Just come up with a 'simple' and interesting problem that I've been struggling to deal with for some time. Consider a filtered probability space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t\in[0,T]},\mathbb{P})$ with usual conditions, where $T>0$ is a constant and the filtration is generated by a Brownian motion $W$ with $W_0=x$ for some $x>0$. Define the first hitting time and the bounded one by $$\tau_w:=\inf\{u>0:W_u=0\} \quad\text{and}\quad \tau:=\tau_w\wedge T.$$ I am interested in the distribution of random variable $$ \int_0^\tau W_u\,du$$ in order to compute the expectation $\mathbb{E}[1/(\delta+\int_0^\tau W_u\,du)]$ for some $\delta>0$. One may also think of finding the conditional distribution

$$ \int_0^T W_u\,du \;\big|\;\tau_w>T$$

due to the idea below.

Ideas (Dimension reduction): I would like to share with you some of my ideas. One can see that this distribution is multimodal from the histogram of simulation results

Histogram of 10^5 simulations

where the first peak is caused by early hittings and the second peak refers to no hittings, as well as from the decomposition

$$\mathbb{P}\bigg(\int_0^\tau W_u\,du\leq a\bigg)=\mathbb{P}\bigg(\int_0^T W_u\,du\leq a\,|\, \tau_w>T\bigg)\,\mathbb{P}(\tau_w>T)+\mathbb{P}\bigg(\int_0^{\tau_w} W_u\,du\leq a\,|\, \tau_w\leq T\bigg)\,\mathbb{P}(\tau_w \leq T).$$ Noting that

$$\int_0^T W_u\,du=\int_0^\tau W_u\,du+\int_\tau^T W_u\,du,$$

the LHS of which is a known Gaussian variable and two integrals on RHS are independent conditional on $\tau$, the first idea follows from

\begin{equation} \begin{aligned} \mathbb{P}\bigg(\int_0^T W_u\,du\leq a\bigg)&=\mathbb{P}\bigg(\int_0^T W_u\,du\leq a\,|\, \tau_w>T\bigg)\,\mathbb{P}(\tau_w>T)\\ &+\mathbb{P}\bigg(\int_0^{\tau_w} W_u\,du+\int_{\tau_w}^T W_u\,du\leq a\,|\, \tau_w<T\bigg)\,\mathbb{P}(\tau_w<T)\\ &=\mathbb{P}\bigg(\int_0^T W_u\,du\leq a\,|\, \tau_w>T\bigg)\,\mathbb{P}(\tau_w>T)\\ &+\int_0^T \int_{-\infty}^a\mathbb{P}\bigg(\int_0^{\tau_w} W_u\,du\leq a-b\,|\, \tau_w=s\bigg)f_b(b|s)f_\tau(s)\,db\,ds \end{aligned} \end{equation}

where $f_\tau$ is the density of the hitting time and $f_b(b|s)$ is the one associated with the $\mathbb{P}(\int_s^TB_u\,du\leq b)$ ($B$ is an independent Brownian motion with $B_s=0$). Since the LHS is a Gaussian CDF and there are two unknown distributions on RHS, if one of them is solved, the other one can be determined ideally. Hence, I pick the first one as stated before. As it's a Riemann integral, it's natural to think about

$$\int_0^T W_u\,du \;\big|\;\tau_w>T\;\approx\;\sum_{j=0}^{n-1} W_{t_j}\,\cdot\Delta \;\big|\;\tau_w>T,$$

with $0=t_0<t_1\cdots<t_n=T$ being a partition of $[0,T]$ of even step size $\Delta$. Given $W_{t_j}$, the distribution of $W_{t_j+1}$ is no longer Gaussian due to the positiveness of the path. The conditional distribution of such 'positive' path can be deduced as follows

\begin{equation} \begin{aligned} f(W_t=a, m_T>-x)&=f(W_t=a, m_t>-x, m_T>-x)\\ &=f(m_T>-x\,|\,W_t=a, m_t>-x)\cdot f(W_t=a, m_t>-x)\\ &=f(m_{T-t}>-(x+a)\,|\,W_0=a)\cdot f(W_t=a, m_t>-x)\\ &=\big(1-2\Phi_{T-t}(-(x+a))\big)\cdot\frac{e^{-\frac{a^2}{2t}}-e^{-\frac{(a+2x)^2}{2t}}}{\sqrt{2\pi t}} \end{aligned} \end{equation}

$$f(W_t=a\,|\,m_T>-x)=\frac{1-2\Phi_{T-t}(-(x+a))}{1-2\Phi_{T}(-x)}\cdot\frac{e^{-\frac{a^2}{2t}}-e^{-\frac{(a+2x)^2}{2t}}}{\sqrt{2\pi t}}$$

for $a>-x$, where $f$ represents the density function, $m$ is the running minimum process and $\Phi_s$ denotes the CDF of the Gaussian variable with mean $0$ and variance $s$. One can also see that the independence of Brownian increments breaks down and the increment now depends on the past only through the latest state value. But, it doesn't seem to be easy to deal with the sum of 'weakly' correlated variables with this specific distribution. Also, the approximation approach can be used to the stochastic integral in

$$\tau W_{\tau}=\int_0^{\tau} W_u\,du+\int_0^{\tau}u\,dW_u,$$

and the simple random walk in order to apply the Donsker invariance principle.

Thanks for your time. Any ideas or comments are highly appreciated and I hope you enjoy this problem.

  • $\begingroup$ Up to known, is the distribution of $\tau_{\omega}$ known? $\endgroup$
    – NN2
    Jun 3, 2021 at 19:11
  • $\begingroup$ @NN2 \tau_w is the Brownian hitting time and its density is known. $\endgroup$
    – FoolAlex
    Jun 4, 2021 at 11:22
  • $\begingroup$ I doubt that the distribution of $\tau_{\omega}$ is known (or even definable). Indeed, what we know is the distribution of the hitting time $\tau_a$ defined as $$\tau_a:=\inf\{u\color{red} {\ge a}:W_u=0\}$$ with $a\color{red}{>}0$. But when $a \to 0$, its density function becomes undefined. $\endgroup$
    – NN2
    Jun 4, 2021 at 11:31
  • $\begingroup$ You can find the density from homepage.univie.ac.at/irene.klein/stoch_6_10.pdf and columbia.edu/~ks20/FE-Notes/4700-07-Notes-BM.pdf for example. I think you may refer to return time, while it is not the case since our BM does not start from 0. $\endgroup$
    – FoolAlex
    Jun 4, 2021 at 12:16


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