Expected Local Time in Brownian Examples

From Occupation Time to Expected Local Time

Brownian local time measures how much of a path accumulates near a specified level. The occupation-density formula connects local time to ordinary time spent in a band around that level. Taking expectations then brings in the Gaussian transition density. There is one important point: occupation averages over shrinking bands converge to local time, but convergence in probability alone does not imply convergence of their expectations. We will supply the additional integrability argument needed to pass to the limit.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. For a fixed starting point $x\in\mathbb{R}$, set $X_t=x+B_t$. Fix a level $a\in\mathbb{R}$. We use the symmetric semimartingale local-time normalization: the occupation formula is $\int_{\mathbb{R}}g(y)L_t^y(X)\,dy=\int_0^t g(X_s)\,ds$ for nonnegative measurable $g$, and Tanaka’s formula is $|X_t-a|=|x-a|+\int_0^t\operatorname{sgn}(X_s-a)\,dB_s+L_t^a(X)$. Here $\operatorname{sgn}(0)=0$. These are standard results from the earlier material on occupation density and Tanaka’s formula; we use the version of Brownian local time that is continuous in the level.

All stochastic integrals below are Itô integrals on this filtered space. Their integrands are progressively measurable and belong to $\mathcal{L}^2$, the class satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, the sign integrand in Tanaka’s formula is bounded and progressively measurable, so its integral is a square-integrable martingale.

Definition: For $\varepsilon>0$, define the occupation average at level $a$ by $A_\varepsilon(t,a)=\dfrac{1}{2\varepsilon}\int_0^t\mathbf{1}_{\{|X_s-a|<\varepsilon\}}\,ds$. By the occupation formula, this is also $\dfrac{1}{2\varepsilon}\int_{a-\varepsilon}^{a+\varepsilon}L_t^y(X)\,dy$. Continuity of $y\mapsto L_t^y(X)$ gives $A_\varepsilon(t,a)\to L_t^a(X)$ almost surely as $\varepsilon\downarrow0$.

The Gaussian transition density of $X_s$, for $s>0$, is $p_s(y-x)=\dfrac{1}{\sqrt{2\pi s}}\exp\!\left(-\dfrac{(y-x)^2}{2s}\right)$. Directly taking expectations in the occupation average gives

\mathbb{E}[A_\varepsilon(t,a)] =\frac{1}{2\varepsilon}\int_0^t\int_{a-\varepsilon}^{a+\varepsilon}p_s(y-x)\,dy\,ds.

The formula suggests that the limit should be $\int_0^t p_s(a-x)\,ds$. The next result proves that this is indeed the expectation of local time at the fixed level, not merely an almost-everywhere statement in the level variable.

The Fixed-Level Expectation Formula

Theorem: For $t>0$, Brownian motion $X_t=x+B_t$, and every fixed $a\in\mathbb{R}$, the local time $L_t^a(X)$ is integrable and $$ \mathbb{E}[L_t^a(X)]=\int_0^t p_s(a-x)\,ds =\int_0^t\frac{1}{\sqrt{2\pi s}}\exp\!\left(-\frac{(a-x)^2}{2s}\right)ds. $$

Proof. Fix $\varepsilon>0$. For $0<s<r$, the pair $(X_s,X_r)$ has a joint Gaussian density whose supremum over $\mathbb{R}^2$ is $1/(2\pi\sqrt{s(r-s)})$. Indeed, its covariance matrix has determinant $s(r-s)$. The rectangle $(a-\varepsilon,a+\varepsilon)^2$ has area $4\varepsilon^2$; therefore $\mathbb{P}(|X_s-a|<\varepsilon,\ |X_r-a|<\varepsilon)\le 4\varepsilon^2/(2\pi\sqrt{s(r-s)})$. The diagonal $s=r$ has two-dimensional Lebesgue measure zero. Expanding $A_\varepsilon^2$, using Tonelli’s theorem for the nonnegative integrand, and splitting the square $[0,t]^2$ into its two triangles gives

\begin{aligned} \mathbb{E}[A_\varepsilon(t,a)^2] &=\frac{2}{4\varepsilon^2}\int_0^t\int_s^t \mathbb{P}(|X_s-a|<\varepsilon,\ |X_r-a|<\varepsilon)\,dr\,ds\\ &\le \frac{1}{\pi}\int_0^t\int_s^t\frac{1}{\sqrt{s(r-s)}}\,dr\,ds =\frac{1}{\pi}\int_0^t\left(\int_0^r\frac{1}{\sqrt{s(r-s)}}\,ds\right)dr =t. \end{aligned}

For the last equality, substitute $s=ru$ in the inner integral; it becomes $\int_0^1[u(1-u)]^{-1/2}du=\pi$. Thus the family $A_\varepsilon(t,a)$ has uniformly bounded second moments and is uniformly integrable. Since it converges almost surely to $L_t^a(X)$, uniform integrability implies convergence in $L^1$, and hence $\mathbb{E}[A_\varepsilon(t,a)]\to\mathbb{E}[L_t^a(X)]$.

It remains to compute the limit of the expectations. For each $s>0$, the average $\dfrac{1}{2\varepsilon}\int_{a-\varepsilon}^{a+\varepsilon}p_s(y-x)\,dy$ tends to $p_s(a-x)$ by continuity of the Gaussian density in $y$. It is bounded above by $1/\sqrt{2\pi s}$, whose integral over $(0,t]$ is finite. Dominated convergence therefore gives the stated formula. This also establishes integrability of $L_t^a(X)$ and completes the proof. $\square$

The proof addresses a subtlety that matters whenever a density is evaluated at one particular level: the occupation formula by itself identifies integrals over levels, and does not directly identify the expected value at every fixed level. Here continuity gives almost-sure convergence of the averages, while the second-moment bound supplies the missing step for expectations.

A Closed Form and an Independent Check

Write $q=|x-a|$, and let $\Phi$ denote the standard normal cumulative distribution function. The density integral has the closed form below. Tanaka’s formula gives an independent route to the same expression.

Corollary: For $t>0$, $$ \mathbb{E}[L_t^a(X)] =\sqrt{\frac{2t}{\pi}}\exp\!\left(-\frac{q^2}{2t}\right)-2q\,\Phi\!\left(-\frac{q}{\sqrt{t}}\right), \qquad q=|x-a|. $$

Proof. First verify that the displayed expression equals the density integral. Let $F(t)$ denote the right-hand side. For $q>0$, differentiating its first term gives $\dfrac{1}{\sqrt{2\pi t}}e^{-q^2/(2t)}+\dfrac{q^2}{\sqrt{2\pi}t^{3/2}}e^{-q^2/(2t)}$. Differentiating its second term gives $-\dfrac{q^2}{\sqrt{2\pi}t^{3/2}}e^{-q^2/(2t)}$, since the standard normal density at $q/\sqrt{t}$ is $e^{-q^2/(2t)}/\sqrt{2\pi}$. The extra terms cancel, leaving $F'(t)=p_t(a-x)$. Also $F(t)\to0$ as $t\downarrow0$ when $q>0$. When $q=0$, $F(t)=\sqrt{2t/\pi}$, so $F'(t)=1/\sqrt{2\pi t}=p_t(0)$ and $F(t)\to0$ as well. Integrating this derivative from zero to $t$ proves the formula.

For the independent check, Tanaka’s formula applies with integrand $H_s=\operatorname{sgn}(X_s-a)$. This process is progressively measurable and $\mathbb{E}\int_0^tH_s^2\,ds\le t$, so the Itô integral is square-integrable. The Itô isometry gives its second moment as $\mathbb{E}\int_0^t H_s^2ds\le t$; its martingale property therefore implies $\mathbb{E}[\int_0^tH_s\,dB_s\mid\mathcal{F}_0]=0$. Taking expectations in Tanaka’s formula yields $\mathbb{E}L_t^a(X)=\mathbb{E}|x+B_t-a|-|x-a|$. Since $x+B_t-a$ is normal with mean $x-a$ and variance $t$, the normal absolute-moment formula gives $\mathbb{E}|x+B_t-a|=\sqrt{2t/\pi}e^{-q^2/(2t)}+q(1-2\Phi(-q/\sqrt t))$. Subtracting $q$ gives precisely the claimed expression. $\square$

Worked Brownian Examples

Worked Example: Started at the Level

Let $X_t=B_t$, so $x=0$, and compute $\mathbb{E}[L_2^0(X)]$. Here $a=0$, so $p_s(0)=1/\sqrt{2\pi s}$. The density formula gives

\mathbb{E}[L_2^0(X)] =\int_0^2\frac{1}{\sqrt{2\pi s}}\,ds =\frac{1}{\sqrt{2\pi}}\,[2\sqrt{s}]_0^2 =\frac{2}{\sqrt{\pi}}.

Check: Tanaka’s formula has initial term $|X_0|=0$, and its stochastic-integral term has mean zero by the square-integrability argument above. Thus $\mathbb{E}[L_2^0(X)]=\mathbb{E}|B_2|$. Since $B_2$ is normal with mean zero and variance $2$, $\mathbb{E}|B_2|=\sqrt{2\cdot2/\pi}=2/\sqrt{\pi}$, agreeing with the density calculation.

Worked Example: Started at Zero, Level One

Let $X_t=B_t$, and find the expected local time at $a=1$ by time $t=1$. The starting point is $x=0$, so $q=|0-1|=1$. The density calculation is

\mathbb{E}[L_1^1(X)] =\int_0^1\frac{1}{\sqrt{2\pi s}}e^{-1/(2s)}\,ds =\sqrt{\frac{2}{\pi}}e^{-1/2}-2\Phi(-1).

Check: In Tanaka’s formula, the expected absolute-value increment is $\mathbb{E}|B_1-1|-1$. For a normal variable with mean $-1$ and variance $1$, the normal absolute-moment formula gives $\mathbb{E}|B_1-1|=\sqrt{2/\pi}e^{-1/2}+1-2\Phi(-1)$. Subtracting the initial absolute value $1$ leaves $\sqrt{2/\pi}e^{-1/2}-2\Phi(-1)$, exactly the same expression. The density integral also shows directly that this expectation is positive.

Worked Example: Started Away from the Level

Let $X_t=2+B_t$, and compute the expected local time at level $a=0$ by time $t=2$. Now $q=|2-0|=2$, and the transition density at level zero is $p_s(-2)$. Thus

\mathbb{E}[L_2^0(X)] =\int_0^2\frac{1}{\sqrt{2\pi s}}e^{-4/(2s)}\,ds =\sqrt{\frac{4}{\pi}}e^{-1}-4\Phi(-\sqrt{2}).

Check: Tanaka’s formula gives $\mathbb{E}L_2^0(X)=\mathbb{E}|2+B_2|-2$. The normal absolute-moment formula, with mean $2$ and variance $2$, gives $\mathbb{E}|2+B_2|=\sqrt{4/\pi}e^{-1}+2(1-2\Phi(-\sqrt2))$. Subtracting $2$ yields $\sqrt{4/\pi}e^{-1}-4\Phi(-\sqrt2)$, which matches the density integral. In particular, the expectation depends on the distance from the starting point to the level, not just on the elapsed time.

Common Pitfall: Passing Expectations Through a Limit

A tempting but invalid argument is: “$A_\varepsilon(t,a)\to L_t^a(X)$ in probability, so $\mathbb{E}A_\varepsilon(t,a)\to\mathbb{E}L_t^a(X)$.” Convergence in probability alone does not imply convergence of expectations. For example, on a probability space with an event $E_n$ of probability $1/n$, the variables $Y_n=n\mathbf{1}_{E_n}$ converge to zero in probability, while $\mathbb{E}Y_n=1$ for every $n$. Thus the limit of the expectations cannot be inferred from convergence in probability alone.

For the local-time averages, the correction is specific and quantitative: the joint Gaussian density yields $\mathbb{E}[A_\varepsilon(t,a)^2]\le t$, uniformly over $\varepsilon>0$. This implies uniform integrability; together with almost-sure convergence, it gives $L^1$ convergence and validates the passage to expectations. Tanaka’s formula then provides a useful independent check, but it does not remove the need to justify a limit if the expectation is obtained through occupation averages.

The density formula also explains the role of the starting point. At the starting level, $p_s(a-x)$ has a $s^{-1/2}$ singularity as $s\downarrow0$, but this singularity is integrable, producing $\sqrt{2t/\pi}$ when $a=x$. Away from the starting level, the exponential factor suppresses the density near time zero. In both cases the expected local time is finite, and the same fixed-level formula applies.

Check Your Understanding

Use the fixed-level formula and, where useful, Tanaka’s formula to answer the following.

For $X_t=B_t$, write the density integral for $\mathbb{E}[L_t^0(X)]$ and evaluate it for general $t>0$.
For $X_t=3+B_t$, identify $q$ when computing expected local time at level $1$. What is the closed-form expression at time $t>0$?
Why does almost-sure convergence of $A_\varepsilon(t,a)$ to $L_t^a(X)$ not, by itself, justify convergence of their expectations?
In the second-moment estimate, why is the diagonal $s=r$ irrelevant to the double integral?
Use Tanaka’s formula to express $\mathbb{E}[L_t^a(X)]$ as a difference of expected absolute values, and state why its stochastic-integral term has expectation zero.