Probability of Brownian motion hitting $n$ integers and the expected hit within a time period

Question

Given an Ito process $x(t):=\sigma B(t)$ with an unknown constant volatility $\sigma$. Is there a way to estimate $\sigma$ by simply counting the number of times the Ito process hits a lattice with interval length $\delta x$? We know $\mathbf E[x(t)^2]=\sigma^2 t$. For a lattice size $\delta x$, $(\delta x)^2=\sigma^2\delta t$ where $\delta t$ is the time for $x(t)$ to traverse a single lattice step of size $\delta x$. On average $\delta t=\frac Tn$ where $T$ is the total time and $n$ is the number of hits. So it is plausible to conjecture $\big(\frac{\delta x}{\sigma}\big)^2=\frac Tn$. How do we make it rigorous and prove it? The following is a rigorous formulation of this conjecture.

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Suppose we have a standard Brownian motion $B(t)$ starting from $0$. $\forall \omega$ in the sample space, let the stopping time sequence $\tau(\omega,T):=\big\{t_i|i\in \overline{\mathbf Z^-}, t_0=0, T\ge t_i>t_{i-1}, t_0=0,\,t_i:=\inf\{t:|B(\omega,t)-B(\omega,t_{i-1})|=1\},\,\forall i\in \mathbf N\big\}$. This is the event that the standard Brownian motion $B$ starting from $0$ hitting integers where any two consecutive integers are distinct. What are the probability $P(n,T)= \text{Pr}(|\tau(\omega,T)|\ge n)$ and $\mathbf E[|\tau(\omega,T)]$?

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1. Let $$\epsilon(\omega, T):=T-\sum_{i=1}^{|\tau(\omega,T)|}t_i(\omega).$$ Take the expectation. If we conjecture $|\tau(\omega,T)|$ is asymptotically independent of $t_i(\omega)$ $\forall i$ as $T\rightarrow\infty$, we have asymptotically $$T\sim \mathbf E[|\tau(\omega,T|]\mathbf E[t_i], \quad T\rightarrow\infty.$$ It can be evaluated by the property of a martingale that $$\mathbf E[t_i]=1.$$

2. I have a rigorous algorithm relying on the Laplace transform, shown below as an answer, for the computation. I am wondering if there is a simpler expression. Particularly, I wonder if there is a martingale method solution in the vein of, say, this expect exit time solution.

Answer 1

Here is my algorithm to compute the required probability and expectation. I would very much like to know a simpler solution.

We know the probability $\Phi(t):=\text{Pr}(\exists s<t \ni|B(s)|=1)$. Let $\phi(t):=\frac{\partial P(t)}{\partial t}$, since the Brownian motion is strongly Markovian, \begin{align} P(n,t)&:=\text{Pr}(|\tau(\omega,t)|\ge n) \\ &= \int_{t_1=0}^t d t_1 \phi(t_1)\int_{t_2=0}^{t-t_1}dt_2 \phi(t_2)\cdots \int_{t_i=0}^{t-\sum_{j=1}^it_j}dt_i\, \phi(t_i)\cdots \int_{t_n}^{t-\sum_{j=1}^{n-1}t_j}dt_n \phi(t_n). \end{align} To facilitate the computation, we take the Laplace transform $\mathscr L$ of the above integral and use $s$ as the variable for the resultant function. \begin{equation} \mathscr L[P(n,t)](n,s) = \frac{\mathscr L[\phi](s)^n}s \end{equation} The expectation $$\mathbf E[n=|\tau(\omega,t)|] = \sum_{n=1}^\infty n(P(n,t)-P(n+1,t)) = \sum_{n=1}^\infty P(n,t),$$ and \begin{equation} \mathscr L\big[\mathbf E[n=|\tau(\omega,t)|]\big](n,s) = \frac1s\frac{\mathscr L[\phi]}{1-\mathscr L[\phi]} \end{equation} \begin{equation} \mathscr L[P(n,t)](n,s) = \frac{\mathscr L[\phi](s)^n}s \end{equation} The expectation $$\mathbf E[n=|\tau(\omega,t)|] = \sum_{n=1}^\infty n(P(n,t)-P(n+1,t)) = \sum_{n=1}^\infty P(n,t),$$ and \begin{equation} \mathscr L[\mathbf E[n=|\tau(\omega,t)|](n,s) = \frac1s\frac{\mathscr L[\phi]}{1-\mathscr L[\phi]} \tag1 \end{equation}

Now following the marvelous suggestion of @SangchulLee in the comment below, consider the random variable $M(t):=e^{\sqrt{2s}B(t)-st}$. Since It is a martingale. By the optional stopping theorem, \begin{align} 1&=\mathbf E[M(0)]=\mathbf E[M(t)]=\mathbf E[M(t_1)] \nonumber\\ &= \int_0^\infty \Big(e^{-\sqrt{2s}-st}\text{Pr}(\inf\{u:B(u)=-1\}\in dt)+e^{\sqrt{2s}-st}\text{Pr}(\inf\{u:B(u)=1\}\in dt)\Big) \nonumber\\ &= \cosh{\sqrt{2s}}\int_0^\infty e^{-st}\phi(t)dt \nonumber\\ &= \cosh{\sqrt{2s}}\;\mathscr L[\phi](s) \tag2 \end{align} since $\text{Pr}(\inf\{u:B(u)=-1\}\in dt)=\text{Pr}(\inf\{u:B(u)=1\}\in dt)=\frac12\phi(t)dt$ by symmetry. Substituting Equation (2) into Equation (1), we have \begin{equation} \mathscr L\big[\mathbf E[n=|\tau(\omega,t)|]\big](n,s) = \frac1{s(\cosh{\sqrt{2s}}-1)}. \end{equation}

Then $\mathbf E[n=|\tau(\omega,t)|]$ is obtained by the inverse Laplace transform.