# Brownian Motion hitting probability for square-root boundaries.

For a standard Brownian Motion $W(t)$, consider the general square-root boundary $g(t) = a + b\sqrt{t}$, where $a>0$ and $b>0$ are given constants. My question is that, for a given time $T > 0$, what is the probability that $W(t)$ hits $g(t)$ before $T$?

In other words, how to compute the following one-side exiting probability? $$P(W(t) \ge a + b\sqrt{t}, \exists t < T ).$$

Considering boundaries of the form $g(t) = a\sqrt{b+t}$ or other general square-root forms is also acceptable if that is easier.

Many thanks!

• Any background for such probability? – Gordon Aug 24 '16 at 16:58

A solution to a similar problem can be found in the Artikle [Brownian Motion Hitting Probabilities for General Two-Sided Square-Root Boundaries] by D. Donchev 2009. It is based on a representation of hitting Probabilieties as a solution to a PDE (by Skorokhod). (I'll write the essence of the prove as i think this can be done)

Consider the stopping time $$\tau=\inf\{t>0:B_t=f(t)\}.$$ Then$$u(t,x)=P_{t,x}(B_{\tau}=f(\tau),\tau<T)$$ satisfies the following PDE: \begin{eqnarray*} &&\frac{\partial u}{\partial t}+\frac{1}{2}\frac{\partial^2 u}{\partial x^2}=0 \\ &&u(t,f(t))=1,\quad u(t,-\infty)=u(T,x)=0. \end{eqnarray*} to have an easier boundary condition we rescale: $u(t,x)=v(t,\frac{x}{f(t)})$ which gives us

\begin{eqnarray*} &&\frac{\partial v}{\partial t}-\frac{f'(t)x}{f(x)}\cdot\frac{\partial v}{\partial x}+\frac{1}{2f(t)^2}\frac{\partial^2 v}{\partial x^2}=0 \\ &&v(t,1)=1,\quad v(t,-\infty)=v(T,x)=0. \end{eqnarray*} plug in the explicit form of the boundary $f(t)=a\sqrt{t+b}$

\begin{eqnarray*} \frac{\partial v}{\partial t}-\frac{x}{2(t+b)}\frac{\partial v}{\partial x}+\frac{1}{2a^2(t+b)}\frac{\partial^2 v}{\partial x^2}=0 \end{eqnarray*}

with a change of time to $t\rightarrow \frac{1}{2}\ln\frac{T+b}{t+b}=:l(t)$ ($v(t,x)=\tilde{v}(l(t),x)$) the problem can be transformed to one with time-independent coefficients

\begin{eqnarray*} &-\frac{\partial \tilde{v}}{\partial t}-x\frac{\partial \tilde{v}}{\partial x}+\frac{1}{2a^2}\frac{\partial^2 v}{\partial x^2}=0 \\ &\tilde{v}(t,1)=1,\quad \tilde{v}(t,-\infty)=\tilde{v}(0,x)=0 \\ &x<1,\quad 0<t<\frac{1}{2}\ln\frac{T+b}{b}. \end{eqnarray*}

By a laplace transform w.r.t. time t an ODE can be obtained (let $\tilde{V}(x,p)$ be the laplacetransform w.r.t. t of $\tilde{v}(t,x)$)

\begin{eqnarray} &\frac{1}{2b^2}\tilde{V}''-x\tilde{V}'-p\tilde{V}=0 \\ &V(1,p)=\frac{1}{p},\quad V(-\infty,p)=0 \end{eqnarray}

The solution of this equation is quite ugly but known (Kamke 1959, Part 3, Section 2, case (10) of Equation 2.273, also wolfram alpha knows the solution-just type it in the webapplication) it involves Hermite polynomials and the Kummer confluent Hypergeometric function.

If you try the boundary $f(t)=b+a\sqrt{t}$ you should be able to obtain a solution in a similar manner (with different time change)

• Thanks so much! I think this is just what I want. I am little confused about the notation. What is that $P_{t,x}$? The probability conditioning on $B(t) = x$ or what? – sjasonma Aug 30 '16 at 19:37
• Thank you! I think I understand most of them, except for some boundary conditions.. For $u$, why isn't it $u(T,x) = 0$? And what I got for $v$ and $\tilde{v}$ are $v(t,-\infty) = v(T,x) = 0$, and for $\tilde{v}(t,-\infty) = \tilde{v}(0,x) = 0$. – sjasonma Aug 31 '16 at 0:20
• And what do you mean by just type it in the web application''? You mean typing the last ODE like this:" \frac{1}{2b^2} V'' - xV - pV = 0, V(1,p) = 1/p, V(-\infty,p) = 0" in wolfram alpha? Sorry I am not very familiar with that web application, I just want to get the solution of it. Thanks a lot! – sjasonma Aug 31 '16 at 0:21