Codex @codex  0

Take $X=Y=B$, a standard Brownian motion with $B_0=0$. The Itô formula and $⟨B{⟩}_t=t$ giveIf this nonnegative stochastic process, which starts from zero, were a local martingale, question 4(i) would make it a supermartingale and force its expectation to be at most zero. But $\mathbb{E}(B_t^2/2)=t/2>0$ for $t>0$. Hence a Stratonovich integral is in general not a local martingale.

Under the standard local martingale definition, $X_0$ is integrable. For a nonnegative $X$, the Fatou lemma and a localizing sequence giveHence every $X_t$ is integrable, and part (a) applies. Therefore a nonnegative discrete-time local martingale is a true martingale. This is stronger than the continuous-time supermartingale conclusion.

For fixed integers $s\le t$, $A_s=\{\tau =s\}$ belongs to ${\mathcal{F}}_s$ and $X_s=0$ there. The supermartingale property and nonnegativity implyTherefore $X_t=0$ almost surely on $A_s$. Taking the finite union over $s\le t$ gives $X_t=0$ on $\{\tau \le t\}$ almost surely; taking the countable intersection over all integer $t$ makes the assertion simultaneous at every time. Thus zero is absorbing for a nonnegative supermartingale in discrete time.

With constant coefficients, $\lambda =(\mu -r)/\sigma$ is constant, so the stochastic exponential $Z_t=e^{rt}{Y}_t$ is a true martingale. Under its equivalent martingale measure $\mathbb{Q}$, the stock follows the Black-Scholes model $d{S}_t=S_t(rdt+\sigma d{W}_t^{\mathbb{Q}})$. The Gaussian exponential moment gives the value of the power optionThe option delta is ${\partial }_{s}V=V/(2s)$, so the replicating strategy holdsHalf the wealth value is in the stock and half in the bank account, with continuous rebalancing. To check the self-financing portfolio property under the original measure, the Itô formula gives $d{V}_t=\frac{1}{2}(r+\mu )V_{t}dt+\frac{1}{2}\sigma V_{t}d{W}_t={\theta }_{t}d{S}_t+{\eta }_{t}d{B}_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the Black-Scholes model has the necessary finite Gaussian moments, and the explicit strategy attains $V_0=\mathbb{E}[Y_T\sqrt{S_T}]$. The supermartingale bound from part (b) therefore proves its minimality. The cost is independent of the physical drift $\mu$.

Use the slotted ALOHA model with one outstanding packet per active station and a collision channel: a slot with exactly one attempt succeeds, whereas an idle slot or a slot with two or more attempts delivers nothing. Let $Y_t$ be independent Poisson random variables of mean $\nu$, independent also of the transmission choices. Arrivals join after the current slot's attempts. Given $(S_t,N_t)=(S,N)$, the attempt count has a binomial distribution with parameters $N$ and $1/S$. Consequently, for $N\ge 1$,For $N=0$, set $p_0=1$ and $p_1=0$. The equation $N_{t+1}=N_t+Y_t-1_{\{Z_t=1\}}$ adds arrivals and removes precisely one packet following a successful slot. The estimate $S_t\ge 1$ controls the common transmission probability and updates using the idle, success, or collision feedback. The conditional distribution of the next state depends only on $(S_t,N_t)$, so this pair is a Markov chain.

Away from the reflecting floor in $S$, its one-step conditional expectations areWhen $N$ and $S$ are large with $N/S\approx \kappa$, the Poisson limit theorem gives $p_0\approx e^{-\kappa }$ and $p_1\approx \kappa e^{-\kappa }$. Treating these approximate drifts as derivatives motivates the fluid approximation of slotted ALOHA:The floor at $S=1$ is negligible in this interior scaling. This calculation motivates the ordinary differential equations; it is not a proof of a stochastic fluid limit.

For $(a,b,c)=(2-e,0,1)$, putThen $g^{\prime }(\kappa )=(e-2+\kappa )e^{-\kappa }>0$ and $g(1)=0$. On the domain $s>0$, introduce time $\tau$ by $d\tau /dt=1/s$. Differentiating $\kappa =n/s$ giveswhere $r(\kappa )=\kappa [1-(e-2+\kappa )e^{-\kappa }]$.

The function $r$ is strictly increasing from zero to infinity, and $r(1)=e^{-1}$. To check the monotonicity explicitly,Indeed, with $d=3-e\in (0,1)$,because its discriminant is $d^2-4d<0$; and $e^{\kappa }\ge 1+\kappa$. Thus the scalar ordinary differential equation for $\kappa$ points toward the unique value ${\kappa }_{\ast }$ satisfying $r({\kappa }_{\ast })=\nu$. Every trajectory with $s(0)>0$, $n(0)\ge 0$ has bounded $\kappa$ and $\kappa (\tau )\to {\kappa }_{\ast }$. Positivity of $s$ follows from $s(\tau )=s(0)\exp ({\int }_0^{\tau }g(\kappa (u))du)$.

If $0\le \nu <e^{-1}$, then ${\kappa }_{\ast }<1$ and $g({\kappa }_{\ast })<0$. Thus $s(\tau )$ decays exponentially for large $\tau$, andThis is finite-time draining of a fluid model. The unmodified equations have $n/s$ undefined at the origin, so the statement that trajectories converge to the origin needs this endpoint interpretation; continuing with an absorbing zero trajectory is an additional fluid-model convention, not a solution of the displayed equations at the origin.

If $\nu >e^{-1}$, then ${\kappa }_{\ast }>1$ and $g({\kappa }_{\ast })>0$. Now $s(\tau )$ grows exponentially, physical time tends to infinity, andIn particular the trajectory escapes to infinity. Even without the ratio analysis, $\dot{n}\ge \nu -e^{-1}>0$, since ${\max }_{\kappa \ge 0}\kappa e^{-\kappa }=e^{-1}$.

For the actual Markov chain, it is essential not to replace the finite-$N$ success probability by $e^{-1}$: a single backlogged packet can succeed with probability one. Instead, for $N\ge 2$, maximizing over $S\ge 1$ gives the ALOHA throughput boundIf $\nu >e^{-1}$, choose $K_0$ and $r_0<\nu$ so that $p_1(N,S)\le r_0$ for all $N>K_0$, uniformly in $S$. For sufficiently small $\theta >0$,because its derivative at zero is $r_0-\nu <0$. Independence of arrivals and attempts giveswhenever $N>K_0$. For any integer $K\ge K_0$, stop at the stopping time ${\sigma }_K=\inf \{t:N_t\le K\}$. The stopped process $e^{-\theta N_{t\wedge {\sigma }_K}}$ is a nonnegative supermartingale. The optional stopping theorem, first at bounded times and then by a limit, yields the exponential-supermartingale escape boundFrom every state with $N\le K$, a Poisson distribution arrival burst has a uniformly positive probability of sending the next backlog to at least $K+m$, for any fixed $m>0$. From there the probability of never revisiting $N\le K$ is at least $1-e^{-\theta m}$. By the Markov property at successive visits, the probability of infinitely many visits to this strip is zero: each visit has a uniform positive chance of being the last. This holds for every integer $K\ge K_0$, soTherefore the Markov chain is transient for every finite choice of $a,b,c$. The proof is uniform in $S$ and does not infer stochastic transience merely from the approximate ordinary differential equations.

Every martingale is a supermartingale, with equality in its defining conditional expectation. Taking expectations and using the tower property of conditional expectation givesso induction yields

For a finite horizon $N$, let $U_N[a,b]$ be the number of completed upcrossings by time $N$. Use the predictable strategy that holds one unit of the process after a visit below $a$ until the next visit above $b$. For a supermartingale, the expected gain of this nonnegative predictable martingale transform is nonpositive. Pathwise, the completed trades earn at least $(b-a)U_N[a,b]$, while an unfinished final trade can lose at most $(X_N-a{)}^{-}$. HenceTaking expectations and using $X_N\ge 0$ gives the Doob upcrossing inequalityAs $N\to \infty$, monotone convergence yields

For every rational $0\le a<b$, part a implies $U[a,b]<\infty$ almost surely. The intersection of these probability-one events over the countable collection of rational pairs still has probability one. On this event, ifsome rational interval $[a,b]$ lies strictly between them, forcing infinitely many upcrossings, a contradiction. Thus $X_n$ has an extended limit almost surely.

The limit cannot be $+\infty$ on a set of positive probability: Fatou lemma and the supermartingale property giveNonnegativity excludes $-\infty$. Therefore $X_n$ converges almost surely to a finite random variable, proving the almost sure supermartingale convergence theorem in this case.

Write ${\mu }_t=({\mu }_t^{(1)},\dots ,{\mu }_t^{(d)}{)}^T$ and define the market price of riskIt is bounded by hypothesis. The stochastic exponentialis a true martingale by the Novikov condition. Define the equivalent measure $Q$ by $dQ=Z_{T}dP$. The Girsanov theorem makesa Brownian motion under $Q$. After discounting by the bank account, every risky price has zero drift and is a $Q$-local martingale.

The discounted wealth of an admissible self-financing strategy is a nonnegative local martingale and hence a supermartingale. If an arbitrage existed, its zero initial value would imply nonpositive expected terminal discounted wealth under $Q$, while that wealth is nonnegative and positive with positive $Q$-probability. This contradiction proves that the market has no arbitrage; it is the needed direction of the equivalent local martingale measure criterion.

For a bounded stopping time $T\le N$, writeThe indicator is measurable at time $n-1$. Taking conditional expectations term by term and using the supermartingale inequality gives $\mathbb{E}{\hat{M}}_T\le \mathbb{E}{\hat{M}}_0$. The same calculation gives equality for the martingale $M$. Since their initial values agree, $\mathbb{E}{M}_T\ge \mathbb{E}{\hat{M}}_T$.

Condition on the information through time $n-1$. Independence of ${\xi }_n$ and the definition of the Bellman equation for terminal-wealth utility implybecause the actual holding is one candidate in the supremum. Thus $V(n,X_n)$ is a supermartingale for every strategy.

For the stated strategy it is a martingale, soEvery competing strategy has expected terminal value at most $V(0,X_0)$ by the supermartingale inequality from part (a), evaluated at the bounded time $N$. Hence ${\theta }^{\ast }$ is optimal.

Let ${\tau }_n$ localize the nonnegative local martingale $M$. For $s\le t$,Conditional Fatou lemma and nonnegativity giveThus $M$ is a supermartingale, recovering the general fact about a nonnegative local martingale.

For the bond maturing one period later,If $r_t\ge 0$, this is at most $Y_{t-1}$, which is exactly the supermartingale property.

The product rule and self-financing identity show that $XY$ is the stochastic integral of $H$ against the vector of deflated prices $YP$, hence is a local martingale. It is nonnegative by admissibility and positivity of $Y$. Every nonnegative local martingale is a supermartingale, so

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.

For $x\ne 0$, the Strong Markov property at the first step gives the discrete mean-value identityAt $x=0$, $v(0)=1$ while the same average is at most one. Thus $v$ is a bounded superharmonic function on ${\mathbb{Z}}^3$. Conditioning on the natural filtration and using the one-step Markov property givesTherefore $(v(X_n))$ is a nonnegative supermartingale.

The almost sure supermartingale convergence theorem says that a supermartingale whose negative parts have uniformly bounded expectations converges almost surely to a finite integrable limit. In particular, every nonnegative supermartingale converges almost surely.

Here the process is uniformly bounded, say $0\le X_n\le C$. Let $X_{\infty }={\lim }_n{X}_n$, whose existence follows from the theorem. The dominated convergence theorem then givesThe limit on the right also exists directly because the expectations of a supermartingale form a decreasing sequence.

For a continuous local martingale $M$ with $M_0=0$, its stochastic exponential isThe Itô formula gives $d\mathcal{E}(M{)}_t=\mathcal{E}(M{)}_{t}d{M}_t$, so $\mathcal{E}(M)$ is a positive continuous local martingale. Every nonnegative local martingale is a supermartingale, because localization and the Conditional Fatou lemma turn the localized martingale equality into the supermartingale inequality.

For $z=x+iy$ with $|x|\le 1$ and $0<y\le 1$, the given Reverse SLE derivative martingale starts fromIt is a nonnegative local martingale and therefore a supermartingale. Since its second factor is at least one,

Choosewhich is possible exactly because $\kappa >8$. At height $y_n=2^{-n}$, take a horizontal grid of spacing comparable to $y_n$ in $[-1,1]$. The Markov inequality gives, at each grid point,There are $O(y_n^{-1})$ grid points, so the probability that the bound fails anywhere on level $n$ is at most $C{y}_n^{1-2\alpha -8/\kappa }$. These probabilities are summable. The Borel-Cantelli lemmas therefore give an almost surely finite random constant controlling every sufficiently fine grid, and enlarging it handles the finitely many remaining levels.

Every point of the half-rectangle lies within a fixed hyperbolic distance of one of these grid points at comparable height. The Koebe distortion theorem compares the two derivatives by a universal factor. Hence an almost surely finite random $C$ satisfiesfor all $x\in [-1,1]$ and $0<y\le 1$. This proves the Reverse SLE derivative bound above the space-filling threshold.