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!
 A: 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)
See also


*

*Novikov 1971 'On Stopping Times for the Wiener Process'

*Novikov, Frishling, Kordzakhia 1999 'Approximations of Boundary crossing Probabilities for a Brownian motion'

