Why Absolute Value Is Not a Direct Ito Example

The Issue Is the Hypothesis at Zero

For a twice continuously differentiable function $f$, the Brownian Ito formula includes both a stochastic integral and a second-derivative correction. Absolute value looks like a natural test function, but it has a corner at zero: it is not differentiable there, much less twice continuously differentiable. If the Brownian path can reach zero, the classical formula cannot be applied directly to $f(x)=|x|$. This is a failure of the theorem's hypothesis, not a small technical adjustment to its formula.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a standard Brownian motion $B_t$, with $B_0=0$. Brownian motion is understood relative to this filtration. On a fixed interval $[0,T]$, write $\mathcal{L}^2$ for the progressively measurable processes $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For such an integrand, the Ito integral $\int_0^t H_s\,dB_s$ is a square-integrable martingale.

Result (Brownian Ito formula): The standard Brownian Ito formula, proved in the earlier material on Ito's formula, says that if $f\in C^2(\mathbb{R})$ and the terms are integrable, then $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds. $$ The stochastic integral here is defined when $f'(B_s)$ is progressively measurable and belongs to $\mathcal{L}^2$ on the time interval in question. Under weaker integrability assumptions, the formula is used after localization.

The correction term comes from Brownian quadratic variation, $[B]_t=t$, or in differential notation $dB_t\,dB_t=dt$. The remaining multiplication rules are $dB_t\,dt=0$ and $dt\,dt=0$. These rules do not remove the need for the function to satisfy the hypotheses of Ito's formula.

Proposition: The classical Brownian Ito formula cannot be applied directly to $f(x)=|x|$ on any open interval containing zero, because $f$ is not differentiable at zero. It can be applied to the appropriate smooth branch when the process is stopped in a region strictly separated from zero.

Proof. For $h>0$, the difference quotient of $f(x)=|x|$ at zero is $(|h|-|0|)/h=1$. For $h<0$, it is $(|h|-|0|)/h=-1$. The two one-sided limits differ, so $f'(0)$ does not exist. In particular, $f$ is not in $C^2$ on any open interval containing zero, as required by the stated classical formula. Away from zero, the function is smooth: it equals $x$ on $(0,\infty)$ and $-x$ on $(-\infty,0)$. If a stopped process remains in one of those regions, the corresponding smooth function can be used along that process. $\square$

Why the Naive Sign Calculation Fails

A tempting calculation is to write $f'(x)=\operatorname{sgn}(x)$, set $f''(x)=0$, and conclude that $|B_t|=\int_0^t\operatorname{sgn}(B_s)\,dB_s$. This is not an application of the classical formula: the derivative at zero does not exist, and the second derivative needed by the formula is not defined there. The next example gives a direct contradiction to the proposed identity.

Worked Example: Testing the Naive Formula at Time One

Define $\operatorname{sgn}(x)=1$ for $x>0$, $-1$ for $x<0$, and $0$ for $x=0$. The process $\operatorname{sgn}(B_s)$ is progressively measurable and bounded by $1$, so $$ \mathbb{E}\int_0^1 \operatorname{sgn}(B_s)^2\,ds\leq 1. $$ Thus it belongs to $\mathcal{L}^2$, and its Ito integral is a square-integrable martingale. In particular, its expectation at time $1$ is zero.

But $B_1$ has distribution $N(0,1)$, so direct integration of its density gives $$ \mathbb{E}|B_1|=\frac{2}{\sqrt{2\pi}}\int_0^\infty x e^{-x^2/2}\,dx =\frac{2}{\sqrt{2\pi}}\cdot 1 =\sqrt{\frac{2}{\pi}}. $$ The integral in the middle equals $1$, since the antiderivative of $xe^{-x^2/2}$ is $-e^{-x^2/2}$. Hence the candidate identity would equate a random variable with expectation $\sqrt{2/\pi}>0$ to one with expectation $0$, which is impossible.

Check: At a general time $t>0$, the same density calculation gives $\mathbb{E}|B_t|=\sqrt{2t/\pi}$, while the proposed stochastic integral still has expectation zero. The contradiction is therefore not specific to the choice $t=1$.

The failure does not mean that absolute value cannot be studied using stochastic calculus. It means that the missing behavior at zero cannot be obtained by pretending the classical second derivative exists there. A formula that describes that behavior requires additional ideas and is not the classical $C^2$ Ito formula.

Stopping Strictly Away from Zero

Stopping can make an absolute-value calculation valid by ensuring that the process stays on one smooth branch. For example, start from $x>0$ and stop when the process first reaches a positive level $a<x$. Up to and including that stopping time the process is at least $a$, so it never reaches the point where absolute value has a corner.

Stopped-branch formula: Let $X_t=x+B_t$, where $x>a>0$, and define $\tau_a=\inf\{t\geq0:X_t\leq a\}$, with $\inf\varnothing=\infty$. For every finite $T$ and $t\leq T$, $$ |X_{t\wedge\tau_a}|=x+\int_0^t \mathbf{1}_{\{s<\tau_a\}}\,dB_s. $$ The integrand is progressively measurable and belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^T\mathbf{1}_{\{s<\tau_a\}}^2\,ds\leq T$.

Proof. On the event $\tau_a>t$, continuity and the definition of $\tau_a$ give $X_t>a>0$. On the event $\tau_a\leq t$, continuity gives $X_{\tau_a}=a>0$. Thus $X_{t\wedge\tau_a}\geq a$ in either case and its absolute value equals itself. Apply the Brownian Ito formula to the smooth function $g(y)=y$, for which $g'(y)=1$ and $g''(y)=0$, at the stopped process. Equivalently, the stopped Brownian increment is $\int_0^t\mathbf{1}_{\{s<\tau_a\}}\,dB_s$; changing the indicator at the single stopping time does not change the integral. This yields the displayed identity. The formula is an application to the smooth branch $g(y)=y$, not a direct application to $|y|$ at zero. $\square$

Worked Example: Absolute Value Before Hitting One

Take $X_t=2+B_t$ and $\tau=\inf\{t\geq0:X_t\leq1\}$. For any fixed $T>0$, continuity implies $X_{t\wedge\tau}\geq1$. Consequently $|X_{t\wedge\tau}|=X_{t\wedge\tau}$. Using $g(y)=y$, with $g'(y)=1$ and $g''(y)=0$, gives $$ |X_{t\wedge\tau}|=2+\int_0^t\mathbf{1}_{\{s<\tau\}}\,dB_s. $$ The integrand is progressively measurable and square-integrable because $\mathbb{E}\int_0^T\mathbf{1}_{\{s<\tau\}}^2\,ds\leq T$.

For a check, the integral is a square-integrable martingale, so its conditional expectation at an earlier time $u\leq t$ is its value at $u$; in particular its expectation is zero. Therefore $\mathbb{E}|X_{t\wedge\tau}|=2$. This is consistent with the identity: the stopped process is the constant starting value plus a mean-zero stopped Brownian increment. At the stopping boundary itself, $X_\tau=1$, still strictly positive.

The negative branch works in the same way, using $g(y)=-y$ while the stopped process remains negative. The essential condition is not that the process is stopped at a particular named level; it is that the entire stopped range stays a positive distance from zero, so a single smooth branch describes absolute value there.

A Nearby Function That Does Satisfy Ito's Hypothesis

A useful contrast is $f(x)=|x|^3$. Although this function is written using absolute value, it is twice continuously differentiable at zero. Its derivative is $f'(x)=3x|x|$, including $f'(0)=0$. For $x\ne0$, differentiating again gives $f''(x)=6|x|$; at zero, the derivative of $f'$ is $$ \lim_{h\to0}\frac{3h|h|-0}{h}=\lim_{h\to0}3|h|=0. $$ Thus $f''(0)=0$, and $f''(x)=6|x|$ holds for every $x$, with a continuous second derivative.

Worked Example: Applying Ito's Formula to the Absolute Cube

Apply Ito's formula to $f(B_t)=|B_t|^3$. The derivatives are $f'(x)=3x|x|$ and $f''(x)=6|x|$. The formula, including its second-derivative correction, is $$ |B_t|^3=\int_0^t 3B_s|B_s|\,dB_s+\frac12\int_0^t 6|B_s|\,ds =\int_0^t 3B_s|B_s|\,dB_s+3\int_0^t|B_s|\,ds. $$ The stochastic integrand is progressively measurable. It belongs to $\mathcal{L}^2$, because $ (3B_s|B_s|)^2=9B_s^4$ and $$ \mathbb{E}\int_0^T9B_s^4\,ds=9\int_0^T3s^2\,ds=9T^3<\infty. $$ Here $\mathbb{E}[B_s^4]=3s^2$, the fourth moment of a centered normal variable with variance $s$.

Taking expectations is valid because the stochastic integral is square-integrable and has expectation zero. Since $\mathbb{E}|B_s|=\sqrt{2s/\pi}$, the right-hand side has expectation $$ 3\int_0^T\sqrt{\frac{2s}{\pi}}\,ds =3\sqrt{\frac{2}{\pi}}\cdot\frac{2}{3}T^{3/2} =2\sqrt{\frac{2}{\pi}}T^{3/2}. $$ Independently, integration against the $N(0,T)$ density gives $\mathbb{E}|B_T|^3=2\sqrt{2/\pi}\,T^{3/2}$, agreeing with the formula. The check also shows why one must retain the correction term: it supplies the nonzero expectation.

Common Pitfall: Assigning a Second Derivative at the Corner

A common wrong calculation is to write $f'(x)=\operatorname{sgn}(x)$, declare $f''(x)=0$, and omit the correction term. Choosing any value for $\operatorname{sgn}(0)$ does not fix the problem: it cannot make the two one-sided derivatives of $|x|$ agree. The calculation is invalid precisely where the process may visit zero. The stopped example is valid for a different reason: it stays strictly on one side, and the calculation uses the smooth function $g(y)=y$, not a fabricated second derivative of absolute value.

When deciding whether to use the classical formula, check the function's differentiability on an open set containing the process values under consideration. If the process reaches a nonsmooth point, stopping strictly before that point may permit a smooth-branch calculation, but it does not justify extending that calculation through the point. If the function is genuinely $C^2$, as $|x|^3$ is, calculate its derivatives at the potentially problematic point and verify the stochastic integrand's square-integrability before applying Ito's formula.

Check Your Understanding

For each question, distinguish failure of the smoothness hypothesis from a valid formula applied on a stopped region.

What are the right and left difference quotients of $|x|$ at zero, and what do they imply about the classical Ito formula?
Why does $\operatorname{sgn}(B_s)$ belong to $\mathcal{L}^2$ on every finite time interval?
For $X_t=x+B_t$ with $x>a>0$, explain why $X_{t\wedge\tau_a}$ remains strictly positive when $\tau_a=\inf\{t:X_t\leq a\}$.
Compute the first and second derivatives of $|x|^3$ at zero and away from zero.
For the absolute-cube example, verify that the Ito integrand has finite expected integrated square on $[0,T]$.