If I were to try to brute force compute the following integral:
$$F(t_{i-1}, x_{i-1})= e^{-r(t_{i}-t_{i-1})}\int_{-\infty}^{\infty} dx_{i} \cdot G(x_{i}, t_{i}|x_{i-1}, t_{i-1})F(t_i, x_{i}),$$
where $G$ is a propagator defined by: $$G(x_{i}, t_{i}|x_{i-1},t_{i-1}) = \frac{1}{\sqrt{2\pi\sigma^{2}(t_{i}-t_{i-1})}}\exp\left(-\frac{[(x_{i}-x_{i-1})-\mu \cdot(t_{i}-t_{i-1})]^{2}}{2\sigma^{2}(t_{i}-t_{i-1})}\right),$$
where $x \in (-\infty, \infty)$ and $t\in (t_0, t_N)$, and the remaining symbols are constants. The path integral is:
$$F(t_0, x_0) = \int \int\int...\int \int_{-\infty}^{\infty}e^{-r(t_{N}-t_{N-1})} \cdot G(x_{N}, t_{N}|x_{N-1}, t_{N-1})F(t_N, x_{N}) dx_{N} dx_{N-1} ... dx_1,$$
in the limit of $N\rightarrow \infty$. The function $F(t_N, x_N)$ is known and the goal is to find $F(t_0, x_0)$. I want to compute this directly.
I would discretize the time and $x$ domains into a regular grid consisting of $N_x×N_t$ grid points. Then I would compute the integral for the points $x_{N-1}$ using the trapezoidal rule. This would involve computing $N_x$ integrals, one for every grid point at $t_{N-1}$. Then I would move to time $t_{N-2}$ etc until I arrive at $t_0$.
How do I determine what the error in this should be? I know that the error of the trapezoidal rule is $O(N_x^{-2})$. At each $t_i$, we perform this integration $N_x$ times, meaning that the error should compound $N_x$ times. Taking into account that, we also have to compute the integral for $N_t$ times, meaning that the error is compounded $N_x^{N_t}$ times. The total compound error should be on the order of $O(N_x^{N_t-2})$.
This would be massive and really should never converge, however, I wrote code that computes the path integral naively like this, and it seems that I do get convergence to the correct value $F(t_0, x_0)$. Whats more, if I keep increasing $N_x$, I get better and better convergence, even though according to my result I really shouldn't.
How do I actually determine the actual compound error in this calculation?
