Positive-Part Tanaka Formulas and Sign Conventions

From Absolute Value to Positive and Negative Parts

The absolute-value Tanaka formula separates Brownian motion into a stochastic integral and a local-time term. The positive and negative parts follow by combining that formula with the elementary identities $x^+=(|x|+x)/2$ and $x^-=(|x|-x)/2$. The combination is simple, but the conventions matter: the integrands can be written with strict or non-strict indicators, and the formulas remain valid when the process is at zero.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ with respect to $(\mathcal{F}_t)$, with $B_0=0$. For a finite horizon $T$, all stochastic integrals below are Ito integrals of progressively measurable integrands $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this is the class $\mathcal{L}^2$. The indicators and sign functions of $B_s$ used below are progressively measurable and bounded, so they satisfy this condition.

Definition: The positive and negative parts of a real number are $x^+=\max(x,0)$ and $x^-=\max(-x,0)$. Define $\operatorname{sgn}(x)=1$ for $x>0$, $\operatorname{sgn}(x)=-1$ for $x<0$, and $\operatorname{sgn}(0)=0$. We use the symmetric Brownian local-time normalization in which the absolute-value Tanaka formula is $|B_t|=\int_0^t\operatorname{sgn}(B_s)\,dB_s+L_t^0$.

Here $L_t^0$ is Brownian local time at zero, continuous and increasing in $t$, with $L_0^0=0$. The absolute-value Tanaka formula is the standard result established in the earlier material on Tanaka’s formula; its normalization fixes the coefficient of local time in the identities derived here. In particular, the coefficient will be one half, not one.

The Positive-Part Tanaka Formula

Theorem: For every $t\ge0$, almost surely, $$ B_t^+=\int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s+\frac12L_t^0. $$ The stochastic integral is unchanged if $\mathbf{1}_{\{B_s>0\}}$ is replaced by $\mathbf{1}_{\{B_s\ge0\}}$.

Proof. Apply $x^+=(|x|+x)/2$ with $x=B_t$, and substitute the absolute-value Tanaka formula and $B_t=\int_0^t1\,dB_s$: $$ B_t^+=\frac12\left(\int_0^t\operatorname{sgn}(B_s)\,dB_s+L_t^0+\int_0^t1\,dB_s\right) =\int_0^t\frac{1+\operatorname{sgn}(B_s)}2\,dB_s+\frac12L_t^0. $$ With the stated convention for the sign at zero, the integrand is $1$ on $\{B_s>0\}$, $0$ on $\{B_s<0\}$, and $1/2$ on $\{B_s=0\}$. It differs from $\mathbf{1}_{\{B_s>0\}}$ only on the zero set. For $s>0$, $B_s$ has a normal distribution and $\mathbb{P}(B_s=0)=0$. Tonelli’s theorem therefore gives $$ \mathbb{E}\int_0^T\mathbf{1}_{\{B_s=0\}}\,ds =\int_0^T\mathbb{P}(B_s=0)\,ds=0. $$ The Ito isometry implies that the integral of this zero-set indicator has second moment zero. Thus the zero-set contribution vanishes almost surely, and the displayed formula follows. The same argument shows that the integrals with indicators of $\{B_s>0\}$ and $\{B_s\ge0\}$ agree. Since the indicator integrands are bounded, each integral is square-integrable; the Ito integral of an $\mathcal{L}^2$ integrand is a martingale, so these integrals are true martingales, not merely local martingales. $\square$

The Negative-Part Tanaka Formula

Theorem: For every $t\ge0$, almost surely, $$ B_t^-=-\int_0^t\mathbf{1}_{\{B_s<0\}}\,dB_s+\frac12L_t^0 =-\int_0^t\mathbf{1}_{\{B_s\le0\}}\,dB_s+\frac12L_t^0. $$ The equality between the two integral conventions is an almost-sure equality of the corresponding Ito integrals.

Proof. Apply $x^-=(|x|-x)/2$ and substitute the same absolute-value formula: $$ B_t^-=\frac12\left(\int_0^t\operatorname{sgn}(B_s)\,dB_s+L_t^0-\int_0^t1\,dB_s\right) =\int_0^t\frac{\operatorname{sgn}(B_s)-1}{2}\,dB_s+\frac12L_t^0. $$ The integrand is $-1$ when $B_s<0$, $0$ when $B_s>0$, and $-1/2$ when $B_s=0$. It equals $-\mathbf{1}_{\{B_s<0\}}$ away from zero; its difference from that integrand is supported on $\{B_s=0\}$. The zero-set integral vanishes by the Ito-isometry argument in the preceding proof. Replacing $<0$ by $\le0$ also changes the integrand only on that zero set. The integrands are bounded and hence belong to $\mathcal{L}^2$ on every finite horizon. Their Ito integrals are square-integrable martingales by the Ito isometry and the martingale property of the Ito integral. This proves both versions. $\square$

These formulas use the same local time $L_t^0$, with coefficient $1/2$ in each. The stochastic-integral signs differ: the negative-part formula has a minus sign in front of the integral with a negative-state indicator. This sign is required by $B_t=B_t^+-B_t^-$.

Worked Checks of the Formulas

Worked Example: Checking the Identities at Zero

Suppose a path is evaluated at a time $t$ when $B_t=0$, and write $\ell=L_t^0$. Substitution into the positive-part formula gives $$ 0=\int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s+\frac{\ell}{2}, \qquad \int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s=-\frac{\ell}{2}. $$ Substitution into the negative-part formula gives $$ 0=-\int_0^t\mathbf{1}_{\{B_s<0\}}\,dB_s+\frac{\ell}{2}, \qquad \int_0^t\mathbf{1}_{\{B_s<0\}}\,dB_s=\frac{\ell}{2}. $$ Adding these two integral values gives $0$, consistent with $B_t=0$ and with the fact that the positive- and negative-state indicators sum to $1$ away from the zero set. As a numerical algebra check, if $\ell=2$, the integrals in these identities must be $-1$ and $1$; substituting gives $0=-1+1$ in the positive-part formula and $0=-1+1$ in the negative-part formula. This substitution checks the algebra; it does not assert that the event $B_t=0,\ L_t^0=2$ has positive probability. Being at zero at the terminal time does not imply that local time accumulated earlier is zero.

Worked Example: Strict and Non-Strict Indicators

Compare the positive-part integrals with strict and non-strict indicators on $[0,3]$. Their difference is $$ D=\int_0^3\left(\mathbf{1}_{\{B_s\ge0\}}-\mathbf{1}_{\{B_s>0\}}\right)dB_s =\int_0^3\mathbf{1}_{\{B_s=0\}}\,dB_s. $$ For $s>0$, $\mathbb{P}(B_s=0)=0$, so $$ \mathbb{E}\int_0^3\mathbf{1}_{\{B_s=0\}}\,ds =\int_0^3 0\,ds=0. $$ By the Ito isometry, $\mathbb{E}[D^2]=0$, which implies $D=0$ almost surely. Thus both versions produce the same integral. The time $s=0$, at which $B_0=0$, does not affect a time integral because a single time has Lebesgue measure zero. Applying the same calculation to the negative-state indicators shows that $\int_0^3\mathbf{1}_{\{B_s\le0\}}\,dB_s=\int_0^3\mathbf{1}_{\{B_s<0\}}\,dB_s$.

Check: the argument uses the Ito isometry, not a claim that Brownian motion never hits zero. Brownian motion can hit zero; the relevant fact is that its zero set has zero Lebesgue measure almost surely on a finite time interval.

Worked Example: The Martingale Parts on a Two-Unit Interval

Define $M_t^+=\int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s$ and $C_t=\int_0^t\mathbf{1}_{\{B_s<0\}}\,dB_s$. Both integrands are bounded and progressively measurable, so for $T=2$ they satisfy $$ \mathbb{E}\int_0^2\mathbf{1}_{\{B_s>0\}}^2\,ds\le2, \qquad \mathbb{E}\int_0^2\mathbf{1}_{\{B_s<0\}}^2\,ds\le2. $$ Consequently $M^+$ and $C$ are square-integrable martingales with mean zero. By the Ito isometry and symmetry of the normal distribution, $\mathbb{P}(B_s>0)=\mathbb{P}(B_s<0)=1/2$ for $s>0$. Hence $$ \mathbb{E}[(M_2^+)^2]=\int_0^2\frac12\,ds=1, \qquad \mathbb{E}[C_2^2]=\int_0^2\frac12\,ds=1. $$ The martingale term in the negative-part formula is $-C$, which also has mean zero and variance $1$. The strict and non-strict indicator versions have the same values by the preceding example.

Check: the indicators of the positive and negative states sum to $1$ except when $B_s=0$, and that exceptional set contributes zero to the Ito integral. Thus $M_2^++C_2=B_2$ almost surely. The two integrals have zero covariance by the Ito isometry for products of integrands, since $\mathbf{1}_{\{B_s>0\}}\mathbf{1}_{\{B_s<0\}}=0$ for every $s$. Their variances therefore add to $1+1=2=\mathbb{E}[B_2^2]$, an independent check.

Common Convention Errors

A frequent error is to assign the full local-time term to the positive part. The coefficient comes directly from the algebra: $B_t^+=(|B_t|+B_t)/2$, so the $L_t^0$ term in $|B_t|$ is also divided by two. Writing $B_t^+=\int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s+L_t^0$ is therefore incorrect under the stated normalization. The corresponding negative-part formula also has coefficient $1/2$, not $1$.

Another error is to treat the choice of indicator at zero as if Brownian motion could not visit zero. It can visit zero, but the Ito integrals are insensitive to changes in their integrands on a set of times with zero Lebesgue measure almost surely. Here that fact follows from $\mathbb{E}\int_0^T\mathbf{1}_{\{B_s=0\}}ds=0$ and the Ito isometry. The formulas do not say that local time vanishes when $B_t=0$; local time records accumulated contact with zero, not just the process’s value at the final time.

Quantity	Formula	Indicator convention
Positive part	$B_t^+=\int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s+\tfrac12L_t^0$	$>0$ or $\ge0$
Negative part	$B_t^-=-\int_0^t\mathbf{1}_{\{B_s<0\}}\,dB_s+\tfrac12L_t^0$	$<0$ or $\le0$
Difference	$B_t=B_t^+-B_t^-$	Zero-set integral contributes nothing

Check Your Understanding

Use the stated local-time normalization and distinguish pointwise algebra from equality of Ito integrals.

Starting from $x^+=(|x|+x)/2$, identify the coefficient of $L_t^0$ in the positive-part formula.
Starting from $x^-=(|x|-x)/2$, derive the sign of the stochastic integral in the negative-part formula.
Why do the integrals with $\mathbf{1}_{\{B_s>0\}}$ and $\mathbf{1}_{\{B_s\ge0\}}$ agree almost surely?
If $B_t=0$, what values must the positive- and negative-state stochastic integrals have in terms of $L_t^0$?
Which integrability condition ensures that the indicator stochastic integrals are square-integrable martingales on $[0,T]$?

Quantity	Formula	Indicator convention
Positive part	\(B_t^+=\int_0^t\mathbf{1}_{\{B_s>0\}}\,dB_s+\tfrac12L_t^0\)	\(>0\) or \(\ge0\)
Negative part	\(B_t^-=-\int_0^t\mathbf{1}_{\{B_s<0\}}\,dB_s+\tfrac12L_t^0\)	\(<0\) or \(\le0\)
Difference	\(B_t=B_t^+-B_t^-\)	Zero-set integral contributes nothing