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Let $(W_t)_{t \ge 0}$ be a Brownian Motion in $\mathbb R^d$ for $d \in \{1,2\}$ (so that the process is recurrent).

Consider the occupation time of Brownian motion given of a set $A \subset \mathbb{R}^d$ (of positive Lebesgue measure) $$ O_A(T) = \int_{0}^T 1_{A}(W_t)dt $$

For fixed $A$, it is clear that $O_A(T)$ diverges to $\infty$ almost surely as $T\to \infty$.

However, I was wondering whether if, for two disjoint sets $A_1,A_2$ (say with same Lebesgue measure) if the variable $$O_{A_1}(T)-O_{A_2}(T)$$ stays bounded as $T$ goes to infinity. My intuition being that a cancellation similar to the one that allows us to define the Green function takes place here.

Furthermore, I would be interested if one can obtain good integrability control of this random variable as $T \to \infty$. Say, control arbitrary integer moments or possibly even exponential ones. The question could also be posed in terms of local times, but I found it simpler to pose this way.

Any references or remarks are welcome. Here I am asking for $A$'s very general, but any particular choice would $A_1$ and $A_2$ would also be very enlightening.

EDIT (24/Jan/24): My language might have been imprecise on the above, so just as a further clarification.

By "good integrability control of this random variable as $T \to \infty$. Say, control arbitrary integer moments or possibly even exponential ones. ", I mean to understand how moments such as $$ (O_{A_1}(T)- O_{A_2}(T))^k , |O_{A_1}(T)- O_{A_2}(T)|^k, e^{\theta|O_{A_1}(T)- O_{A_2}(T)|} $$ behave as $T \to \infty$ (where $k \in \mathbb{N},\mathbb{R}^+$).

This allows for the moments to diverge as $T \to \infty$, but I would like to understand the order of such divergence in terms of $T$.

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I answer in the case of dimension 1.

Call $(L_t^x)_{t \ge 0}^{x \in \mathbb{R}}$ the local time process of $W$ and $(\tau_r)_{r \ge 0}$ the inverse of local time at $0$.

Then, by Markov property, the process $((L_{\tau_r}^x)^{x \in \mathbb{R}})_{r \ge 0}$ (with values in the space of all non-negative continuous functions on $\mathbb{R}$ with compact support) have independent and stationary increments.

Moreover, $\mathbb{E}[L_{\tau_r}^x] = r$ for all $r$ and $x$ (it follows from Ray - Knight theorem, but it can also be proved directly with stochastic calculus). Integrating with regard to $x$, and using $$O_A(t) = \int_A 1_{A}(x) L_t^x dx,$$ we get that for every Borel subset $A$ of $\mathbb{R}$, $\mathbb{E}[O_A(\tau_r)] = r\lambda(A)$.

If $A_1$ and $A_2$ are Borel subsets with the same finite Lebesgue measure, such that $A_1 \triangle A_2$ has positive Lebesgue measure, the the process $(O_{A_1}(\tau_r)-O_{A_2}(\tau_r))_{r \ge 0}$ is centered, not deterministic, and has independent and stationary increments.

Hence this process cannot be bounded, although law of large numbers applies and central limit theorem also (the random variables are square integrable). Therefore, the process $(O_{A_1}(t)-O_{A_2}(t))_{t \ge 0}$ cannot be bounded.

EDIT : getting an upper bound of moments at any stopping time $\tau$, including constant times.

If $A_1$ and $A_2$ are Borel subsets with the same finite Lebesgue measure, $O_{A_1}(\tau)-O_{A_2}(\tau)$ can be written as this Lebesgue measure times the integral of $L_\tau^x-L_\tau^y$ with regard to $d\pi(x,y)$, where $\pi$ is any coupling of the uniform measures on $A_1$ and $A_2$. Hence it suffices to bound above the moments of $L_\tau^x-L_\tau^y$.

By Tanaka's formula, for $t \ge 0$, $$L_t^x-L_t^y = |B_t-x|-|B_t-y|-|x|+|y| + M^{x,y}_t,$$ where $M^{x,y}$ is the martingale defined by $$M^{x,y}_t = \int_0^t (\mathrm{sign}(B_s-x)-\mathrm{sign}(B_s-y)) dB_s.$$ Thus for every stopping time $\tau$ (including constant times) $$|L_\tau^x-L_\tau^y| \le 2|x-y| + |M^{x,y}_\tau|.$$ Given $p \ge 1$, we derive $$\Vert L_\tau^x-L_\tau^y \Vert_p^p \le \Big( 2|x-y| + \Vert M^{x,y}_\tau \Vert_p\Big)^p.$$ To get an upper bound, we use Barlow - Yor inequalities. $$\Vert M^{x,y}_\tau \Vert_p^p \le C_p\mathbb{E}[\langle (M^{x,y}_\tau)^{p/2} \rangle],$$ where $C_p$ is some positive constant.

Note that its quadratic variation is given by $$\langle M^{x,y} \rangle_t = \int_0^t (\mathrm{sign}(B_s-x)-\mathrm{sign}(B_s-y))^2 ds = 4\int_0^t \mathbb{1}_{[\min(x,y) \le B_s \le \max(x,y)]} ds,$$ since
$$\int_0^t \mathbb{1}_{[B_s \in \{x,y\}} ds = 0.$$ A trivial upper bound is $\langle M^{x,y} \rangle_t \le 4t$. A less trivial upper bound is $$\mathbb{E}[\langle M^{x,y} \rangle_t] = \int_0^t \mathbb{P}[B_s \in \{x,y\}] ds \le \int_0^t \frac{1}{\sqrt{2\pi s}} |x-y| ds = \frac{\sqrt{2s}}{\sqrt{\pi}} |x-y|.$$

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  • $\begingroup$ Hi there, thanks for your answer! I really appreciate it. Perhaps I should have been more explicit, and therefore, I will update the question. However, by "good integrability control of this random variable as $T \to \infty$", I meant to understand how those moments behave as $T \to \infty$. This allows for the moments to diverge as $T \to \infty$, but I would like to understand the order of such divergence in terms of $T$. $\endgroup$
    – Kernel
    Commented Jan 24 at 9:32
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    $\begingroup$ @Kernel I completed the answer to give a method to bound above the moments. $\endgroup$ Commented Jan 31 at 8:28
  • $\begingroup$ Thanks! That was very enlightening :) $\endgroup$
    – Kernel
    Commented Jan 31 at 11:26

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