Integrating the error function in a calculation related to Brownian motion

Question

I wish to calculate the probability that a standard linear Brownian motion $B(t)$, $t\ge 0$, will be at time $t_0$ inside the interval $[a,b]$, and at time $t_1$ in the interval $[c,\infty)$. To do so, I have done the following calculation: suppose the event that it happens is $A$. So:

\begin{equation} \mathbb{P}(A) = \frac{1}{2} \int_a^b\frac{1}{\sqrt{2\pi t_0}}e^{-\frac{x^2}{2t_0}} \left(1 - \text{erf}\left(\frac{c-x}{\sqrt{2(t_1-t_0)}}\right)\right)dx \end{equation}

Explanation: the first term in the integral is the pdf of the normal distribution with mean 0 and variance $t_0$; conditioning on that, the second term (including the $1\over 2$ left to the integral) is $1$ minus the cdf of the normal distribution with mean $x$ and variance $t_1-t_0$ at point $c$.

I believe my calculation is right. However, I don't know how to solve that integral, or even if there is a "solution". If not, what numeric methods can I apply here?

Answer 1

I don't think there's a closed-form solution. For numerical purposes, you might use "integrate" in R, or any standard method, like Simpson's method since the integrand is a standard function.

For example:

 a <- 0
 b <- 2 
 c <- -2
 t0 <- 1
 t1 <- 2

 f1 <- function(x){
    dnorm(x,0,sqrt(t0))*(1-pnorm(c-x)) 
 }  

 integrate(f1,a,b)