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A martingale residual is observed minus model-expected event count:where ${\hat{\Lambda }}_i$ is the fitted individual cumulative hazard. Under an adequate model it estimates the terminal value of a counting-process martingale.

Fit a model omitting the continuous explanatory variable $z$, plot ${\hat{M}}_i$ against $z_i$, and add a flexible smooth curve. A curve fluctuating around zero without structure supports omission. A monotone or curved trend indicates that event incidence still depends on $z$, suggesting inclusion of $z$ or a nonlinear transformation of it. The residuals are highly skewed, so the smoothed trend is more informative than an assumption of Gaussian scatter.

Let $\tau =T_r\wedge T_R$. The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r<|z|<R$, so Itô formula shows thatis a bounded martingale. By the optional sampling theorem for a supermartingale applied to this martingale,where $p={\mathbb{P}}_x(T_r<T_R)$. Solving gives the planar Brownian annulus hitting probability

Put ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$ and $\tau =T_0\wedge T_y$. Before $\tau$, the integer-valued increment belongs to $\{-1,0,1\}$. The martingale property givesTheir sum is at least $1/2$, so each conditional probability is at least $1/4$. From any state in $\{1,\dots ,y-1\}$, a run of at most $y$ upward moves reaches $y$ and has conditional probability at least $4^{-y}$. Applied in successive blocks of $y$ steps, this givesso $\tau <\infty$ almost surely.

The stopped process $X_{n\wedge \tau }$ takes values in $[0,y]$, hence is a bounded martingale and has uniform integrability. The optional sampling theorem for a supermartingale givesbecause $X_{\tau }$ is zero or $y$. Therefore

Let $L=N-H\cdot M$. It is a continuous square-integrable martingale, and the assumed bracket identity givesfor every continuous square-integrable martingale $K$. Choose $K=L$ and use part (i):Thus $L_{\infty }=0$ almost surely, and the conditional expectation property gives $L_t=\mathbb{E}[L_{\infty }\mid {\mathcal{F}}_t]=0$ for every $t$. Hence $N=H\cdot M$ up to indistinguishability of stochastic processes.

The martingale product identity says that $N_t{K}_t-[N,K{]}_t$ is a martingale. Passing to the terminal values of the square-integrable martingales and using $N_0=K_0=0$ givesThe quadratic covariation identity for a stochastic integral isApplying the same product identity to $H\cdot M$ and $K$ therefore gives

Define the deterministic function $f(t)=\mathbb{E}[M_t^2]$. The independent increments from part (i) show that $f$ is increasing and that $M_t^2-f(t)$ is a martingale. Mean-square continuity follows from path continuity and the Gaussian laws, so $f$ is continuous.

The Itô formula also says that $M_t^2-[M{]}_t$ is a local martingale. Their difference $[M{]}_t-f(t)$ is therefore a continuous finite-variation process that is also a local martingale. By the theorem that a continuous finite-variation local martingale is constant, and because the difference starts at zero,for all $t\ge 0$ almost surely.

Fix $0\le r\le t$. Since $M$ is a martingale,Every finite vector consisting of $M_{t+s}-M_t$ and past values $M_{r_1},\dots ,M_{r_n}$ has a multivariate normal distribution. Therefore uncorrelated jointly normal variables are independent, so the increment is independent of every finite vector of past values. A Monotone class theorem then extends this to independence from ${\mathcal{F}}_t=\sigma (M_r:0\le r\le t)$. This is the independent increments of a Gaussian martingale.

False. Let $Z\sim N(0,1)$ and define $X_t=tZ$. This is a centered continuous Gaussian process. Its natural filtration satisfies $Z=X_s/s\in {\mathcal{F}}_s^X$ for every $s>0$, and hence, for $0<s<t$,with positive probability. Thus $X$ is not a martingale and does not belong to the stated martingale class.

The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r_1<|z|<r_2$. The Itô formula therefore makes $\log |B_{t\wedge \tau }|$ a bounded martingale. The optional sampling theorem for a supermartingale giveswhere $p=\mathbb{P}(|B_{\tau }|=r_1)$. Solving this linear equation yields the planar Brownian annulus hitting probability

Let $\rho =t\wedge {\tau }_0\wedge {\tau }_1$ and apply the Itô formula to $u(t-s,B_s)$ for $0\le s\le \rho$. The heat equation cancels the drift, so the stopped process is a bounded martingale. The optional sampling theorem for a supermartingale gives $u(t,x)={\mathbb{E}}_x[u(t-\rho ,B_{\rho })]$. On the three mutually exclusive terminal events, the initial and Dirichlet boundary conditions identify this value asThis is the probabilistic representation of the heat equation with time-dependent Dirichlet data.

The Strong Markov property and part (b) show that, before $\tau ={\tau }_x\wedge {\tau }_y$,Thus $F(V_{t\wedge \tau })$ is a bounded martingale. From the given stochastic differential equation,The Itô formula says that the drift of $F(V_t)$ isIt must vanish. Dividing by $(1-v{)}^2/2$ and using the algebraic identity supplied in the question gives

For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (b) therefore identifies $X_n$ with $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]$, so $(X_n,{\mathcal{F}}_n)$ is a martingale.

The atom formula also provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The Martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.

Let $(M_k,{\mathcal{F}}_k{)}_{k=0}^N$ be a martingale, let $a<b$, and let $U_N[a,b]$ count the completed upcrossings of $[a,b]$ by time $N$. Doob upcrossing inequality statesEquivalent conventions for a process with a prescribed initial holding add the corresponding endpoint term.

Let $D\subset {\mathbb{R}}^d$ be bounded, choose $R$ with $D\subset B(0,R)$, and let $T$ and ${\tau }_R$ be the respective exit times. Then $T\le {\tau }_R$. Sinceis a martingale, the optional sampling theorem for a supermartingale at ${\tau }_R\wedge n$ givesMonotone convergence theorem now gives

For $u\in C^2(\overline{D})$, Itô formula shows thatis a martingale. Take expectations and let $t\to \infty$. The function $u$ is bounded on the compact set $\overline{D}$, while $\Delta u$ is bounded and ${\mathbb{E}}_{x}T<\infty$ by part (a). The dominated convergence theorem therefore gives Dynkin formula for Brownian motion

If $S\le T$ are stopping times and $A\in {\mathcal{F}}_S$, then $U=S{1}_A+T{1}_{A^c}$ is a stopping time. The key identity isThis uses the ordering $S\le T$: without it, $A$ need not be known when $T$ occurs. Such pastings test the martingale identity through expectations at bounded stopping times.

Let $V_t=\mathbb{E}[P_t{P}_t^{\top }\mid {\mathcal{F}}_{t-1}]$ be positive definite. Market completeness restricts each parent atom to at most $n$ successors, while positive definiteness forces exactly $n$ independent successor price vectors. The positive martingale deflator supplied by the fundamental theorem of asset pricing then has the unique one-step ratio $Z_t=P_t^{\top }{V}_t^{-1}{P}_{t-1}>0$. Its product deflates the prices into martingales.

For a finite-valued predictable process $K$, this is a stopping time and increases to infinity. The stopped martingale transform uses coefficients $K_s{1}_{\{s\le {\sigma }_j\}}$, which are predictable and bounded by $j$. Thus a transform of a discrete-time martingale is a local martingale without a global boundedness assumption. The stop is before, not after, the first large coefficient is used.

A martingale can diverge to $+\infty$ almost surely when its increments are finite but not uniformly bounded. For independent Bernoulli random variables with $\mathbb{P}(Y_n=1)=2^{-n}$,has centered integrable increments. The First Borel-Cantelli lemma makes the rare negative jumps occur only finitely often, after which every increment equals one. The expectation remains zero at every finite time despite this pathwise divergence.

For $S_t=S_0{e}^{\mu t+\sigma W_t}$, shifting Brownian drift bymakes $S_t=S_0{e}^{(r-{\sigma }^2/2)t+\sigma W_t^Q}$ under an equivalent martingale measure $Q$. The discounted stock is then a martingale.