Why the Tanaka Equation Has No Strong Solution

The Equation and the Meaning of “Strong”

Use a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) satisfying the usual conditions, carrying a Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). The Tanaka equation considered here is the Ito stochastic differential equation \(dX_t=h(X_t)\,dB_t,\ X_0=0\), where \(h(x)=1\) for \(x\ge0\) and \(h(x)=-1\) for \(x<0\). Its integrand \(H_t=h(X_t)\) is progressively measurable and bounded, so on every finite interval \([0,T]\) it belongs to \(\mathcal{L}^2\): \(\mathbb{E}\int_0^T H_t^2\,dt=T<\infty\). The integral is an Ito integral, not a Stratonovich integral.

Weak solutions can be constructed by reflecting \(B\) at its running minimum and assigning a random sign to each excursion away from zero. The issue is that the signs cannot be read off from the driving path. We first identify exactly what the path does determine.

Reflection Determines the Absolute Value

For a continuous function \(w:[0,T]\to\mathbb{R}\) with \(w(0)=0\), define its reflection and regulator by \(a_t=\max(0,-\inf_{0\le s\le t}w_s)\), \(\Gamma(w)_t=w_t+a_t\). The regulator \(a\) is continuous and nondecreasing, \(\Gamma(w)\ge0\), and \(a\) can increase only when \(\Gamma(w)=0\). The following deterministic fact says these properties specify the reflection uniquely.

Proof. Since \(R_s=w_s+L_s\ge0\), we have \(L_s\ge-w_s\) for every \(s\le t\). Monotonicity gives \(L_t\ge L_s\), and \(L_t\ge0\), so \(L_t\ge\max(0,-\inf_{s\le t}w_s)=a_t\). To prove the reverse inequality, suppose \(L_t>a_t\) for some \(t\). Choose a number \(c\) with \(a_t<c<L_t\). Continuity and \(L_0=0\) imply that \(L\) reaches level \(c\) by time \(t\). Since \(L\) increases only where \(R=0\), it must reach that level at a time \(u\le t\) with \(R_u=0\). Then \(L_u=-w_u\le a_t\), contradicting \(L_u=c>a_t\). Thus \(L_t=a_t\), and \(R_t=w_t+a_t=\Gamma(w)_t\). \(\square\)

For a weak solution \(X\), the absolute-value formula uses symmetric local time. Write \(\operatorname{sgn}_0(x)=1\) if \(x>0\), \(-1\) if \(x<0\), and \(0\) if \(x=0\). Tanaka’s formula, with local time \(L_t^0(X)\) in its usual normalization, states \(|X_t|=|X_0|+\int_0^t\operatorname{sgn}_0(X_s)\,dX_s+L_t^0(X)\). The local-time term is continuous and nondecreasing, and it increases only when \(X=0\).

Proof. The integrand \(h(X)\) is progressively measurable and bounded, so the Ito integral defining \(X\) is square-integrable on every finite interval. The Ito isometry gives \(\mathbb{E}[X_t^2]=\mathbb{E}\int_0^t h(X_s)^2\,ds=t\). More generally, the Ito integral is a true martingale here: for \(s\le t\), the integral over \([s,t]\) has conditional expectation zero given \(\mathcal{F}_s\), first for adapted simple integrands by the Brownian increment property and then for \(h(X)\) by the defining \(L^2\) limit. Its quadratic variation is \([X]_t=\int_0^t h(X_s)^2\,ds=t\). By Lévy’s characterization, \(X\) is Brownian motion relative to the filtration. In particular, \(\mathbb{P}(X_s=0)=0\) for every \(s>0\). Tonelli’s theorem gives \(\mathbb{E}\int_0^T\mathbf{1}_{\{X_s=0\}}\,ds=0\), so the zero set has Lebesgue measure zero almost surely.

Apply Tanaka’s formula and substitute \(dX_s=h(X_s)\,dB_s\). For \(X_s\ne0\), \(\operatorname{sgn}_0(X_s)h(X_s)=1\). The zero set has zero time measure, so the Ito isometry shows that \(\int_0^t\operatorname{sgn}_0(X_s)h(X_s)\,dB_s=B_t\): the difference from \(\int_0^t1\,dB_s\) has expected squared value \(\mathbb{E}\int_0^t\mathbf{1}_{\{X_s=0\}}\,ds=0\). Thus \(|X_t|=B_t+L_t^0(X)\), with a nondecreasing regulator supported on \(\{|X|=0\}\). The deterministic proposition gives \(|X|=\Gamma(B)\). \(\square\)

Why the Brownian Driver Does Not Determine the Signs

The open intervals on which \(|X|>0\) are its excursions away from zero. On each such interval, continuity forces \(X\) to have one constant sign. Reflection determines the magnitude of every excursion from \(B\), but it does not determine whether that excursion is positive or negative.

The conditional statement is the decisive point: observing the full driver still leaves the sign at a fixed positive time undetermined. To see why it rules out a strong solution, suppose \(X\) were strong. Then \(X_q\) would be measurable with respect to the completed filtration generated by \(B\) up to time \(q\), and therefore with respect to \(\sigma(B_s:s\ge0)\). Since \(|X_q|>0\) almost surely, \(\operatorname{sgn}(X_q)\) is also measurable with respect to that sigma-field. A bounded random variable measurable with respect to a sigma-field equals its conditional expectation given that sigma-field. The excursion-sign theorem instead says that this conditional expectation is zero, whereas \(|\operatorname{sgn}(X_q)|=1\) almost surely. This contradiction proves the result.

Proof. Assume a strong solution exists and fix \(q>0\). The reflection proposition gives \(|X_q|=\Gamma(B)_q\), which is positive almost surely; equivalently, the reflected Brownian motion has no atom at zero at a fixed positive time. Strong adaptedness makes \(\operatorname{sgn}(X_q)\) measurable with respect to the entire Brownian path. Therefore \(\mathbb{E}[\operatorname{sgn}(X_q)\mid\sigma(B_s:s\ge0)]=\operatorname{sgn}(X_q)\) almost surely. The excursion-sign theorem gives that the same conditional expectation is zero. This would imply \(\operatorname{sgn}(X_q)=0\) almost surely, contradicting \(|X_q|>0\) almost surely. Hence no strong solution exists. \(\square\)

What This Does—and Does Not—Say

The conclusion is not that the equation lacks solutions. It has weak solutions: one can take a Brownian driver, reflect it to obtain the magnitude, and independently mark each excursion with a fair sign. The resulting process satisfies the equation in the Ito sense. The obstruction is specifically to making those signs adapted functions of the given Brownian path.

Nor does the result identify weak existence with uniqueness in law, or with pathwise uniqueness. A strong solution is a solution built from the specified driver; uniqueness in law concerns distributions across possible weak constructions. The reflection identity settles the magnitude, but a separate argument is needed for any claim about the full joint law. Here the excursion-sign theorem supplies exactly the conditional randomness needed for the no-strong-solution conclusion.

Common mistake. A tempting but incorrect calculation is to say that \(X=|X|=\Gamma(B)\), so \(X\) must be a function of \(B\). The correct identity is \(|X|=\Gamma(B)\), not \(X=\Gamma(B)\). For an excursion where \(\Gamma(B)>0\), the two possibilities \(X=\Gamma(B)\) and \(X=-\Gamma(B)\) have opposite signs but the same absolute value. Reflection determines the height of that excursion, not its sign.