Applying Ito's Formula Before a Singular Boundary

Why Stop Before the Boundary?

For $f(x)=\log(1+x)$ and $g(x)=1/(1+x)$, the point $x=-1$ is singular: the functions are not defined there, and their derivatives become unbounded as $x$ approaches it from above. Brownian motion can reach $-1$. Applying Ito’s formula without controlling the distance to that point would therefore use the functions outside their domain and could introduce unbounded stochastic integrands. The remedy is to stop the process while it is still a positive distance from the boundary.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. All stochastic integrals below are Ito integrals. For a fixed horizon $T<\infty$, their integrands are progressively measurable and belong to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, a bounded progressively measurable integrand on $[0,T]$ satisfies this condition.

Definition: For $n\ge2$, let $\tau_n=\inf\{t\ge0:B_t\le -1+1/n\}$, with $\inf\varnothing=\infty$. Let $\tau=\inf\{t\ge0:B_t\le-1\}$. The times $\tau_n$ are stopping times, increase with $n$, and satisfy $\tau_n\le\tau$. The stopped process $B_{t\wedge\tau_n}$ never goes below $-1+1/n$.

Here $t\wedge\tau_n=\min(t,\tau_n)$. Since Brownian paths are continuous, at a finite $\tau_n$ the process first reaches the level $-1+1/n$. Thus $1+B_{s\wedge\tau_n}\ge1/n$ for all $s$. This lower bound keeps both functions and the derivatives used below finite.

Stopped Identities

Theorem: For every $n\ge2$ and $t\ge0$, the following identities hold almost surely: $$ \log(1+B_{t\wedge\tau_n}) =\int_0^{t\wedge\tau_n}\frac{1}{1+B_s}\,dB_s -\frac12\int_0^{t\wedge\tau_n}\frac{1}{(1+B_s)^2}\,ds, $$ and $$ \frac{1}{1+B_{t\wedge\tau_n}} =1-\int_0^{t\wedge\tau_n}\frac{1}{(1+B_s)^2}\,dB_s +\int_0^{t\wedge\tau_n}\frac{1}{(1+B_s)^3}\,ds. $$

Proof. On the region $x>-1$, set $f(x)=\log(1+x)$. Its derivatives are $f'(x)=1/(1+x)$ and $f''(x)=-1/(1+x)^2$. Ito’s formula for a twice continuously differentiable function of Brownian motion is $f(B_u)=f(B_0)+\int_0^u f'(B_s)\,dB_s+\tfrac12\int_0^u f''(B_s)\,ds$, as long as the function is defined along the path. To apply it safely, one may first use a twice continuously differentiable extension that agrees with $f$ on $[-1+1/n,\infty)$, then evaluate the formula at $u=t\wedge\tau_n$. The stopped integrand is $\mathbf{1}_{\{s\le\tau_n\}}/(1+B_s)$; on $[0,T]$ its absolute value is at most $n$, so its squared integral has expectation at most $n^2T$. The drift integrand is bounded by $n^2$. Since $B_0=0$, $f(B_0)=0$, giving the first identity with the displayed $\tfrac12 f''$ correction.

For the second identity, take $g(x)=(1+x)^{-1}$. Then $g'(x)=-(1+x)^{-2}$ and $g''(x)=2(1+x)^{-3}$. The Ito correction is $\tfrac12g''(B_s)\,ds=(1+B_s)^{-3}\,ds$. The stopped stochastic integrand has absolute value at most $n^2$, so its expected squared integral on $[0,T]$ is at most $n^4T$; the drift integrand is at most $n^3$. Both integrals are therefore well-defined, and $g(B_0)=1$ gives the second identity. The same extension-and-stopping argument applies because the stopped path stays in the region where the functions agree with their original definitions. $\square$

The indicator versions of the stochastic integrands are progressively measurable. They are bounded on every finite horizon, so the Ito integrals are square-integrable martingales: their second moments are bounded by the corresponding Ito-isometry estimates above, and for $u\le v$, their conditional increments have expectation zero given $\mathcal{F}_u$. The identities do not assert that the analogous unstopped integrals are square-integrable up to $\tau$.

Worked Applications

Worked Example: Logarithm Stopped at a Fixed Safety Level

Let $\rho=\inf\{t\ge0:B_t\le-1/2\}$. Find the Ito identity for $\log(1+B_{t\wedge\rho})$, and check the correction term and integrability.

On the stopped path, $1+B_s\ge1/2$. For $f(x)=\log(1+x)$, the derivatives are $f'(x)=(1+x)^{-1}$ and $f''(x)=-(1+x)^{-2}$. At $B_s$, the Ito terms are therefore $(1+B_s)^{-1}\,dB_s$ and $-\tfrac12(1+B_s)^{-2}\,ds$. Since $f(B_0)=\log(1)=0$,

\log(1+B_{t\wedge\rho}) =\int_0^{t\wedge\rho}\frac{1}{1+B_s}\,dB_s -\frac12\int_0^{t\wedge\rho}\frac{1}{(1+B_s)^2}\,ds.

Before and at $\rho$, the stochastic integrand is at most $2$ in absolute value, and the drift integrand $(1+B_s)^{-2}$ is at most $4$. Hence the stochastic integrand has expected squared integral at most $4T$ on $[0,T]$. Check: at the stopping level, $B_\rho=-1/2$ if $\rho<\infty$, so the left side is $\log(1/2)=-\log 2$. The formula’s drift has a negative sign because $f''(x)<0$; a positive correction would contradict the explicit second derivative.

Worked Example: Reciprocal Stopped at the Same Level

Use the same $\rho$ to derive and check the reciprocal identity.

For $g(x)=(1+x)^{-1}$, $g'(x)=-(1+x)^{-2}$, while $g''(x)=2(1+x)^{-3}$. Thus the stochastic term has a minus sign and the correction is $\tfrac12g''(B_s)\,ds=(1+B_s)^{-3}\,ds$. Since $g(B_0)=1$,

\frac{1}{1+B_{t\wedge\rho}} =1-\int_0^{t\wedge\rho}\frac{1}{(1+B_s)^2}\,dB_s +\int_0^{t\wedge\rho}\frac{1}{(1+B_s)^3}\,ds.

The stochastic integrand is bounded by $4$, so its expected squared integral on $[0,T]$ is at most $16T$. The drift integrand is bounded by $8$. Check: if $\rho<\infty$, the left side at $t\ge\rho$ is $1/(1-1/2)=2$. The Ito correction is positive because $g''(x)>0$ throughout $x>-1$, consistent with the derivative calculation.

Worked Example: Stopping at a Level Closer to the Singularity

Take $\tau_4=\inf\{t\ge0:B_t\le-3/4\}$. Write the stopped identities and give explicit bounds that ensure their stochastic integrals exist on $[0,T]$.

Here $1+B_s\ge1/4$ before and at $\tau_4$. Substituting $n=4$ into the stopped identities gives

\log(1+B_{t\wedge\tau_4}) =\int_0^{t\wedge\tau_4}\frac{1}{1+B_s}\,dB_s -\frac12\int_0^{t\wedge\tau_4}\frac{1}{(1+B_s)^2}\,ds, $$ $$ \frac{1}{1+B_{t\wedge\tau_4}} =1-\int_0^{t\wedge\tau_4}\frac{1}{(1+B_s)^2}\,dB_s +\int_0^{t\wedge\tau_4}\frac{1}{(1+B_s)^3}\,ds.

For the logarithmic identity, the stochastic integrand is bounded by $4$, giving the squared-integral expectation bound $16T$. For the reciprocal identity it is bounded by $16$, giving $256T$. The drift integrands are bounded by $16$ and $64$, respectively. Check: if $\tau_4<\infty$, the two left sides at $t\ge\tau_4$ are $\log(1/4)=-\log 4$ and $1/(1/4)=4$. These endpoint values are finite because the stopping level is still strictly above $-1$.

What Happens at the Singular Boundary?

The levels $-1+1/n$ decrease toward $-1$, so $\tau_n$ increases toward $\tau$. For a fixed time $t<\tau$, continuity of the Brownian path implies that its minimum on $[0,t]$ is strictly greater than $-1$. Consequently, for all sufficiently large $n$, $\tau_n>t$, and the stopped identities then give the ordinary Ito identities up to that time.

This local conclusion does not allow substitution of $t=\tau$. If $\tau<\infty$, continuity gives $B_\tau=-1$, so $\log(1+B_\tau)$ is not finite and $1/(1+B_\tau)$ is undefined. Also, the bounds on the stopped integrands grow with $n$: for example, the reciprocal stochastic integrand can be as large as $n^2$. The square-integrability estimates for each fixed $n$ do not provide a bound uniform in $n$. Thus these formulas justify calculations strictly before the boundary, not a finite identity at the boundary itself.

Takeaway: Stopping at $\tau_n$ keeps the process inside the smooth domain and bounds the derivatives used in Ito’s formula. The resulting identities hold for every fixed $n$; they do not automatically extend to the singular hitting time $\tau$.

Common Pitfall: Ignoring the Stopping Time

A tempting but invalid calculation is to write $d\log(1+B_t)=(1+B_t)^{-1}dB_t$ and then integrate all the way to $\tau$. This omits the Ito correction and presumes the reciprocal integrand remains controlled as $B_t$ approaches $-1$. The correct calculation before $\tau_n$ includes $-\tfrac12(1+B_t)^{-2}\,dt$, and its integrand bounds depend on $n$. Likewise, for the reciprocal, omitting the correction would incorrectly remove the positive term $(1+B_t)^{-3}\,dt$.

The multiplication rules explain the corrections: $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. They are shorthand for the quadratic-variation calculation already built into Ito’s formula; they do not remove the need to check the function’s domain and the stopped integrands’ integrability.

Check Your Understanding

For each question, keep the distinction between a fixed safe stopping level and the singular boundary in view.

For $f(x)=\log(1+x)$, compute $f'(x)$ and $f''(x)$, then state the stopped identity using $\tau_n$.
For $g(x)=1/(1+x)$, explain why the Ito correction has a positive sign.
For $\tau_3=\inf\{t\ge0:B_t\le-2/3\}$, bound the absolute value of the stochastic integrand in the reciprocal identity.
Why does square-integrability for each fixed $\tau_n$ not by itself justify taking $n$ to infinity in the reciprocal stochastic integral?
If $\tau<\infty$, what are $B_\tau$ and the status of $\log(1+B_\tau)$?