Worked Problems on Local Time and Reflection

Putting Local Time and Reflection Together

Local time turns the amount of time a Brownian path spends near a level into a quantity attached to that level. Reflection gives a complementary interpretation at a boundary: the regulator that prevents a path from becoming negative is the local time accumulated at zero. In mixed problems, these connections are useful only if the level shift and local-time normalization are kept consistent.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with its usual augmented filtration. An Ito integrand below is progressively measurable and belongs to $\mathcal{L}^2$ on each finite time interval: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. We use the symmetric Brownian local time normalization specified by Tanaka’s formula. In particular, for $X_t=x+B_t$ and $a\in\mathbb{R}$,

|X_t-a|=|x-a|+\int_0^t \operatorname{sgn}(X_s-a)\,dB_s+L_t^a(X), \qquad \operatorname{sgn}(0)=0.

For this normalization, the occupation-density formula is $\int_0^t g(X_s)\,ds=\int_{\mathbb{R}}g(a)L_t^a(X)\,da$ for nonnegative measurable $g$. These identities are the starting points for the calculations below.

Expected Local Time: Keep the Level Shift

Theorem: For $X_t=x+B_t$, $a\in\mathbb{R}$, and $t\ge0$, the expected local time at level $a$ is $$ \mathbb{E}[L_t^a(X)]=\mathbb{E}|x+B_t-a|-|x-a|. $$ For $t=0$, both sides are zero. For $t>0$, the right-hand side is nonnegative and is zero when $t=0$.

Proof. Apply the displayed Tanaka formula and take expectations. The integrand $H_s=\operatorname{sgn}(x+B_s-a)$ is progressively measurable and satisfies $\mathbb{E}\int_0^t H_s^2\,ds\le t$, so it belongs to $\mathcal{L}^2$. By the Ito isometry, its Ito integral is square-integrable. It is a martingale: for $u\le v$, the conditional expectation of the integral increment $\int_u^v H_s\,dB_s$, given $\mathcal{F}_u$, is zero, first for adapted step integrands by independence and centering of Brownian increments, and then for $H$ by the $L^2$ definition of the integral. Thus its expectation is zero. Taking expectations in Tanaka’s identity gives the formula. $\square$

For an explicit evaluation, let $\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}$ and let $\Phi(z)=\int_{-\infty}^z\phi(u)\,du$. If $Z$ is standard normal, then for $m\in\mathbb{R}$ and $t>0$, splitting the integral at $z=-m/\sqrt{t}$ gives $\mathbb{E}|m+\sqrt{t}Z|=2\sqrt{t}\,\phi(m/\sqrt{t})+m(2\Phi(m/\sqrt{t})-1)$. The formula is even in $m$, as an absolute-value expectation must be.

Worked Example: Local Time at a Nonzero Level

Take $X_t=B_t$, $a=1$ and $t=1$. Compute $\mathbb{E}[L_1^1(X)]$.

Here $m=x-a=0-1=-1$, so $\mathbb{E}|B_1-1|=2\phi(1)+(-1)(2\Phi(-1)-1)=2\phi(1)+1-2\Phi(-1)$. The initial absolute value in the expected-local-time formula is $|x-a|=1$. Therefore

\mathbb{E}[L_1^1(B)] =2\phi(1)+1-2\Phi(-1)-1 =2\phi(1)-2\Phi(-1) \approx 0.1666.

Check: substituting $\phi(1)\approx0.2420$ and $\Phi(-1)\approx0.1587$ into the final expression gives $2(0.2420)-2(0.1587)=0.1666$, which is positive as local time’s expectation must be. The level shift matters: the terminal absolute value is $|B_1-1|$, not $|B_1|$.

Occupation Time as an Integral of Local Time

For Brownian motion, the occupation-density formula has $d[B]_s=ds$, since its quadratic variation is $[B]_s=s$. Thus the time spent in a measurable set $A$ is $\int_A L_t^a(B)\,da$, up to the usual equality almost surely. Taking expectations is legitimate for nonnegative integrands by Tonelli’s theorem. In particular, since $B_s$ is symmetric about zero for every $s>0$, the expected time spent above zero is half the observation time.

Worked Example: Expected Time Above Zero

Compute the expected occupation time $\mathbb{E}\int_0^2\mathbf{1}_{\{B_s>0\}}\,ds$, and check it using expected local time.

For each $s>0$, $B_s\sim N(0,s)$, so $\mathbb{P}(B_s>0)=1/2$. The single time $s=0$ does not affect a time integral. Hence

\mathbb{E}\int_0^2\mathbf{1}_{\{B_s>0\}}\,ds =\int_0^2\mathbb{P}(B_s>0)\,ds =\int_0^2\frac12\,ds=1.

For the local-time check, apply occupation density to $g(a)=\mathbf{1}_{\{a>0\}}$, then take expectations and use Tonelli: $\mathbb{E}\int_0^2\mathbf{1}_{\{B_s>0\}}\,ds=\int_0^\infty\mathbb{E}L_2^a(B)\,da$. The latter is $1$, because the occupation-density identity gives this same integral and the preceding computation evaluates the expected occupation time. This check uses the normalization with coefficient $1$ in Tanaka’s formula; inserting an extra factor of $2$ would give an incorrect occupation time.

A related density calculation follows directly by taking expectations in the occupation-density formula and using the Gaussian density of $B_s$. For nonnegative $g$, Tonelli gives $\int g(a)\mathbb{E}L_t^a(B)\,da=\int_0^t\int g(a)(2\pi s)^{-1/2}e^{-a^2/(2s)}\,da\,ds$. Therefore, for almost every $a$, and in the standard continuous version for every $a$, $\mathbb{E}L_t^a(B)=\int_0^t(2\pi s)^{-1/2}e^{-a^2/(2s)}\,ds$. At $a=0$, this integral equals $\sqrt{2t/\pi}$, agreeing with Tanaka’s formula, since $\mathbb{E}|B_t|=\sqrt{2t/\pi}$.

Reflection: The Regulator Is Boundary Local Time

For a continuous input path $x+B_t$ starting at $x\ge0$, the Skorokhod reflection map on the half-line defines $K_t=\max(0,-\inf_{0\le s\le t}(x+B_s))$ and $Y_t=x+B_t+K_t$. Then $Y_t\ge0$, $K_0=0$, $K$ is continuous and nondecreasing, and $K$ can increase only when $Y=0$. The process $Y$ is reflected Brownian motion and $K$ is its regulator.

Proposition: For the reflected process $Y_t=x+B_t+K_t$ just defined, with the symmetric local-time normalization in Tanaka’s formula, $K_t=L_t^0(Y)$ for every $t\ge0$, almost surely.

Proof. Since $Y\ge0$, $|Y_t|=Y_t$. Tanaka’s formula at zero gives $Y_t=Y_0+\int_0^t\operatorname{sgn}(Y_s)\,dY_s+L_t^0(Y)$, where $Y_0=x$. The integral with respect to $Y$ splits into its Brownian and finite-variation parts: its finite-variation term is $\int_0^t\operatorname{sgn}(Y_s)\,dK_s=0$, because $K$ increases only where $Y_s=0$, and $\operatorname{sgn}(0)=0$. The Brownian part is $\int_0^t\mathbf{1}_{\{Y_s>0\}}\,dB_s$. For each $s>0$, $Y_s$ has a density, so $\mathbb{P}(Y_s=0)=0$; Tonelli implies $\int_0^t\mathbf{1}_{\{Y_s=0\}}\,ds=0$ almost surely. The integrand $\mathbf{1}_{\{Y_s=0\}}$ is bounded and progressively measurable, hence belongs to $\mathcal{L}^2$. The Ito isometry now gives $\int_0^t\mathbf{1}_{\{Y_s=0\}}\,dB_s=0$ almost surely. Consequently, $\int_0^t\mathbf{1}_{\{Y_s>0\}}\,dB_s=B_t$. Substitution into Tanaka’s formula yields $Y_t=x+B_t+L_t^0(Y)$. Comparing with $Y_t=x+B_t+K_t$ proves the equality for each $t$; continuity gives simultaneous equality for all $t$ outside one null set. $\square$

Worked Example: Boundary Local Time and the Reflected Density

Start the reflected process at $x=0$ and take $t=1$. Use the reflected transition density to check its total mass and boundary condition, then find its mean and the expected regulator.

For $y\ge0$, the transition density is $p_t(x,y)=\phi_t(y-x)+\phi_t(y+x)$, where $\phi_t(z)=(2\pi t)^{-1/2}e^{-z^2/(2t)}$. At $x=0,t=1$, this becomes $p_1(0,y)=2\phi(y)$. Its mass on the state space is $\int_0^\infty2\phi(y)\,dy=2(1/2)=1$. The reflecting boundary condition is zero derivative in $y$ at $y=0$. For general $x\ge0$, differentiating the two Gaussian terms gives $\partial_y p_t(x,0)=(x/t)\phi_t(x)-(x/t)\phi_t(x)=0$.

The mean from the density is $\mathbb{E}Y_1=\int_0^\infty y\,2\phi(y)\,dy=2\phi(0)=\sqrt{2/\pi}$, using $\int_0^\infty y\phi(y)\,dy=\phi(0)$. The proposition gives $K_1=L_1^0(Y)$. Also $K_1=Y_1-B_1$, and $\mathbb{E}B_1=0$, so

\mathbb{E}K_1=\mathbb{E}L_1^0(Y)=\mathbb{E}Y_1-\mathbb{E}B_1 =\sqrt{\frac{2}{\pi}}-0=\sqrt{\frac{2}{\pi}}.

Check: the density calculation gives $\mathbb{E}Y_1=\sqrt{2/\pi}$; Tanaka’s expected-local-time formula for a Brownian motion started at zero gives $\mathbb{E}L_1^0(B)=\mathbb{E}|B_1|=\sqrt{2/\pi}$. The agreement confirms the normalization. The reflected process and the driving Brownian motion are different processes, but their expected boundary and level-zero local times agree here.

Common Pitfalls and What to Check

Dropping the level shift. A common incorrect calculation is to write $\mathbb{E}L_t^a=\mathbb{E}|x+B_t|-|x-a|$. With $x=0,a=1,t=1$, this gives $\sqrt{2/\pi}-1$, approximately $-0.2021$. This cannot be an expected local time. The correct Tanaka expression is $\mathbb{E}|x+B_t-a|-|x-a|$, which gives $2\phi(1)-2\Phi(-1)$, approximately $0.1666$.

Confusing local-time conventions. Under the convention used here, the absolute-value Tanaka formula has $+L_t^a$, with no factor of $1/2$, and the occupation formula uses $L_t^a\,da$. A different convention may rescale local time; formulas from different conventions cannot be combined without adjusting the constant.

Checking only the interior density. A candidate reflected density must have total mass one on $y\ge0$ and satisfy the reflecting boundary condition $\partial_y p_t(x,0)=0$. A Gaussian expression that meets only one of these checks is not enough. Here the two image terms together supply both checks.

Check Your Understanding

Keep the level shift and the normalization fixed as you work through these questions.

For $X_t=2+B_t$, use Tanaka’s formula to express $\mathbb{E}L_t^2(X)$ using an absolute-value expectation.
For standard Brownian motion, compute $\mathbb{E}\int_0^3\mathbf{1}_{\{B_s<0\}}\,ds$.
Explain why the Ito integral in Tanaka’s formula has expectation zero, and identify the integrability condition used.
For reflected Brownian motion started at zero, use $K_t=L_t^0(Y)$ and the reflected density to find $\mathbb{E}K_t$.
Differentiate $p_t(x,y)=\phi_t(y-x)+\phi_t(y+x)$ at $y=0$ and verify the reflecting boundary condition.