Further Results on Stochastic Integrals

Itō’s formula becomes especially useful when we choose a function to cancel a drift term. This converts a question about a diffusion into a martingale identity, often with an ordinary differential equation determining the function. The examples below use this idea for exit probabilities and exponential functionals, then study running maxima and the local-time term that appears when the function is an absolute value.

Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.

Scale functions and exit probabilities

Suppose \(X_t=B_t+\int_0^t b(X_s)\,ds\), where \(B_0=0\), \(b:\mathbb R\to\mathbb R\) is continuous, and the adapted continuous solution is defined for all times under consideration. For \(F\in C^2\), Itō’s formula gives

\[F(X_t)=F(0)+\int_0^tF'(X_s)\,dB_s+ \int_0^t\left(\frac12F''+bF'\right)(X_s)\,ds.\]

We cancel the drift by solving \(F''+2bF'=0\) with \(F(0)=0\) and \(F'(0)=1\). These two normalisations give

\[F'(x)=\exp\left(-2\int_0^x b(z)\,dz\right),\qquad F(x)=\int_0^x\exp\left(-2\int_0^y b(z)\,dz\right)dy.\]

Thus \(F\) is strictly increasing and \(F(X)\) is a continuous local martingale. Fix \(c<0<d\), let \(T_c=\inf\{t:X_t=c\}\), \(T_d=\inf\{t:X_t=d\}\), and \(T=T_c\wedge T_d\). On \(\{T=\infty\}\), every unit-time increment of \(B\) satisfies

\[|B_n-B_{n-1}|\leq d-c+\sup_{x\in[c,d]}|b(x)|=:K.\]

The independent standard normal increments cannot all satisfy this bound: the probability is \(\lim_{n\to\infty}\mathbb P(|Z|\leq K)^n=0\). Hence \(T<\infty\) almost surely. Before \(T\), \(F(X)\) is bounded by its values on \([c,d]\), so its stopped local martingale is a bounded martingale. Optional sampling and dominated convergence give

\[0=F(c)\mathbb P(T_c<T_d)+F(d)\mathbb P(T_d<T_c), \qquad\mathbb P(T_c<T_d)=\frac{F(d)}{F(d)-F(c)}.\]

Exponential weights

Let \(f\in C^2(\mathbb R)\), \(g\in C(\mathbb R)\), and define

\[X_t=f(B_t)\exp\left(-\int_0^t g(B_s)\,ds\right).\tag{1}\]

The time integral has finite variation on compact intervals because \(g(B_s)\) is continuous. If desired, its variation decomposition uses \(g^+=\max(g,0)\) and \(g^-=\max(-g,0)\), with \(g=g^+-g^-\). Applying the product rule yields

\[dX_t=e^{-\int_0^tg(B_s)ds}f'(B_t)\,dB_t+ e^{-\int_0^tg(B_s)ds}\left(\frac12 f''(B_t)-g(B_t)f(B_t)\right)dt.\]

Consequently \(X\) is a local martingale if \(f''=2gf\). The converse also holds for Brownian motion on the whole line. Uniqueness of the continuous local-martingale/finite-variation decomposition forces the displayed drift integral to vanish at every time. Its continuous integrand is therefore zero at every time outside one null set. At any fixed positive time, Brownian motion has a strictly positive density on the line, so continuity implies \(f''/2-gf=0\) everywhere.

For a useful expectation formula, assume now that \(g\geq0\) and that its support is contained in \([a,b]\subset(0,\infty)\). Specify the normalisation \(f(x)=1\) for \(x\leq a\). The ordinary differential equation with \(f(a)=1,f'(a)=0\) has a unique solution; equivalently, on \([a,b]\),

\[f(x)=1+2\int_a^x(x-y)g(y)f(y)\,dy.\]

Successive substitution converges uniformly on bounded intervals (the \(n\)-th iterated integral is bounded by a constant times \(C^n(x-a)^{2n}/(2n)!\)). It gives existence, and the same estimate applied to a difference gives uniqueness. All terms are nonnegative, hence \(f\geq1\), \(f'\geq0\). Beyond \(b\), extend by \(f(x)=f(b)+f'(b)(x-b)\); the result is \(C^2\) and solves the equation globally.

For a level \(z>0\), let \(T_z=\inf\{t:B_t=z\}\). Brownian recurrence, or the reflection identity \(\mathbb P(T_z\leq t)=2\mathbb P(B_t\geq z)\to1\), gives \(T_z<\infty\) almost surely. The stopped process \(X^{T_z}\) is bounded by \(\sup_{x\leq z}f(x)<\infty\), since the exponential factor is at most one. It is therefore a uniformly integrable martingale, and

\[1=\mathbb{E}X_{T_z}=f(z)\mathbb{E}\exp\left(-\int_0^{T_z}g(B_s)\,ds\right),\qquad \mathbb{E}\exp\left(-\int_0^{T_z}g(B_s)\,ds\right)=\frac1{f(z)}.\]

The sign condition on \(g\) is used in this boundedness argument. Without it, the same local martingale identity alone does not justify passing to the unbounded stopping time.

Running maxima and drawdowns

Let \(M\) be a continuous local martingale with \(M_0=0\), and \(S_t=\sup_{s\leq t}M_s\). Pathwise, the Stieltjes measure \(dS\) is supported on \(\{M=S\}\). Indeed, if \(M_t<S_t\), continuity supplies a neighbourhood on which the previously attained maximum cannot be exceeded, so \(S\) is locally constant there. The open set \(\{M<S\}\) is a countable union of such intervals. Thus, for every locally bounded Borel function \(h\),

\[\int_0^t(S_s-M_s)h(s)\,dS_s=0.\tag{2}\]

For \(\varphi\in C^1(\mathbb R_+)\), put \(\Phi(x)=\int_0^x\varphi(y)\,dy\). The ordinary chain rule for \(S\), the product rule, and the support property give

\[d[(S-M)\varphi(S)]=\varphi(S)\,dS-\varphi(S)\,dM+(S-M)\varphi'(S)\,dS =d\Phi(S)-\varphi(S)\,dM.\]

\[(S_t-M_t)\varphi(S_t)=\Phi(S_t)-\int_0^t\varphi(S_s)\,dM_s.\tag{3}\]

Taking \(\varphi(x)=\lambda e^{-\lambda x}\), with \(\lambda>0\), gives the local martingale

\[e^{-\lambda S_t}[1+\lambda(S_t-M_t)] =1-\int_0^t\lambda e^{-\lambda S_s}\,dM_s.\tag{4}\]

Assume \(\langle M\rangle_\infty=\infty\). The Brownian time-change theorem gives \(M_t=\beta_{\langle M\rangle_t}\), so \(\liminf_{t\to\infty}M_t=-\infty\). Therefore the drawdown time \(T_a=\inf\{t:S_t-M_t=a\}\) is finite for every \(a>0\). Before \(T_a\), the expression in (4) is between zero and \(1+\lambda a\). Its stopped process is a bounded martingale. Continuity and dominated convergence give

\[1=(1+\lambda a)\mathbb{E} e^{-\lambda S_{T_a}},\qquad \mathbb{E} e^{-\lambda S_{T_a}}=(1+\lambda a)^{-1}.\]

This is the Laplace transform of an exponential random variable of rate \(1/a\), so \(S_{T_a}\) has that law. The drawdown need not be monotone; its bound before the first hitting time and its continuity are the properties used here.

Squared Bessel processes

Suppose \(X\geq0\) is a continuous adapted solution of \(dX_t=\alpha\,dt+2\sqrt{X_t}\,dB_t\), with \(X_0=x\geq0\), \(\alpha\geq0\). Let \(f\geq0\) be continuous on \([0,1]\). There is a unique positive solution of

\[\varphi''=2f\varphi,\qquad\varphi(0)=1,\qquad\varphi'(1)=0.\]

For existence, solve backwards with \(\psi(1)=1,\psi'(1)=0\). Its Volterra equation is \(\psi(t)=1+2\int_t^1(s-t)f(s)\psi(s)\,ds\). The convergent successive-substitution argument above gives \(\psi\geq1\). Set \(\varphi=\psi/\psi(0)\). Any solution with derivative zero at (1) is a scalar multiple of \(\psi\), proving uniqueness. We have \(\varphi>0\), \(\varphi'\leq0\). Define \(u=\varphi'/(2\varphi)\); direct differentiation gives

\[u'+2u^2=\frac{\varphi''}{2\varphi}=f,\qquad u\leq0,\qquad u(1)=0.\]

Apply Itō’s formula to \(Y_t=u(t)X_t-\int_0^t f(s)X_s\,ds\):

\[dY_t=2u(t)\sqrt{X_t}\,dB_t+\left(\alpha u(t)-2u(t)^2X_t\right)dt.\]

It follows that

\[Z_t=\exp(Y_t)\varphi(t)^{-\alpha/2} =\exp\left(u(0)x+2\int_0^tu(s)\sqrt{X_s}\,dB_s-2\int_0^tu(s)^2X_s\,ds\right)\]

is a local martingale. Since \(u\leq0\), \(f,X\geq0\), and \(\alpha\geq0\), it satisfies \(0\leq Z_t\leq\varphi(1)^{-\alpha/2}\) on \([0,1]\). It is therefore a bounded martingale. Taking expectations at time (1) gives

\[\mathbb{E}\exp\left(-\int_0^1 f(s)X_s\,ds\right) =\varphi(1)^{\alpha/2}\exp\left(\frac{x\varphi'(0)}2\right).\tag{5}\]

If \(f\equiv\lambda>0\), then \(\varphi(t)=\cosh(\sqrt{2\lambda}(1-t))/\cosh(\sqrt{2\lambda})\). Thus

\[\mathbb{E}\exp\left(-\lambda\int_0^1X_s\,ds\right) =\cosh(\sqrt{2\lambda})^{-\alpha/2} \exp\left[-\frac{x\sqrt{2\lambda}}2\tanh(\sqrt{2\lambda})\right].\tag{6}\]

For Brownian motion \(\beta\) starting at \(y\), Itō’s formula gives \(d(\beta_t^2)=dt+2\beta_t\,d\beta_t\). Set \(B_t=\int_0^t\operatorname{sgn}(\beta_s)\,d\beta_s\). Fubini and the Gaussian density show \(\int_0^t\mathbb1_{\{\beta_s=0\}}ds=0\) almost surely, so \(\langle B\rangle_t=t\); Lévy’s characterization makes \(B\) Brownian. Hence \(\beta^2\) is the case \(x=y^2,\alpha=1\) of the preceding equation, which gives its integrated-square Laplace transform.

Tanaka’s formula and excursion counts

Let \(B_0=0\) and approximate the absolute value by \(g_\varepsilon(x)=\sqrt{x^2+\varepsilon^2}\). Itō’s formula gives

\[g_\varepsilon(B_t)=\varepsilon+M_t^\varepsilon+A_t^\varepsilon, \quad M_t^\varepsilon=\int_0^t\frac{B_s}{\sqrt{B_s^2+\varepsilon^2}}\,dB_s, \quad A_t^\varepsilon=\frac12\int_0^t\frac{\varepsilon^2}{(B_s^2+\varepsilon^2)^{3/2}}\,ds.\]

Set \(\operatorname{sgn}(0)=0\) and \(\beta_t=\int_0^t\operatorname{sgn}(B_s)\,dB_s\). The bounded integrands converge pointwise to the sign function. The Itō isometry and dominated convergence, followed by Doob’s inequality, give for every \(T<\infty\)

\[\mathbb{E}\sup_{t\leq T}|M_t^\varepsilon-\beta_t|^2 \leq4\mathbb{E}\int_0^T\left|\frac{B_s}{\sqrt{B_s^2+\varepsilon^2}}-\operatorname{sgn}(B_s)\right|^2ds\longrightarrow0.\]

Since \(0\leq g_\varepsilon(x)-|x|\leq\varepsilon\), the increasing processes \(A^\varepsilon\) converge uniformly on compact intervals in probability to the continuous process \(L_t=|B_t|-\beta_t\). Choose a subsequence converging uniformly almost surely on every integer-length interval. The limit is increasing, starts at zero, and satisfies Tanaka’s formula

\[|B_t|=\int_0^t\operatorname{sgn}(B_s)\,dB_s+L_t.\]

The measure \(dL\) is supported on \(\{B=0\}\). Indeed, on any interval \([u,v]\) where \(|B_s|\geq\delta>0\),

\[0\leq A_v^\varepsilon-A_u^\varepsilon\leq\frac{(v-u)\varepsilon^2}{2\delta^3}\longrightarrow0.\]

Using rational \(u,v,\delta\), the same almost sure subsequence proves this simultaneously on a countable family of intervals. Continuity then makes \(L\) constant on every excursion interval away from zero. Fubini gives \(\langle\beta\rangle_t=\int_0^t\mathbb1_{\{B_s\ne0\}}ds=t\), so Lévy’s characterization makes \(\beta\) Brownian motion.

The identity \(|B|=\beta+L\geq0\) implies \(L_t\geq\sup_{s\leq t}(-\beta_s)\). For the reverse inequality, let \(r\) be the last zero of \(B\) in \([0,t]\). Such a zero exists because \(B_0=0\), and it is attained by continuity. If \(r<t\), \(L\) is constant on \((r,t]\); if \(r=t\), no constancy argument is needed. In both cases \(L_t=L_r=-\beta_r\leq\sup_{s\leq t}(-\beta_s)\). Therefore

\[L_t=\sup_{s\leq t}(-\beta_s),\qquad \mathbb P(L_t\leq a)=2\Phi(a/\sqrt t)-1\quad(a\geq0,t>0).\]

Here \(\Phi\) is the standard normal distribution function. The probability is zero for \(a<0\); the formula follows from the reflection principle. In particular \(L_t/\sqrt t\) has the law of \(|Z|\), with \(Z\) standard normal.

For an excursion-count approximation, define \(S_1^\varepsilon=0\), \(T_n^\varepsilon=\inf\{t\geq S_n^\varepsilon:|B_t|=\varepsilon\}\), and \(S_{n+1}^\varepsilon=\inf\{t\geq T_n^\varepsilon:B_t=0\}\). Let \(N_t^\varepsilon=\#\{n:T_n^\varepsilon\leq t\}\). There are finitely many completed crossings on a compact interval: otherwise uniform continuity would be contradicted by infinitely many disjoint fluctuations of size \(\varepsilon\). The process

\[H_s^\varepsilon=\sum_{n\geq1}\mathbb1_{(S_n^\varepsilon,T_n^\varepsilon]}(s)\]

is predictable and bounded by one. Endpoint choices do not affect Brownian integrals. During each completed outgoing segment, Tanaka’s formula gives an increase of \(|B|\) equal to \(\varepsilon\). During a return segment, \(L\) is constant. Summing these identities gives

\[L_t+\int_0^tH_s^\varepsilon\operatorname{sgn}(B_s)\,dB_s =\varepsilon N_t^\varepsilon+r_t^\varepsilon,\qquad0\leq r_t^\varepsilon\leq\varepsilon.\]

More precisely, \(r_t^\varepsilon=|B_t|\) if \(t\) lies in the next outgoing segment, and \(r_t^\varepsilon=0\) if it lies in a return segment. The latter segment can travel beyond \(\varepsilon\); it contributes no remainder because its stochastic integral accounts for that movement. Since \(H_s^\varepsilon\leq\mathbb1_{\{|B_s|\leq\varepsilon\}}\) apart from irrelevant endpoints, the isometry yields

\[\|\varepsilon N_t^\varepsilon-L_t\|_2 \leq\left(\int_0^t\mathbb P(|B_s|\leq\varepsilon)\,ds\right)^{1/2}+\varepsilon\longrightarrow0.\]

For \(s>0\), the integrand tends to zero by the Gaussian density and is bounded by one; dominated convergence proves the last step. Finally, Brownian scaling gives \(N_t^1\stackrel d=N_1^{1/\sqrt t}\). Applying this approximation at time (1), with \(\varepsilon=1/\sqrt t\), proves

\[\frac{N_t^1}{\sqrt t}\xrightarrow[t\to\infty]{d}L_1\stackrel d=|Z|.\]

An explicit excursion-count error bound

The Gaussian density makes the preceding convergence quantitative. For \(s>0\),

\[\mathbb P(|B_s|\leq\varepsilon) \leq\frac{2\varepsilon}{\sqrt{2\pi s}},\qquad \int_0^t\mathbb P(|B_s|\leq\varepsilon)\,ds \leq2\sqrt{\frac2\pi}\,\varepsilon\sqrt t.\]

Consequently the isometry and \(0\leq r_t^\varepsilon\leq\varepsilon\) give

\[\|\varepsilon N_t^\varepsilon-L_t\|_2 \leq\left(2\sqrt{\frac2\pi}\,\varepsilon\sqrt t\right)^{1/2}+\varepsilon.\]

At \(\varepsilon=1\), division by \(\sqrt t\) proves the direct comparison

\[\left\|\frac{N_t^1-L_t}{\sqrt t}\right\|_2 \leq\left(2\sqrt{\frac2\pi}\right)^{1/2}t^{-1/4}+t^{-1/2}\longrightarrow0.\]

Because \(L_t/\sqrt t\) has the same folded-normal law for every \(t>0\), this gives a second proof of the displayed large-time limit by Slutsky’s lemma. The square root in the isometry estimate is essential: the expected occupation time bounds the squared \(L^2\) norm of the stochastic integral.

Le Gall, Jean-François. 2016. Brownian Motion, Martingales, and Stochastic Calculus. 1st ed. 2016. Graduate Texts in Mathematics 274. Cham: Springer International Publishing : Imprint: Springer. https://doi.org/10.1007/978-3-319-31089-3.

Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.