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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 an L2-bounded continuous martingale and ordered possibly infinite stopping times $\sigma \le \tau$, interpret the terminal stopped value as $M_{\infty }$. The Doob L2 maximal inequality bounds the expected squared path supremum. Localization of $M^2-[M]$ then gives $\mathbb{E}[M{]}_{\infty }<\infty$, so this square-minus-bracket local martingale has integrable supremum and is uniformly integrable. Apply the optional stopping theorem to it and to $M$. Expanding $(M_{\tau }-M_{\sigma }{)}^2$ makes the cross term vanish conditionally, proving the identity. Bounded ${\mathcal{F}}_{\sigma }$-measurable factors can be multiplied into either side.

A continuous local martingale $M$ with $M_0=0$ and deterministically bounded quadratic variation on every finite horizon is a square-integrable martingale, with $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t$. Localization, the Doob L2 maximal inequality, and the resulting integrable bound on the squared supremum justify removing the localizing sequence.

For a nonnegative submartingale $X$, the maximal tail bound is $\lambda \mathbb{P}({\max }_{k\le n}{X}_k\ge \lambda )\le \mathbb{E}[X_n;{\max }_{k\le n}{X}_k\ge \lambda ]$. Applied to the absolute value of a martingale and integrated over $\lambda$, it gives the Doob Lp maximal inequality; at $p=2$ this is the Doob L2 maximal inequality.

Let $X$ be a continuous martingale on $[0,1]$ with $X_0=0$ and $|X_t|\le C$. For the dyadic squared-increment sums, including the unfinished increment at time $t$, set $A_t^{(n)}={\sum }_k(X_{k{2}^{-n}\wedge t}-X_{(k-1)2^{-n}\wedge t}{)}^2$. The associated martingale transform $M_t^{(n)}=(X_t^2-A_t^{(n)})/2$ satisfies $\mathbb{E}(M_1^{(n)}{)}^2\le C^4$ and $\mathbb{E}(A_1^{(n)}{)}^2\le 10C^4$. Discrete martingale orthogonality, the path modulus of continuity and the Cauchy-Schwarz inequality show that the terminal transforms are Cauchy in $L^2$. The Doob L2 maximal inequality then gives $\mathbb{E}{\sup }_{t\le 1}|A_t^{(n)}-A_t^{(m)}{|}^2\to 0$. The sums need not be increasing in time before taking the limit. This provides an elementary construction of quadratic variation under boundedness.

For a L2-bounded continuous martingale, the terminal and maximal norms satisfyThe first inequality follows from almost sure convergence. Apply the Doob L2 maximal inequality on $[0,T]$ and then the monotone convergence theorem as $T\to \infty$ for the second. These are equivalent norms on the space modulo indistinguishability of stochastic processes; neither inequality needs a zero initial value.

For a continuous local martingale starting at zero, finite expected terminal quadratic variation is equivalent to being an L2-bounded continuous martingale. Stop at bounded path and bracket levels and use $\mathbb{E}(M_t^{\tau }{)}^2=\mathbb{E}[M{]}_{t\wedge \tau }$. Uniform integrability removes the localization; Doob L2 maximal inequality justifies convergence of squares. The terminal identity is $\mathbb{E}{M}_{\infty }^2=\mathbb{E}[M{]}_{\infty }$. For nonzero initial values one must additionally assume $M_0\in L^2$.

For a density-dependent Markov jump process with finitely many jump vectors, bounded nonnegative rates and Lipschitz drift $F(x)={\sum }_{\ell }\ell {\beta }_{\ell }(x)$, an initial state converging in probability to $x_0$ gives uniform convergence in probability on every finite horizon to the ODE $\dot{x}=F(x)$, $x(0)=x_0$. The compensated martingale has expected maximal squared size $O(N^{-1})$ by the Doob L2 maximal inequality; Gronwall's inequality transfers that estimate to the solution error.

// Target: probability-and-statistics.bigb

For a left-continuous locally bounded adapted process $H$ and a continuous semimartingale $X$, the elementary predictable left-endpoint approximations tend pointwise to $H$. Decompose $X$ into a continuous local martingale and a continuous finite-variation process. After localization, dominated convergence with respect to the variation measure handles the finite-variation integral. The Itô isometry and Doob L2 maximal inequality handle the martingale integral. The omitted partial last interval is bounded by the local bound on $H$ times the modulus of continuity of $X$. This proves uniform convergence on compacts in probability and is the grid-sum foundation of the Itô product rule.

For $dX=\sigma (X)dB+b(X)dt$, $X_0=0$, and ${\sigma }^2+b^2\le A(1+X^2)$, the Doob L2 maximal inequality and the Cauchy-Schwarz inequality imply, for $t\le 1$, $F(t)\le 8A{\int }_0^t(1+F(s))ds$, where $F(t)=\mathbb{E}{\sup }_{s\le t}|X_s{|}^2$. The Gronwall inequality yields $F(t)\le e^{8At}-1$. Apply the calculation first to stopped paths; truncating locally Lipschitz functions and using this uniform bound gives exit probabilities at most $(e^{8A}-1)/n^2$, proving nonexplosion on the unit time interval.

For Brownian motion started at zero, let $f_n$ equal $|x|$ outside $(-1/n,1/n)$ and $n{x}^2/2+1/(2n)$ inside. Its derivative is the clipped sign function $g_n(x)=\max \{-1,\min \{1,nx\}\}$. The Itô integral $M_t^{(n)}={\int }_0^t{g}_n(B_s)d{B}_s$ converges uniformly on bounded time intervals in $L^2$ to a continuous martingale $M$, by the Itô isometry and Doob L2 maximal inequality: the squared maximal error is at most $8\sqrt{2T/\pi }/n$. The Itô formula gives $\frac{n}{2}{\int }_0^t{1}_{\{|B_s|<1/n\}}ds=f_n(B_t)-f_n(0)-M_t^{(n)}$. Along $n=k^2$, the maximal errors are summable, so First Borel-Cantelli lemma gives uniform almost-sure convergence. The ratio-one interpolation for shrinking-window occupation integrals extends it to every integer $n$. The limit is $|B_t|-M_t=L_t^0$, the local time of a semimartingale in the Tanaka formula convention. The unnormalized factor $n$ gives twice this limit.

On each finite horizon the displayed Radon-Nikodym derivative changes Wiener measure into the law of $dX=dB-\mu Xdt$. Stop at coordinate magnitude n so the Novikov condition applies. Under each stopped tilted measure the coordinate solves the drifted equation, and the Gronwall inequality with the Doob L2 maximal inequality bounds its maximal second moment uniformly in n. The probability of reaching n therefore tends to zero; this proves that the unstopped density has expectation one. No unrestricted large-time Novikov estimate is needed.

True. The unit-time increments $Z_k=B_k-B_{k-1}$ are independent and identically distributed random variables with standard normal distribution, hence integrable with mean zero. The strong law of large numbers gives $B_n/n\to 0$ almost surely at integer times.

To control the intervals between integers, put $M_n={\sup }_{0\le u\le 1}|B_{n+u}-B_n|$. Stationary increments and the Doob L2 maximal inequality give $\mathbb{E}{M}_n^2\le 4$. Thus the Markov inequality yieldsThe Borel-Cantelli lemma, followed by a countable intersection over positive rational $\epsilon$, implies $M_n/n\to 0$ almost surely. For $n\le t\le n+1$,Both terms tend to zero on the same probability-one event, proving $\boxed{B_t/t\to 0\text{ almost surely}}$ for continuous time as well.

For a simple predictable process, define its stochastic integral byCommon refinement of the deterministic partitions makes this definition independent of the representation. Each summand is a martingale increment multiplied by information already available at its left endpoint, so the resulting process is a continuous square-integrable martingale starting at zero. Its terminal value is the same sum with $t$ replaced by $t_n$.

Write ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $j<k$, then $Z_{t_j}{\Delta }_{j}M{Z}_{t_k}$ is ${\mathcal{F}}_{t_k}$-measurable. By the conditional expectation identity for increments of a martingale,The conditional bracket isometry for stopped martingale increments givesFor completeness, this identity follows by applying the optional stopping theorem to $M$ and $M^2-[M]$ and expanding the square. The latter process is a true uniformly integrable martingale here: the Doob L2 maximal inequality bounds $\mathbb{E}{\sup }_t|M_t{|}^2$, stopped square identities and Fatou's lemma give $\mathbb{E}[M{]}_{\infty }<\infty$, and its absolute supremum is bounded by ${\sup }_t|M_t{|}^2+[M{]}_{\infty }$. Consequently all the stopped identities pass to the limit.

Multiplying the conditional identity by $Z_{t_k}^2$, taking expectations, and using the vanished cross terms proves the Itô isometry:

To extend the stochastic integral, use the finite quadratic-variation measure ${\nu }_M(C)=\mathbb{E}\int 1_{C}d[M]$ on the predictable sigma-algebra. The density of simple predictable processes for finite measures gives approximations to every $H\in L^2({\nu }_M)$. The Itô isometry makes the corresponding terminal integrals Cauchy in $L^2$, and the Doob L2 maximal inequality makes the whole processes Cauchy in expected squared supremum. Their limit defines a continuous martingale, and the isometry persists. For a locally bounded predictable process, choose increasing stopping times tending to infinity on which it is bounded, also restrict the time horizon, and perform this construction after stopping. The isometry shows that the stopped definitions agree on overlaps, so they patch into a uniquely defined continuous local martingale $H\cdot M$.

First stop the solution on leaving $[-n,n]$, at ${\tau }_n$, and set $F_n(t)=\mathbb{E}{\sup }_{u\le t}|X_{u\wedge {\tau }_n}{|}^2$. The stochastic integral representation, the Doob L2 maximal inequality and the Cauchy-Schwarz inequality give, for $0\le t\le 1$,Here stopping ensures square integrability before the calculation, and $8{\sigma }^2+2t{b}^2\le 8({\sigma }^2+b^2)$ explains the constant. The Gronwall inequality gives $F_n(t)\le e^{8At}-1$. For globally Lipschitz functions as coefficients, the global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a nonexplosive solution. Letting $n$ increase and using Fatou's lemma proves the slightly stronger maximal second-moment bound under linear growth:

For coefficients that are only locally Lipschitz functions, let ${\pi }_n(x)=\max (-n,\min (x,n))$ and replace the coefficients by $\sigma ({\pi }_n(x))$ and $b({\pi }_n(x))$. These are globally Lipschitz functions, agree with the originals on $[-n,n]$, and obey the same linear growth condition for an SDE, since $|{\pi }_n(x)|\le |x|$. The globally defined solutions agree until their common exits from $[-n,n]$ by pathwise uniqueness. The uniform preceding estimate and Markov's inequality implyPatch the solutions before their increasing exit times. Their limiting lifetime exceeds $1$ almost surely by this bound. Thus a pathwise unique strong solution exists throughout $[0,1]$ even for locally Lipschitz coefficients.

Use coordinates $1,2,3$ for paper, scissors and stone. Let $X^N$ take values in the lattice simplex $S_N=\{(n_1,n_2,n_3)/N:n_i\ge 0,{\sum }_i{n}_i=N\}$. A uniformly selected unordered pair consists of types $i$ and $j$, $i\ne j$, with probability $2n_i{n}_j/(N(N-1))$. Losing copies the winner, so the possible jumps of the continuous-time Markov chain areTies make no change. For arbitrary event rate $\lambda$, the respective transition rates are $2\lambda N{x}_1{x}_3/(N-1)$, $2\lambda N{x}_2{x}_1/(N-1)$ and $2\lambda N{x}_3{x}_2/(N-1)$. ChooseThen the rates are exactly $N{x}_1{x}_3$, $N{x}_2{x}_1$, $N{x}_3{x}_2$. The infinitesimal generator of this cyclic imitation chain iswhere $\mathcal{J}=\{e_1-e_3,e_2-e_1,e_3-e_2\}$ and the three $\beta$ functions are, in that order, $x_1{x}_3,x_2{x}_1,x_3{x}_2$. Its drift is precisely

Here is a precise fluid limit with its error control. Suppose $X_0^N\to x_0$ in probability, and let $x$ solve $\dot{x}=F(x)$, $x(0)=x_0$. For every finite $T$,Indeed, the martingale decomposition of a density-dependent jump process givesThis follows by compensating each transition count by its integrated rate; the compensated counts are square-integrable martingales, with variance equal to the expected compensator. Since $\Vert \ell {\Vert }^2=2$ and $x_1{x}_2+x_2{x}_3+x_3{x}_1\le 1/3$ on the simplex, the Doob L2 maximal inequality, applied coordinatewise, givesThe polynomial drift is Lipschitz on the compact simplex, with some finite constant $L$. It preserves the sum of the coordinates, and each coordinate solves an equation of the form ${\dot{x}}_i=x_i{h}_i(x)$, so nonnegativity is preserved. Thus the ordinary differential equation stays in the simplex and exists for all time. The Gronwall inequality giveswhich proves the claim by Markov's inequality. This proves the required version directly, rather than invoking an unspecified approximation theorem.

For an interior trajectory, differentiating the logarithm of the product of numbers givesThe same identity for the product itself holds on the boundary, so $x_1(t)x_2(t)x_3(t)$ is conserved. A strictly positive initial product keeps the deterministic trajectory away from the boundary. Except at $(1/3,1/3,1/3)$ these interior trajectories are periodic: the strictly concave function ${\sum }_i\log x_i$ has smooth compact level curves surrounding its unique maximum, and the vector field is nonzero and tangent to each such curve.

The finite Markov chain behaves differently at long times. For $P(x)=x_1{x}_2{x}_3$, the change in $P$ during the first jump isThe other two terms are cyclic permutations. Their first-order contributions cancel, leaving the exact product decay in a cyclic imitation chain identityFor $N\ge 3$, while all three counts are positive, their product is at least $N-2$, so $P(X_t^N)\ge (N-2)/N^3$. A missing type cannot reappear. If $\tau$ is the first loss of a type, thenOnce only two types remain, one always wins, so its count increases until an absorbing monochromatic vertex is reached, almost surely in finite time. The cases of one or two initially present types follow the same argument without the first stage. Every finite chain eventually becomes monochromatic, whereas an interior deterministic solution retains its positive product. The fluid limit holds on each fixed finite interval; it does not justify exchanging the limits $N\to \infty$ and $t\to \infty$.

A simple predictable process can be writtenwhere each ${\xi }_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. Define its integral to start at zero:Refining the partition leaves this sum unchanged. Every summand is a martingale, by conditioning its future increment on the information at $t_j$. Bounded coefficients and the $L^2$ bound on $M$ make the integral an L2-bounded continuous martingale. Its terminal value is the same sum with $M_{\infty }$ substituted at any infinite endpoint; $L^2$ martingale convergence supplies this limit.

We need the conditional square identity, rather than assuming independent increments. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_t|M_t{|}^2<\infty$. Localize the square identity for $M^2-[M]$ at level and bracket stopping times. The stopped identity and this maximal bound give $\mathbb{E}[M{]}_{\infty }<\infty$ by monotone convergence. Nowis integrable, so localization passes to conditional expectations and $M^2-[M]$ is a uniformly integrable martingale. Expanding a squared increment and using $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ provesThe infinite endpoint follows from $L^2$ convergence and integrability of the bracket. The initial value $M_0$ need not be zero: the integral itself is zero-starting.

In the square of the terminal integral, all cross terms vanish. For $j<k$, the $j$th increment and both coefficients in its cross term are measurable at $t_k$, whereas the conditional mean of the $k$th increment there is zero. For the diagonal terms, the preceding conditional identity can be multiplied by the bounded ${\xi }_j^2$. ThereforeThis proves the Itô isometry for these simple integrands. Ordered stopping-time intervals also work, replacing the conditional identity by optional stopping theorem, but deterministic intervals suffice for the approximation requested here.

For that approximation, put $\mu (A)=\mathbb{E}{\int }_0^1{1}_A(\omega ,s)ds$ on the predictable sigma-algebra. This is a finite measure. That sigma-algebra is generated, apart from the zero-time slice of measure zero, by rectangles $A\times (s,t]$ with $A\in {\mathcal{F}}_s$. The indicator of every such rectangle is a bounded simple predictable process.

Here is a density proof. Let $V$ be the closed linear span of these indicators in $L^2(\mu )$, and let $\mathcal{D}$ contain the predictable sets whose indicators belong to $V$. The whole space belongs to $\mathcal{D}$; complements stay in $\mathcal{D}$; and disjoint countable unions do also, because their finite partial indicator sums converge in $L^2$ under the finite measure. Thus $\mathcal{D}$ is a Dynkin system containing the generating rectangle pi-system. The Pi-lambda theorem makes it the entire predictable sigma-algebra. Bounded predictable simple functions therefore belong to $V$. Truncating any $L^2(\mu )$ function and then approximating its truncated range by finitely many levels shows that every such function belongs to $V$. A finite sum of rectangle indicators can be rewritten on one common deterministic time partition, with the coefficient on each interval measurable at its left endpoint; it is exactly an allowed simple process.

Consequently choose $H^n$ of this form with $\Vert H^n-H{\Vert }_{L^2(\mu )}<1/n$. This provesIt is density of simple predictable processes for finite measures, with a proof of the actual approximation step.

Finally apply the preceding isometry to $M_t=B_{t\wedge 1}$, whose bracket is $t\wedge 1$. The elementary integrals $I_n={\int }_0^1{H}_s^{n}d{B}_s$ satisfyDefine the Itô integral as the $L^2$ limit of these elementary integrals. The same identity shows independence of the approximating sequence, linearity, mean zero, and $\mathbb{E}({\int }_0^1{H}_{s}d{B}_s{)}^2=\mathbb{E}{\int }_0^1{H}_s^{2}ds$. Applying the Doob inequality to the difference processes also gives uniform-in-time $L^2$ convergence on $[0,1]$, with a continuous martingale version of the integral.

If $X$ is a P-Brownian motion, its starting value is deterministic, so Q5(a) applies. The correction $[X,M]$ has finite variation and does not alter quadratic variation; Q1(b) gives $[\tilde{X}{]}_t=t$ under Q. The Lévy characterization of Brownian motion therefore identifies $\tilde{X}$ as a Q-Brownian motion with the same starting value.

For the likelihood, use Wiener measure ${\mathbb{Q}}^x$ as the reference measure and write its coordinate as ${\omega }_t=x+W_t$. Setand let $L=\mathcal{E}(M)$ be its stochastic exponential. We must prove it is a true martingale; unverified large-horizon Novikov estimates would not suffice.

Fix $T$ and stop at ${\tau }_n=\inf \{s:|{\omega }_s|\ge n\}$, with $n>|x|$. The bounded integrand satisfies the Novikov condition, so $L_{T\wedge {\tau }_n}$ defines a probability $P_n$ on ${\mathcal{F}}_T$. By the result just established,is Brownian under $P_n$ on this horizon. Therefore $d{\omega }_s=d{W}_s^{(n)}-\mu 1_{\{s\le {\tau }_n\}}{\omega }_{s}ds$. The Gronwall inequality givesThe Doob L2 maximal inequality then bounds its squared expectation by $C_T=2e^{2|\mu |T}(x^2+4T)$, uniformly in $n$. In particular $P_n({\tau }_n\le T)\le C_T/n^2$. On $\{{\tau }_n>T\}$ the stopped and unstopped densities agree, soContinuous Brownian paths are bounded on $[0,T]$, and the indicators increase to one. The monotone convergence theorem proves $\mathbb{E}{L}_T=1$. Since a nonnegative local martingale is a supermartingale, constant expectation at every time makes $L$ a true martingale.

Under its tilted measure, ${\hat{W}}_t=W_t+\mu {\int }_0^t{\omega }_{s}ds$ is Brownian and $d{\omega }_t=d{\hat{W}}_t-\mu {\omega }_{t}dt$. The explicit linear solution gives uniqueness of this law, identifying it with ${\mathbb{P}}^x$. Consequently the Ornstein-Uhlenbeck likelihood relative to Wiener measure isThis is a finite-horizon density statement; it does not assert absolute continuity of the full infinite-horizon path laws.

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.