )t}$. If $\beta =0$, the displayed power expression is understood in its limit: $\varphi (s)=e^{(\alpha /\gamma )(s-1)}$, a Poisson distribution with mean $\alpha /\gamma$.

$T$ is the interval between generations or censuses, and $r$ the prey multiplication without predators. The factor $e^{-aP}$ can be the probability of escaping a Poisson number of encounters with mean $aP$, so $a$ measures attack intensity. The factor $s$ converts prey resources into predator recruitment, and $b$ modifies the fraction of prey contributing no recruitment. At $b=1$ predators recruit from attacked prey; $b<1$ adds background recruitment, even at $P=0$. A population interpretation valid at every nonnegative state needs $b\le 1$, since otherwise the second equation gives negative recruitment near $P=0$.

Set $Y_n={\sup }_{m\ge n}|f_m|$. It is an extended-real measurable function and decreases to $Y={\mathrm{limsup}}_m|f_m|$. If $f_n\to 0$ almost surely, then $Y_n\to 0$ almost surely. For any $ϵ>0$, the decreasing sets $\{Y_n>ϵ\}$ have intersection contained, up to a null set, in the exceptional set for convergence. Continuity from above of the probability measure gives $P(Y_n>ϵ)\to 0$.

Conversely, if $Y_n\to 0$ in probability, then $P(Y>ϵ)\le P(Y_n>ϵ)\to 0$ for every positive $ϵ$. Taking rational $ϵ>0$ gives $Y=0$ almost surely, so $|f_n|\to 0$ almost surely. This proves both directions, even when some individual $Y_n$ take the value infinity.

One complete cycle consists of ten stops of mean one minute and ten journeys of mean four minutes, so its mean duration is fifty minutes. Over the eight-hour shift, the elementary renewal approximation gives $480/50=9.6$ completed circuits on average.

For the end-of-shift calculation use the long-run occupation probabilities of the twenty successive phases. Each stop phase has probability $1/50$, and each travel phase probability $4/50$. Alighting stop $j$ is therefore selected with probability $1/50+4/50=1/10$, from either stopping there or travelling toward it. If that stop is the station, no further stops are passed. If it is stop $j\ge 2$, the intermediate stops are $j+1,\dots ,10$, numbering $10-j$. Thus the requested approximate mean isThe approximation treats eight hours as long compared with a cycle and ignores the initial-phase transient. Exponential holding times justify the twenty-phase Markov description; independent cycle times justify the renewal rate.

An arbitrage is a portfolio with nonpositive initial cost, nonnegative terminal payoff almost surely and a positive payoff with positive probability, allowing unused initial cash to be invested in the riskless asset. Equivalently it is a zero-cost portfolio with those terminal properties. An equivalent martingale measure $Q$ has the same null sets as the physical measure and satisfies $E_Q[S_1^i]=(1+r)S_0^i$ for every traded asset, with appropriate integrability.

For an agent with differentiable increasing concave utility and an optimal portfolio, let $W^{\ast }$ be its terminal wealth. The first-order conditions for variations in risky holdings are $E[U^{\prime }(W^{\ast })(S_1^i-(1+r)S_0^i)]=0$. If $U^{\prime }(W^{\ast })>0$ and integrable, normalize it to the probability density $Z=U^{\prime }(W^{\ast })/E[U^{\prime }(W^{\ast })]$. The first-order conditions make $Q$ with $dQ=ZdP$ an equivalent martingale measure. A small additional payoff $H$ then has marginal priceThis follows by setting the first variation of expected utility for buying a small claim at price $p$ equal to zero.

A zero-cost stock holding $\theta$ has terminal payoff $\theta (S_1-1)$. If $a\le 1$, shorting the share and holding a bond gives $1-S_1\ge 0$ and a strictly positive payoff almost surely. If $a>1$, both signs of $S_1-1$ occur with positive probability, so no nonzero holding is an arbitrage. Hence arbitrage freedom is equivalent to $a>1$.

The equivalent martingale measures are exactly the densities $Z(s)>0$ almost everywhere on $(0,a)$ satisfyingThey exist for $a>1$: exponential tilts of the uniform law have means ranging continuously from zero to $a$, so one has mean one.

Under the usual utility assumptions $U^{\prime }>0$, $U^{\prime \prime }<0$, and existence of an admissible optimum, $J(\theta )=E[U(w+\theta (S_1-1))]$ is strictly concave and $J^{\prime }(0)=U^{\prime }(w)(a/2-1)$. Its decreasing derivative implies that the optimizer is positive exactly when $a>2$, negative when $a<2$, and zero when $a=2$, provided the feasible interval contains zero in its interior. The hypothesis $C^2$ alone is insufficient: a linear increasing utility can give an unbounded objective rather than a finite optimizer. Concavity and existence are implicit in the standard utility formulation.

For $a=2$, $\theta =0$ makes marginal utility constant, so the pricing measure is the original uniform law andFor any equivalent pricing measure, the call expectation is positive and strictly less than $E_Q{S}_1/2=1/2$, since $(s-1{)}^{+}<s/2$ for $0<s<2$. Both endpoints can be approached. Symmetric densities concentrated in $[1-\delta ,1+\delta ]$ give prices tending to zero; symmetric densities concentrated near zero and two give prices tending to one half. Mix either density with an arbitrarily small amount of the uniform law to keep it everywhere positive; symmetry keeps its mean one. The set of positive mean-one densities is convex and the call price is linear, so every intermediate value is attained. Thus the exact price range is $(0,1/2)$, with neither endpoint attained.

With $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$, the matrix of $S_z\cos \theta +S_x\sin \theta$ is $(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Multiplication by $(c,s{)}^T$ and $(-s,c{)}^T$ gives eigenvalues $+\hslash /2$ and $-\hslash /2$, using the double-angle identities.

The one-particle transformation is the matrix $R=\left(\begin{array}{cc}c& -s\\ s& c\end{array}\right)$, with determinant one. The antisymmetric tensor $|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩$ transforms by this determinant under $R\otimes R$, so the singlet is unchanged. The total $S_z$ kills it. Its invariance under these rotations implies that total $S_z\cos \theta +S_x\sin \theta$ also kills it for every $\theta$, so total $S_x$ kills it. The spin commutator then gives total $S_y$ acting as zero. Therefore $S^2|\chi ⟩=0$ and it has total spin zero.

Rotating only particle B producesThe measurement probabilities are thereforeEach arrow represents the corresponding value $\pm \hslash /2$. The sequential measurements commute because they act on different particles. The two parallel outcomes leave triplet states with total spin one and $S^2=2{\hslash }^2$. Each antiparallel product state is a superposition of the spin-zero singlet and the spin-one zero-projection triplet, and is not an $S^2$ eigenstate. Outcomes of zero probability at a special angle do not actually occur.

Independence and the Poisson probabilities give the log-likelihoodIts score and curvature areIf some $x_i\ne 0$, this is strictly concave. A finite maximum is therefore the unique zero of the score when such a zero exists. Newton iteration givesStart from a finite value, compute the score and curvature, and iterate until both the score and step are small. A line search that shortens a step until the likelihood increases improves robustness. Strict concavity identifies any converged stationary point as the global maximum. A finite root need not exist, for example when all observations are zero and all $x_i>0$, in which case the likelihood supremum is at $\beta \to -\infty$. If all $x_i=0$, the parameter is unidentifiable.

For a nonnegative integer-valued random variable, its expected value is $\mathbb{E}N={\sum }_{r=0}^{\infty }r{p}_r$, allowing the value $+\infty$. Since $r={\sum }_{n=1}^{r}1$, rearranging nonnegative summands gives the tail-sum formula for expectation:No integrability assumption is needed for the rearrangement. The same formula holds for an extended nonnegative integer-valued variable by the monotone convergence theorem, using $N={\sum }_{n\ge 1}{1}_{\{N\ge n\}}$.

For the first ascent in independent continuous observations, continuity makes ties probability zero. By order symmetry of independent random variables, the first $m$ observations have each of their $m!$ strict orderings with equal probability. For $n\ge 2$, $N\ge n$ means exactly that the first $n-1$ observations are decreasing, soSubtracting consecutive tails giveswith probability zero for $r=0,1$. The events $\{N\ge n\}$ decrease to $\{N=\infty \}$; continuity of probability under decreasing intersections gives $P(N=\infty )={\lim }_{n}1/(n-1)!=0$. Finally,The first observation cannot itself end the run; this is why the tail includes the separate term $P(N\ge 1)=1$.

With a cyclically oriented independence digraph of events, each event is independent of one of the other two. Independence is symmetric, so all three pairs are independent events. This does not imply mutual independence. Put $x=\mathbb{P}(A_1\cap A_2\cap A_3)$. The inclusion-exclusion principle givesbecause $x\le \mathbb{P}(A_1\cap A_2)=a^2$. Thus the oriented case guarantees positive avoidance exactly when $0<a<1/2$.

For sharpness, for any $1/2\le a<1$ prescribe the law of the three event indicators by assigning mass $(1-a{)}^2$ to each of the three singleton success patterns, $(1-a)(2a-1)$ to each doubleton, $1-3a+3a^2$ to the tripleton, and zero to the zero-success pattern. All masses are nonnegative and their sum is one. Each marginal isand each pair intersection has mass $(1-a)(2a-1)+(1-3a+3a^2)=a^2$. These events satisfy the oriented independence conditions and have zero avoidance probability.

The unoriented triangle imposes no independence conditions at all: every vertex is adjacent to every other. The union bound gives positive avoidance if $3a<1$. To see that this is sharp for every $a\ge 1/3$, use uniform measure on the circle of circumference one and take three arcs of length $a$ starting at $0,1/3,2/3$. They cover the circle. Hence the unoriented case guarantees positive avoidance exactly when $0<a<1/3$.

Write $p_i=\mathbb{P}(A_i)$. A strong independence graph of maximum degree one consists of isolated vertices and disjoint pairs. For a component $C$, let $B$ be the event that all events outside $C$ are avoided. Each $A_i$, $i\in C$, is independent of $B$, since the outside vertices are all nonneighbors. The union bound therefore givesRemove the components successively. If every $p_i<1/2$, all these factors are positive, including singleton components, so simultaneous avoidance has positive probability. Conversely, two complementary events of probability $1/2$, joined by one edge, cover the sample space. Consequently $\gamma (1)=1/2$. The argument does not assume joint independence between entire two-event components.

For higher degrees use a dependency-degree obstruction on a rooted tree. Put $d=\Delta -1$ and chooseOn a finite rooted $d$-ary tree take independent bits $B_v$ and define $A_v$ to mean that $B_v=1$ and every child bit is zero. The underlying tree is a strong independence graph: the variables used by $A_v$ are disjoint from all variables used by its nonneighbors. Its maximum degree is at most $d+1$.

Give the leaves success probability $r_0=p$, and, working upwards, useas long as this is less than one. This recursion must reach or exceed one after finitely many steps. Indeed,so the iterates strictly increase. An infinite sequence staying below one would either converge to a fixed point, contradicting the displayed maximum, or tend to one, in which case its next iterate would exceed one. At the first offending step set the root success probability to one instead of the proposed value. Every nonroot event then has probability exactly $p$; the root event has probability $(1-r_j{)}^d\le p$.

Every configuration contains an occurring event: start at the root, whose bit is one; if its event fails, descend to a child whose bit is one, and repeat. A finite tree forces this descent to end at an occurring event. Thus avoidance is impossible. Given any $c>p_{\ast }$, choose $p\in (p_{\ast },\min (c,1))$; all event probabilities are strictly below $c$. This proves

The claimed sufficient condition is false for the printed all-colours requirement. This is a polychromatic hypergraph colouring problem, which is stronger than merely avoiding monochromatic edges.

For a counterexample even with $k=3$, take $V=\{1,2,3,4\}$ and all four triples as edges. Every edge meets the other three, so $m=3$, andAny assignment of three colours to four vertices repeats a colour on a pair. Any triple containing that pair has at most two colours. Thus this uniform hypergraph satisfies the numerical hypothesis but has no required colouring. A single two-vertex edge would already provide a smaller counterexample if $k=2$ is allowed.

Here is the correlation graph of events argument that correctly applies to the printed colouring definition. Colour each vertex independently and uniformly with three colours, and let $B_E$ be the event that edge $E$ misses some colour. For $k\ge 1$, the inclusion-exclusion principle givesJoin two bad events when their edges intersect. This is a dependency graph of events, hence also a correlation graph of events, of maximum degree at most $m$: each $B_E$ depends only on colours on $E$, which are jointly independent of the colours on all disjoint edges. Precisely, the symmetric Lovász local lemma says that events of probability at most $q$ with such a graph have positive simultaneous avoidance probability whenever $eq(m+1)\le 1$. It follows from the asymmetric form $q\le x(1-x{)}^m$, taking $x=1/(m+1)$ for $m\ge 1$ and using $(m/(m+1){)}^m\ge e^{-1}$; the case $m=0$ is direct independence.

Thus a valid replacement sufficient condition isor the simpler, stronger condition $3e(m+1)\le (3/2{)}^k$. Under either replacement, avoiding all $B_E$ gives exactly the requested polychromatic hypergraph colouring. For infinitely many vertices, finite subfamilies can be coloured and compactness of the product of three-point spaces extends this to all constraints; positive simultaneous avoidance probability is only asserted for finite families. The printed $3^k$ bound cannot be justified by substituting monochromatic-event probabilities for missing-colour probabilities.

Partition $N=mw$ bits into $m$ tribes of size $w$, and let the tribes function be one when some tribe consists entirely of ones. Permuting tribes and permuting bits within a tribe acts transitively on the $N$ coordinates, so this is a transitive increasing event. Under independent bits of density $p$, its exact success probability isChoose $m=\lfloor (\log 2)2^w\rfloor$. Then $u(1/2)\to 1/2$, while $\log N=w\log 2+O(\log w)$. For fixed $0<x<1$, its unique $x$-quantile isPut $a_x=-\log (1-x)$. Expanding $1-e^{-a_x/m}=a_x/m+O_x(m^{-2})$ givesThus, for fixed $0<\epsilon <1/2$,This positive window prevents an improvement to $o(1/\log N)$ in the Friedgut-Kalai sharp threshold theorem. Moreover, as the fixed error parameter tends to zero, the displayed logarithmic coefficient is $\log (1/\epsilon )+\log \log (1/\epsilon )+o(1)$, so the theorem's logarithmic error dependence is also optimal up to constants. This last comparison takes $w\to \infty$ for each fixed $\epsilon$; the expansion is not being claimed uniformly for arbitrarily tiny $w$-dependent errors.

We prove the following precise form of the Friedgut-Kalai sharp threshold theorem. There is an absolute $C$ such that, for a nontrivial transitive increasing event $A\subseteq \{0,1{\}}^n$, $n\ge 2$, and $0<\epsilon <1/2$, its success probability increases from $\epsilon$ to $1-\epsilon$ over an interval of length at mostIn particular, ${\mu }_p(A)>\epsilon$ and $q-p\ge C\log ((1-\epsilon )/\epsilon )/\log n$, with $p<q<1$, imply ${\mu }_q(A)>1-\epsilon$. Here ${\mu }_p$ denotes independent bits equal to one with probability $p$. Symmetry means invariance under a permutation group acting transitively on the $n$ coordinates, not necessarily invariance under all coordinate permutations.

Use the allowed weighted-cube influence theorem in this exact form: an absolute $c>0$ exists such that, for every $n\ge 2$, $0<p<1$, and zero-one Boolean function of mean $u$ under ${\mu }_p$, some coordinate has pivotal probability at least $cu(1-u)\log n/n$. A coordinate's pivotal event is determined by the other coordinates and means that the two possible values of that coordinate give different outputs. This uniform form is a consequence of the weighted Kahn-Kalai-Linial theorem; it applies to nonsymmetric functions as well.

Write $u(p)={\mu }_p(A)$. Differentiating the finite product expression for $u(p)$, one coordinate at a time, givesThe last equality uses monotonicity, so each summand is the pivotal probability. This derives the Margulis–Russo formula in the normalization being used. Transitivity makes all the influences equal; applying the weighted influence theorem therefore yieldsNontriviality ensures $0<u(p)<1$ throughout $(0,1)$. Dividing and integrating the derivative of the log odds givesAt the two specified quantiles the left side is $2\log ((1-\epsilon )/\epsilon )$. Taking $C=2/c$ proves the window bound and, with a strict starting inequality, the asserted strict final inequality. This is the sharp threshold conclusion, rather than an assumption about a discontinuous finite-$n$ probability curve.

The Harris theorem for square-lattice bond percolation asserts $\theta (1/2)=0$, and hence $p_c\ge 1/2$. Work in the dual square lattice, declaring a dual bond open precisely when its primal bond is closed. At $p=1/2$ this is again independent bond percolation of density one half.

Around a square centered at the origin place four dual rectangles of dimensions $6r$ by $2r$, one on each side, with overlap squares at the corners. Long crossings of all four must meet in the corner squares and their union contains a dual circuit surrounding the central square. The fixed-aspect crossing bound and the Harris lemma give this dual barrier a probability at least $\eta :=c_3^4>0$, independent of $r$.

Take geometrically increasing integer radii, for example with a factor of eight, and shift by half a lattice spacing to lie on the dual lattice. The four-rectangle annuli then have disjoint bond sets. Their barrier events are independent. An infinite primal open path from the origin would have to avoid every closed dual circuit, so the independent annular barriers for percolation giveLetting $j\to \infty$ provesThe extension to smaller $p$ follows by the monotone coupling of Bernoulli percolation. Translation and a countable union also show that no infinite open cluster exists anywhere at $p=1/2$.

The Harris lemma says that, in any finite product of independent Bernoulli coordinates, two increasing events satisfyThe coordinates need not have the same parameter. For completeness, the same statement for increasing functions follows by induction on the number of coordinates. Conditional on the last bit, induction makes the conditional covariance nonnegative; the covariance of the two conditional means is $p(1-p)(a_1-a_0)(b_1-b_0)\ge 0$. The covariance decomposition proves the induction step. Complementing both events gives the inequality for decreasing events too. Finite-cylinder approximation extends it to the countable product setting when needed.

For increasing events $A_1,\dots ,A_r$ of a common probability $a$, their decreasing complements giveTherefore the nth root trick isThis is a lower bound for an individual success probability when a symmetric union is very likely. Without equality of the individual probabilities, the conclusion applies to their maximum. Independence among the $A_j$ is not required.

If $S=1$, the strict inequality in question already fails at time zero, almost surely, since $X_0=0$. Its probability is therefore zero, as required. Assume for the rest that $0<S<1$, so $M_0=S<1$.

Before time one the boundary inequality is equivalent to $M_t<1$. The process $M$ increases continuously between arrivals and has only downward jumps. Its first time $T<1$ of reaching level one must consequently be a continuous crossing, except possibly if an arrival occurs at the very same instant. All possible first crossing times belong to the finite deterministic setA uniform arrival hits none of these times almost surely. Thus $M_T=M_{T-}=1$ on $\{T<1\}$. This finite candidate set also proves the stopping-time property directly: determining whether a crossing has occurred by time $t$ only uses the arrival history up to $t$. The boundary set always contains time one because $X_1=S$, so $T\le 1$.

For a deterministic $r<1$, the martingale is bounded on $[0,r]$ by $S/(1-r)$. The optional stopping theorem for a bounded stopping time therefore givesMoreover $0\le M_{T\wedge r}\le 1$: if $T\le r$ its value is one, and otherwise no crossing has yet occurred. As $r\uparrow 1$, the stopped value tends to one on $\{T<1\}$ and to zero on $\{T=1\}$, because after the last arrival the unstopped process is zero. Dominated convergence now givesThis is the finite-jump boundary crossing identity. It uses uniform integrability of the bounded stopped values, not uniform integrability of the original family, which fails.

Let ${\xi }_j=B_{{\sigma }_1^j}-B_{{\sigma }_1^{j-1}}$. Continuity makes each displacement equal to $+1$ or $-1$. Reflection symmetry of a fresh Brownian motion gives probability one half to each sign, and the Strong Markov property makes these signs independent across successive exits. Thus $B_{{\sigma }_1^n}={\sum }_{j=1}^n{\xi }_j$.

For the normalized position, the characteristic function issince $\log \cos z=-z^2/2+O(z^4)$ near zero. By the Lévy continuity theorem, this provesThis is the Gaussian limit of a Brownian exit skeleton.

Compare along the subsequence $n=2^{2m}$ with ${\tau }_m={\sigma }_{2^{-m}}^{2^{2m}}$. Joint Brownian scaling of the successive exits givesThe right-hand side converges weakly to $N(0,1)$. The preceding part and path continuity give $B_{{\tau }_m}\to B_C$ almost surely on the original Brownian probability space. Hence its weak limit is also the law $N(0,C)$. Uniqueness of the weak limit forces $C=1$, so finally $\mathbb{E}{\sigma }_a=a^2$. Independence of the position and its random clock was not assumed in this comparison.