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Let a simple symmetric random walk start from a deterministic integer $S_0=i$, and put $L_n={\sum }_{k=0}^{n-1}{1}_{\{S_k=0\}}$. Then $M_n=|S_n|-L_n$ is a martingale with $M_0=|i|$. The visit count includes time $0$ if $i=0$.

For every integer $s$ and step $\xi \in \{-1,1\}$, the pathwise identityholds with $\mathrm{sgn}(0)=0$. Subtract the indicator function and sum over time to obtainThe sign function of the current state is a bounded predictable process for the next increment, so the sum is a martingale transform. Alternatively, averaging the two possible next absolute values shows directly that the conditional drift is $1_{\{S_k=0\}}$. Bounds $|S_n|\le |i|+n$ and $L_n\le n$ verify integrability at each finite time.

This is an absolute-value version of the discrete Tanaka formula. The correction records the occasions when the random walk is at the corner of the absolute-value function; away from zero that function is locally linear along a single step.

Once a nonnegative martingale hits zero, it stays zero almost surely. Conditional expectation on $\{M_n=0\}$ gives $\mathbb{E}[M_{n+1}{1}_{\{M_n=0\}}]=0$; nonnegativity forces the next value to vanish there. A countable intersection makes these assertions simultaneous for all times, proving absorption at the first zero.

Let a nonnegative martingale start at one, let $T$ be its first zero, and let $T_R$ be its first hit of $[R,\infty )$. Suppose $\mathbb{E}[(M_{n+1}-M_n{)}^2\mid {\mathcal{F}}_n]\ge {\sigma }^2{1}_{\{T>n\}}$ and $M_{T_R}\le \tau R$ on finite hits, for thresholds $R\ge 1$. Then $M_{T\wedge T_R\wedge n}^2-{\sigma }^2(T\wedge T_R\wedge n)$ is a submartingale. The bounded stopped values have mean one, so their second moment is at most $\tau R$. Consequently $\mathbb{E}(T\wedge T_R)\le (\tau R-1)/{\sigma }^2$. For $\tau \ge 1$ this implies the weaker bound ${\tau }^{2}R/{\sigma }^2$. The threshold restriction prevents an impossible overshoot requirement below the starting value.

For an adapted process with right-continuous paths and a stopping time $T$, $X_t^T=X_{t\wedge T}$ is adapted. At fixed $t$, round $t\wedge T$ upward on finite partitions of $[0,t]$. Every approximating value is ${\mathcal{F}}_t$-measurable, and right continuity gives the limit. Completeness handles a null exceptional set if right continuity holds only almost surely. A martingale assumption and boundedness of $T$ are unnecessary.

For independent standard Gaussian random variables, the partial sums have characteristic functions $\exp (-u^2{v}_n/2)$, where $v_n={\sum }_{k\le n}{a}_k^2$. A finite almost-sure limit implies convergence of these characteristic functions to one continuous at zero; $v_n\to \infty$ would instead give zero at every nonzero $u$. Conversely, bounded $v_n$ makes the partial sums an martingale bounded in L2 space, which converges almost surely and in L2 space.

Let independent increments ${\xi }_k$ equal $4^k$ or $-4^k$ with probabilities $2^{-k-1}$ each, and zero otherwise. Every partial sum $M_n={\sum }_{k\le n}{\xi }_k$ is integrable and is a martingale. The Borel-Cantelli lemma makes only finitely many increments nonzero almost surely, so $M_n$ has a finite almost-sure limit $Y$. Yet $\mathbb{E}|Y|=\infty$: the disjoint events that only the $k$th increment is nonzero have probabilities at least $c{2}^{-k}$ for $c={\prod }_j(1-2^{-j})>0$, and contribute at least $c{2}^k$ each. This does not contradict the martingale convergence theorem, whose integrability conclusion requires a uniform first-moment bound.

// Target: markov-process.bigb

A submartingale with ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable random variable. The positive-part bound also bounds negative parts because $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$. The result specializes to an $L^1$-bounded martingale, and hence to every nonnegative martingale. It does not by itself give convergence in L1; uniform integrability supplies that stronger conclusion.

If a martingale has increments with $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then each one-sided deviation of magnitude $t>0$ from $M_0$ has probability at most $\exp (-t^2/(2{\sum }_i{c}_i^2))$. If all $c_i$ vanish, the martingale is constant. The inequality gives concentration from bounded increments without requiring independent increments.

Suppose $\epsilon =o(1)$ and $\epsilon \ge n^{-1/6}$. The breadth-first exploration of a binomial random graph has deterministic drift $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$ up to $3\epsilon n$ steps. Its weighted martingale has variance $O(\epsilon n)$. For $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$, the Doob L2 maximal inequality makes its maximum smaller than $h{\epsilon }^{2}n/6$ with high probability. Then $A_t>0$ throughout $[h\epsilon n,(2-h)\epsilon n]$, giving a graph component of order $(2-o(1))\epsilon n$.

For the upper expectation bound, let $K=\lceil {\epsilon }^{-3}\rceil$. The dominating binomial branching process has branching survival probability $(2+o(1))\epsilon$ by the binomial branching survival correction. Its branching process conditioned on extinction has mean $1-\epsilon +O({\epsilon }^2+\epsilon /n)$ and total-progeny expected value $O(1/\epsilon )$. Therefore $\mathbb{E}{N}_{\ge K}\le n\rho +O(n/(\epsilon K))$. Since $L_1\le K+N_{\ge K}$ and $K=o(\epsilon n)$, this yields the matching upper bound. Controlling rare large components is necessary to conclude an expected value asymptotic from a typical-size statement.

For any real separable Hilbert space $H$, any orthonormal basis $(e_j)$ and independent standard normal random variables $(Z_j)$, the displayed series converges almost surely for each fixed $h\in H$ and in L2 space. Indeed, the Parseval identity for a Hilbertian basis gives ${\sum }_j⟨h,e_j{⟩}^2=\Vert h{\Vert }^2$, so the partial sums form an L2 space-bounded martingale. The martingale convergence theorem supplies both limits. Every finite linear combination has the normal distribution with variance $\Vert {\sum }_j{u}_j{h}_j{\Vert }^2$. Thus its finite-dimensional characteristic function is $\exp (-\Vert {\sum }_j{u}_j{h}_j{\Vert }^2/2)$, independent of the chosen orthonormal basis. This proves equality of the finite-dimensional distributions of the resulting isonormal Gaussian processes; pathwise equality needs a coupling.

Let $0<\alpha <1$, $X_0=x\in [0,1]$, and let $B_{n+1}$ be a binary random variable satisfying $\mathbb{E}[B_{n+1}\mid {\mathcal{F}}_n]=X_n$. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable and $B_{n+1}$ is ${\mathcal{F}}_{n+1}$-measurable. The updateremains in $[0,1]$ and defines a bounded martingale. For example, take independent uniform random variables $U_{n+1}$ independent of the current filtration and set $B_{n+1}=1_{\{U_{n+1}\le X_n\}}$.

The binary identity $B_{n+1}^2=B_{n+1}$ givesand henceThe L2-bounded martingale convergence theorem gives a terminal random variable $X_{\infty }$ with almost sure convergence and convergence in L2. Iteration followed by the bounded convergence theorem yieldsThus $X_{\infty }$ takes only $0$ and $1$, and preservation of its expected value gives a Bernoulli distribution with parameter $x$. The terminal conditional expectation representation gives the exact mean squared errorThe finite-time paths can stay strictly inside $(0,1)$ while their limit lies entirely at the endpoints. The positive conditional variance in the interior makes permanent convergence to an interior value impossible.

For a compensated jump martingale starting at zero, with jumps bounded in absolute value by $c$ and predictable quadratic variation at most $v$ on the interval, the displayed maximal bound holds. For $0<\theta c<3$, the elementary exponential boundmakes the exponential of $\theta M$ minus that multiple of its predictable bracket a nonnegative supermartingale. Stop at the first crossing of $\eta$ and use its mean bound. Taking $\theta =\eta /(v+c\eta /3)$ yields the one-sided bound; apply the same argument to $-M$ and use the union bound for the displayed two-sided version. The elementary estimate follows from the exponential series and $k!\ge 23^{k-2}$ for $k\ge 2$. If $v=0$, the compensated martingale is constant and the crossing probability is zero; the parameter choice above is needed only for $v>0$.

For $X_t=W_t+\alpha t$ with an independent equiprobable drift $\alpha \in \{-a,a\}$, the likelihood ratio between the two drifts is $e^{2a{X}_t}$. Multiplying the normal distribution densities of successive increments proves this ratio for every observation partition; continuity then gives the full observed-path posterior. The Bayes' theorem gives $q_t=\mathbb{P}(\alpha =a\mid {\mathcal{F}}_t^X)=(1+e^{-2a{X}_t}{)}^{-1}$. Thus the observed drift is $a(2q_t-1)=a\tanh (a{X}_t)$. The process ${\hat{W}}_t=X_t-{\int }_0^{t}a\tanh (a{X}_s)ds$ is an observed-filtration martingale with quadratic variation $t$, so the Lévy characterization of Brownian motion makes it a Brownian motion. The observed state is a scaled diffusion with hyperbolic tangent drift.

With independent up moves of physical probability $p$ and risk-neutral probability $q$, the Radon-Nikodym derivative on the first $n$ periods is the displayed likelihood ratio, where $N_n$ counts up moves. Its conditional one-step multiplier has physical expectation one, so $L_k$ is a positive mean-one martingale. The terminal state-price density is $L_n/B_n$ for deterministic bank-account value $B_n$.

For a zero-start continuous local martingale $M$ with $[M{]}_{\infty }\le C$ deterministically, its stochastic exponential is a uniformly integrable martingale. For each $p>1$, ${\sup }_t\mathbb{E}\mathcal{E}(M{)}_t^p\le e^{p(p-1)C/2}$. Its terminal value is strictly positive, so it defines an equivalent infinite-horizon change of measure by the Girsanov theorem.

For asset dynamics $dS=\mathrm{diag}(S)(\mu dt+\sigma dW)$, let $\lambda ={\sigma }^{-1}\mu$ and $Z=\mathcal{E}(-\int {\lambda }^{T}dW)$. The Itô product rule cancels the drift of each $Z{S}^i$, leaving the displayed stochastic integral. When $\mu$ is bounded and $\sigma {\sigma }^T\ge \epsilon I$, $\lambda$ is bounded and the Novikov condition makes $Z$ a true martingale on every finite horizon. If $\sigma$ is also bounded, each $Z{S}^i/S_0^i$ satisfies the Novikov condition as a stochastic exponential with bounded integrand ${\sigma }_i-\lambda$, so it too is a true martingale.

For a zero-starting solution with $|\sigma |\le C$, the Itô isometry gives $\mathbb{E}{X}_t^2\le C^{2}t$, so $X$ is a true martingale on every finite horizon. The Itô formula represents the displayed process as $2{\int }_0^t{X}_s\sigma (X_s)d{B}_s$, whose expected squared integrand integral is at most $2C^4{t}^2$. Thus it too is a true square-integrable martingale. No continuity or Lipschitz condition on the coefficient is needed once a solution is given.

Every bounded harmonic function on a graph is constant if the connected locally finite graph is recurrent. Along its random walk on a graph, the harmonic values form a bounded martingale and converge almost surely. Recurrence visits each vertex infinitely often, making every vertex value a subsequential limit of the same convergent sequence. Thus every value agrees.

If $u$ is a bounded harmonic function on a domain $D$, continuous on its closure, and the Brownian exit time $T_D$ is almost surely finite from $z$, then $u(z)={\mathbb{E}}_z[u(B_{T_D})]$. The Itô formula and compact localization make $u(B_{t\wedge T_D})$ a bounded martingale. bounded convergence theorem at the exit time proves the representation even if the domain is unbounded. Removing boundedness of $u$ can invalidate this conclusion through loss of uniform integrability.

A discrete-time martingale with a common deterministic bound on all increments either converges finitely or has $\mathrm{liminf}{X}_n=-\infty$ and $\mathrm{limsup}{X}_n=+\infty$, almost surely. Stopping on first crossing below $-A$ gives a lower bound $-A-M$ on the overshoot. Adding $A+M+X_0^{-}$ produces a nonnegative martingale, so the martingale convergence theorem forces finite convergence on the event of never crossing. Apply the same argument to $-X$.