Itos Formula and Brownian Martingales — Examples

For each $m\ge 1$, set $L_0^m=B_0$ and $L_s^m=B_{k{2}^{-m}}$ on $(k{2}^{-m},(k+1)2^{-m}]$, $k\ge 0$. Every $L^m$ is predictable by [F1]. Define ${\beta }_s={\lim }_m{L}_s^m$ where the limit exists finitely and ${\beta }_s=0$ otherwise. The convergence set is predictable by the countable Cauchy criterion, so $\beta$ is predictable. On the common full event of continuous Brownian paths, the left grid points increase to $s$ and ${\beta }_s=B_s$ simultaneously for all $s\ge 0$. This construction makes no exceptional path set part of the definition.

For any fixed integer $r\ge 1$, Gaussian moments in [F2] and Tonelli on the predictable map ${\beta }^{2r}$ give $E{\int }_0^T{\beta }_s^{2r}ds=\frac{(2r-1)!!}{r+1}{T}^{r+1}<\mathrm{\infty }\mathrm{.}$ The integrand 1 has energy T. Thus every polynomial in $\beta$ and time used below is predictable and has finite energy on every finite horizon. To quantify step approximation, use $\mathrm{\mid }{x}^r-y^r\mathrm{\mid }\le r\mathrm{\mid }x-y\mathrm{\mid }(\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }{)}^{r-1}\mathrm{.}$ Cauchy--Schwarz and [F2] show, uniformly for $0\le u\le s\le T$, $E\mathrm{\mid }{B}_s^r-B_u^r{\mathrm{\mid }}^2\le C_{r,T}(s-u)\mathrm{.}$ Indeed the fourth moment of the increment is $3(s-u{)}^2$, while the expectation of $(\mathrm{\mid }{B}_s\mathrm{\mid }+\mathrm{\mid }{B}_u\mathrm{\mid }{)}^{4r-4}$ is uniformly bounded by Gaussian even moments; when r=1 this factor is 1.

Fix $0<t\le T$ and take $m=2^k$, $h=t\mathrm{/}m$, $t_j=jh$, $D_j=B_{t_{j+1}}-B_{t_j}$. The predictable step process with coefficients $B_{t_j}^r$ converges to ${\beta }^r$ in $L^2(dt\otimes P)$, since step 2.1 bounds its squared error integral by $C_{r,T}th$. Its integral is ${\sum }_j{B}_{t_j}^r{D}_j$: truncate the finitely many coefficients to bounded values, apply [F3], and let the truncation bound tend to infinity. The coefficient errors tend to zero in $L^2$ by [F4]; independence gives $E\mathrm{\mid }({\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}_j)D_j{\mathrm{\mid }}^2=hE\mathrm{\mid }{\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}}_j{\mathrm{\mid }}^2$, so the finite sums converge in $L^2$ too. This includes r=0 with coefficient 1 without truncation. Consequently ${\sum }_j{B}_{t_j}^r{D}_j⟶{\int }_0^t{\beta }_s^{r}d{B}_s\text{in }{L}^2(P)\mathrm{.}$

The weighted centered quadratic error $Q_k={\sum }_j{B}_{t_j}^{n-2}(D_j^2-h)$ has mean zero. Different summands are orthogonal in $L^2$: for i<j the earlier summand and $B_{t_j}^{n-2}$ are known at $t_j$, and the remaining centered increment has conditional mean zero by [F2] and [F5]. All products are integrable by Gaussian moments and Cauchy--Schwarz. The variance is therefore $E{Q}_k^2=2h^2{\sum }_{j}E{B}_{t_j}^{2n-4}\le C_{n,T}th⟶0,$ interpreting the power as 1 when n=2. On the common continuity event, the sums ${\sum }_{j}h{B}_{t_j}^{n-2}$ converge to ${\int }_0^t{\widehat{B}}_s^{n-2}ds$ by continuity and the left Riemann sums. The latter integral exists on every normalized path and is measurable by joint measurability and the parameter-integral statement of [F4], using positive and negative parts.

Expand each power increment by the finite binomial identity and sum: $B_t^n-B_0^n=n{\sum }_j{B}_{t_j}^{n-1}{D}_j+\left(\genfrac{}{}{0px}{}{n}{2}\right){\sum }_j{B}_{t_j}^{n-2}{D}_j^2+{\sum }_{\mathrm{\ell }=3}^n\left(\genfrac{}{}{0px}{}{n}{\mathrm{\ell }}\right){\sum }_j{B}_{t_j}^{n-\mathrm{\ell }}{D}_j^{\mathrm{\ell }}\mathrm{.}$ For each $\mathrm{\ell }\ge 3$, independence, Gaussian moments and Cauchy--Schwarz give $E{\sum }_j\mathrm{\mid }{B}_{t_j}{\mathrm{\mid }}^{n-\mathrm{\ell }}\mathrm{\mid }{D}_j{\mathrm{\mid }}^{\mathrm{\ell }}\le C_{n,T}m{h}^{\mathrm{\ell }\mathrm{/}2}=C_{n,T}t{h}^{\mathrm{\ell }\mathrm{/}2-1}⟶0.$ This uses the even moment of order $2\mathrm{\ell }$ to bound the absolute moment of order $\mathrm{\ell }$; the factor of degree n minus ell has bounded moments on [0,T], and is 1 when ell=n. The remainder is empty for n=2. Since $B_0=0$ almost surely, steps 3.1 and 4.1 prove the desired identity at fixed t by uniqueness of limits in probability. Explicitly $L^1$ errors and $L^2$ errors tend to zero in probability by Markov's inequality applied to their absolute values and squares; almost-sure Riemann-sum convergence implies convergence in probability by dominated convergence of exceedance indicators. If two candidate limits differ by more than epsilon, at least one approximation error exceeds epsilon/2, so their difference vanishes almost surely.

By [F3] choose continuous versions for the countably many powers' stochastic integrals. The Lebesgue terms along $\widehat{B}$ are continuous on every path, and the original B is continuous on a common full event. Intersect that event with the countably many equalities from step 5.1 at rational t for all integer n and with the stochastic-integral continuity events. Continuity extends every identity to all real times on this measurable full event. This establishes the process convention in the Example without asserting measurability of the entire all-time equality set.

A second finite telescope yields $t{B}_t={\sum }_j{t}_j{D}_j+{\sum }_{j}h{B}_{t_{j+1}}$ almost surely, since $B_0=0$ (its coefficient is in any case zero). The first sum converges in $L^2$ to ${\int }_0^{t}sd{B}_s$ by [F3], because the deterministic left-step times converge uniformly to s. The second converges on the continuity event to ${\int }_0^t{\widehat{B}}_{s}ds$. The same uniqueness and rational-continuity argument as step 5.1 and step 6.1 gives $t{B}_t={\int }_0^{t}sd{B}_s+{\int }_0^t{\widehat{B}}_{s}ds$ on a full event at all times. Subtract three times this equality from the n=3 identity and use [F3]'s linearity to obtain $B_t^3-3t{B}_t=3{\int }_0^t({\beta }_s^2-s)d{B}_s$. The n=2 formula similarly gives the square identity.

The two original polynomial processes are adapted because B is adapted. Gaussian moments give their square integrability at each finite time. Their deterministic-time equality to the continuous square-integrable martingales in step 7.1 therefore transfers the conditional martingale identity by [F5]; adaptation is checked separately. The integral coefficient energies are finite on every horizon: the square coefficient has energy $4{\int }_0^{T}sds$, and the cubic coefficient has energy $9{\int }_0^{T}E(B_s^2-s{)}^{2}ds=18{\int }_0^T{s}^{2}ds$, by [F2]. Continuity holds on the Brownian continuity event.

At t=0 both identities for n>=2 vanish on the common zero-start event. The n=2 coefficient is 1 and its zero power means the constant function 1. Separately, the n=1 identity is $B_t={\int }_0^{t}1d{B}_s$ on a common full event, and the constant integrand has finite energy T; one does not substitute a meaningless $0\cdot B^{-1}$ term. The n=0 constant function has zero increment and is outside the displayed range. Full AC supplies the countable-choice assumption in the Riemann-to-Lebesgue bridge of [F4], and is inherited through [F5] and the countable integral construction; no claim about divergent negative powers is needed.

First identity: apply [F1] to $g(t,y)=x\exp ((\mu -{\sigma }^2\mathrm{/}2)t+\sigma y)$ along $\widehat{B}$, which is indistinguishable from $B$ and has drift $0$ and diffusion $1$. Then ${\mathrm{\partial }}_{t}g=(\mu -{\sigma }^2\mathrm{/}2)g$, ${\mathrm{\partial }}_{y}g=\sigma g$ and ${\mathrm{\partial }}_{yy}^{2}g={\sigma }^{2}g$, so $d{X}_t=\mu X_{t}dt+\sigma X_{t}d{B}_t$ up to indistinguishability. Positivity and continuity hold everywhere by the explicit definition. The composition defining $X$ is adapted, so [F1] also makes $X$, $\mu X$ and $\sigma X$ progressive and predictable. If $C_T$ is the upper bound in [F2], then ${\int }_0^T\mathrm{\mid }\mu X_s\mathrm{\mid }ds\le \mathrm{\mid }\mu \mathrm{\mid }T{C}_T$ and ${\int }_0^T(\sigma X_s{)}^{2}ds\le {\sigma }^{2}T{C}_T^2$ on every path. Together with $X_0=x$ and the displayed decomposition these verify the Ito-class assertion explicitly.

Second identity: because $X$ was defined by a positive exponential, taking the ordinary logarithm gives the everywhere pathwise identity $\log X_t=\log x+(\mu -{\sigma }^2\mathrm{/}2)t+\sigma {\widehat{B}}_t\mathrm{.}$ Since $\widehat{B}$ and $B$ are indistinguishable, this is exactly $d\log X_t=(\mu -{\sigma }^2\mathrm{/}2)dt+\sigma d{B}_t$ up to indistinguishability. The explicit bounds of [F2] also verify directly that $X$ stays in a compact subinterval of $(0,\mathrm{\infty })$ and its logarithm is bounded on every finite horizon; no localization or SDE existence theorem is being smuggled into the argument.

Boundary and consistency cases: at $t=0$ the identities read $X_0=x$ and $\log X_0=\log x$; for $\sigma =0$ the formulas reduce to the deterministic exponential and its logarithm; for $\mu =0$ the drift of $\log X$ is $-{\sigma }^2\mathrm{/}2$; and $x>0$ is required for the logarithm. The formulas concern the explicitly defined process, not existence for an SDE, and AC enters only through [F3].

Fix $0<b<a$, $\theta >0$ and an integer $n\ge 1$. Set $m=2^n$, $t_j=jT\mathrm{/}m$, and use the original adapted process on this grid. Define $J$ as the first index $j\in \{0,\dots ,m\}$ with $B_{t_j}>b$, or $m$ if there is no such index. For $j<m$, the event $\{J\le j\}$ is the finite union ${\bigcup }_{k\le j}\{B_{t_k}>b\}$ and is in ${\mathcal{F}}_{t_j}$; the event for $j=m$ is the whole space. Thus $J$ is a bounded discrete stopping index for the grid filtration. By [F1] and [F2], $E{Z}_{t_J}=1$. This variable is measurable and integrable, being a finite sum of integrable grid values times indicators.

Let $E_n=\{{\max }_{0\le j\le m}{B}_{t_j}>b\}$. On $E_n$ the selected value satisfies $B_{t_J}>b$ and $t_J\le T$, whence $Z_{t_J}\ge \exp (\theta b-{\theta }^{2}T\mathrm{/}2)$. Positivity therefore gives $P(E_n)\le \exp ({\theta }^{2}T\mathrm{/}2-\theta b)\mathrm{.}$ No continuous-time hitting time or finiteness of an unbounded hitting time has entered.

Write $M_T={\sup }_{t\le T}{\widehat{B}}_t$. This is the supremum over the countable union of the nested dyadic grids: for any $t$ in the interval there are grid times tending to it, and continuity gives convergence of the path values. The supremum is finite, since a continuous function on a compact interval is bounded. Measurability also follows from the countable supremum. The normalized grid events ${\widehat{E}}_n=\{{\max }_j{\widehat{B}}_{t_j}>b\}$ increase to $\{M_T>b\}$ and have the same probabilities as $E_n$, since the original and normalized paths agree on the common full event. Consequently [F4] and step 2.1 give $P(M_T>b)\le \exp ({\theta }^{2}T\mathrm{/}2-\theta b)$. Normalizing on another full event gives the same $M_T$ on their full intersection, so its distribution is independent of the choice.

Choose $\theta =b\mathrm{/}T$ in step 3.1, the positive minimizer of the quadratic, to get $P(M_T>b)\le e^{-b^2\mathrm{/}(2T)}$. Since $\{M_T\ge a\}\subseteq \{M_T>b\}$ for every $0<b<a$, take the explicit sequence $b_k=a(1-1\mathrm{/}k)$, $k\ge 2$, and let $k$ tend to infinity in the numerical upper bounds. Continuity of the exponential gives $P(M_T\ge a)\le e^{-a^2\mathrm{/}(2T)}$. This last argument does not assume that a dyadic grid attains the continuous maximum or that $M_T$ has no atoms.

The parameter range is $a,T>0$. At $a=0$ the probability is one and the limiting bound is one; for $a<0$ the probability is also one, but the displayed formula would be less than one and is not asserted. At $T=0$ and $a>0$ the probability is zero and division by $T$ is not used. With $T$ fixed, the bound tends to zero as $a\to \mathrm{\infty }$; with $a>0$ fixed, it tends to one as $T\to \mathrm{\infty }$ and to zero as $T\downarrow 0$. The real exponential martingale has random magnitude; its integrability follows from its Gaussian unit mean, not a deterministic modulus. Only finite-grid optional sampling is used, so no uniform-integrability assertion for an unbounded stopped family is needed. AC has exactly the interface uses in [F5].

The two functions are space-time harmonic: by [F2] both have vanishing Laplacian and no time dependence, so the lifetime-local assertion [F1] applies with the relatively open set $U=[0,\mathrm{\infty })\times {\mathbb{R}}^2$, whose lifetime is infinity. Hence $B_t^1{B}_t^2=p(B_t)$ and $(B_t^1{)}^2-(B_t^2{)}^2=q(B_t)$ are continuous local martingales.

Time-capped stopping: fix $R>0$ and for each integer $n\ge 1$ set $K_n=[0,n]\times \{y:\mathrm{\mid }y\mathrm{\mid }\le R\}$. This is a compact subset of $U=[0,\mathrm{\infty })\times {\mathbb{R}}^2$ containing $(0,0)$ in its relative interior; the exit time in [F1] is exactly ${\tau }_{K_n}=n\wedge {\tau }_R$. The harmonic theorem makes this a stopping time and supplies its stopped integral identity and square-integrable martingale. Also $\{{\tau }_R\le t\}=\{{\tau }_{K_n}\le t\}$ whenever $n>t$, so ${\tau }_R$ is a stopping time. Given any finite horizon $T$, choose an integer $n>T$; then $t\wedge {\tau }_{K_n}=t\wedge {\tau }_R$ for every $0\le t\le T$. Thus the compact-stopped process and integral from [F1] coincide with the disc-stopped ones throughout that horizon. This proves the disc-stopped martingale property for every pair of finite times without treating a spatial disc as compact space-time.

Explicit form of the stopped integrals: from [F1] and the stopping identity, $p(B_{t\wedge {\tau }_R})={\int }_0^t{1}_{[0,{\tau }_R]}(s)(B_s^{2}d{B}_s^1+B_s^{1}d{B}_s^2),$ $q(B_{t\wedge {\tau }_R})={\int }_0^t{1}_{[0,{\tau }_R]}(s)(2B_s^{1}d{B}_s^1-2B_s^{2}d{B}_s^2)\mathrm{.}$ On each finite horizon the integrands agree with the bounded predictable compact-stopped integrands from step 2.1, including the endpoint indicator; their squared Euclidean norms are bounded by $R^2$ and $4R^2$. Thus each scalar component has finite expected energy and the finite sums are square-integrable martingales. The identities hold up to indistinguishability: intersect the probability-one identities for integer horizons.

Boundary and consistency cases: at $t=0$ both processes start at $0$; as integer $R\to \mathrm{\infty }$, every continuous path is bounded on each compact time interval, so ${\tau }_R\to \mathrm{\infty }$ and the stopped processes eventually equal the unstopped ones on that interval; the proof here asserts square-integrability after disc stopping using bounded gradients; unboundedness on the plane alone is not an obstruction to a true martingale, and no such obstruction is claimed; for $B$ starting at the origin the disc contains the starting point for every $R>0$; and AC enters only through [F4].

Applying Dynkin: for each integer $n\ge 1$, [F0] shows that the stopping time $\tau \wedge n$ is bounded, so [F1] applied to the function $f$ of [F2] gives $E_x[f(B_{\tau \wedge n}^x)]=f(x)+E_x{\int }_0^{\tau \wedge n}\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}(B_s^x)ds$. On the common event $B_0^x=x$, one has $B_s^x\in [-a,b]$ for $s\le \tau$. Since $f=u$, $f^{\mathrm{\prime }\mathrm{\prime }}=u^{\mathrm{\prime }\mathrm{\prime }}=-2$ on a neighbourhood of $[-a,b]$, the right-hand side equals $u(x)-E_x(\tau \wedge n)$.

Left-hand limit: on $\{\tau <\mathrm{\infty }\}$ one has $B_{\tau \wedge n}^x\to B_{\tau }^x\in \{-a,b\}$ by continuity of the path, hence $f(B_{\tau \wedge n}^x)\to 0$; on $\{\tau =\mathrm{\infty }\}$ (a null set by [F3]) the sequence stays bounded and the conclusion is not needed. Since $f$ is bounded, dominated convergence gives $E_x[f(B_{\tau \wedge n}^x)]\to 0$.

Conclusion: combining steps 1.1 and 2.1, $u(x)-E_x(\tau \wedge n)\to 0$, so $E_x(\tau \wedge n)\to u(x)$; by monotone convergence of the nondecreasing sequence $(\tau \wedge n)$ the limit of the expectations is $E_x\tau$, hence $E_x\tau =(x+a)(b-x)$. At $x=0$ this is $ab$.

Boundary and consistency cases: for $x\to -a$ or $x\to b$ the formula tends to $0$, consistent with the starting point being at the boundary; for $a=b$ and $x=0$ it gives $a^2$; the cutoff agrees with $u$ on a neighbourhood of the whole closed interval, so the computation is unaffected by the modification; the exit time is finite almost surely by [F3], and the argument does not need $E\tau <\mathrm{\infty }$ in advance because monotone convergence allows the value $+\mathrm{\infty }$ and the computation identifies it as finite; the bounded-stopping hypothesis of Dynkin's formula is met by $\tau \wedge n$ at each $n$; and the inherited uses of AC are exactly those recorded in [F5], while all Brownian data remain hypotheses.

Stopping identity at a bounded continuous time: fix $T>0$, put $\rho =\tau \wedge T$, and for $n\ge 1$ round $\rho$ upward to the grid $\{jT{2}^{-n}:0\le j\le 2^n\}$, obtaining ${\rho }_n$. At a grid point $u<T$, $\{{\rho }_n\le u\}=\{\rho \le u\}\in {\mathcal{F}}_u$, so ${\rho }_n$ is a bounded stopping index for the sampled discrete martingale. Discrete optional sampling gives $E_x{M}_{{\rho }_n}=E_x{M}_0=e^{-2\mu x}$. It also identifies $M_{{\rho }_n}$ as a conditional expectation of the fixed integrable variable $M_T$ at the grid stopped sigma-algebra, so [F4] makes $(M_{{\rho }_n}{)}_n$ uniformly integrable. Continuity gives $M_{{\rho }_n}\to M_{\rho }$ almost surely, hence in probability; [F4] upgrades this to $L^1$, proving $E_x{M}_{\tau \wedge T}=e^{-2\mu x}$. This uses the discrete theorem only on finite grids and proves the continuous bounded-time passage explicitly.

Define $X_{\tau }$ by evaluation on $\{\tau <\mathrm{\infty }\}$ and as $0$ otherwise, and define $M_{\tau }=e^{-2\mu X_{\tau }}$. These are measurable: bounded-time evaluations are limits of the finite-grid evaluations in step 1.1, and the finite-exit value is their eventual value as integer horizons increase. Since $\tau <\mathrm{\infty }$ almost surely by [F3], $M_{\tau \wedge T}\to M_{\tau }$ as $T\to \mathrm{\infty }$; the deterministic bound in [F5] gives $E_x{M}_{\tau }=e^{-2\mu x}$ by dominated convergence.

The value at the exit: $X_{\tau }\in \{-a,b\}$ almost surely by continuity and the definition of $\tau$ as the first exit, so $M_{\tau }=e^{-2\mu X_{\tau }}$ equals $e^{2\mu a}$ on $\{X_{\tau }=-a\}$ and $e^{-2\mu b}$ on $\{X_{\tau }=b\}$. Writing $p:=P_x(X_{\tau }=b)$, the identity of step 2.1 becomes $e^{-2\mu x}=p{e}^{-2\mu b}+(1-p)e^{2\mu a}$.

Solving: $p={\displaystyle \frac{e^{-2\mu x}-e^{2\mu a}}{e^{-2\mu b}-e^{2\mu a}}}={\displaystyle \frac{1-e^{-2\mu (x+a)}}{1-e^{-2\mu (a+b)}}}$, multiplying numerator and denominator by $e^{-2\mu a}$; the denominator is nonzero because $\mu \mathrm{\ne }0$ and $a+b>0$ make the two endpoint exponentials distinct.

The case $\mu =0$: then $X=Z$ is Brownian motion started at $x$, and Two-sided Brownian exit probability gives $P_x(X_{\tau }=b)=(x+a)\mathrm{/}(a+b)$ directly; this agrees with the limit of the formula of step 4.1 as $\mu \to 0$.

Boundary and consistency cases: for $x\to -a$ the probability tends to $0$ and for $x\to b$ it tends to $1$, consistent with the starting point being at the boundary; for $\mu \to 0$ step 4.1 has a removable singularity with limit $(x+a)\mathrm{/}(a+b)$; for $\mu <0$ the same computation applies with the sign carried through; the stopping time is not bounded, and the passage to the limit was justified by the uniform boundedness of $M_{\tau \wedge t}$ from [F5] rather than by assuming uniform integrability of an unbounded family; the exit time is finite almost surely by the law of the iterated logarithm; and AC supplies the conditional-expectation interface and the inherited Brownian and LIL choice requirements, as declared in the Given hypotheses.