MathOverflow Asked by as1 on November 9, 2021
I am interested in the expected exit time of a one-dimensional Brownian particle from a stochastically evolving interval as follows.
Context:
If $L_t$ and $R_t$ denote the distance to the left and right boundaries at time $t$ respectively, then the process $(L,R)$ has generator
begin{multline}
Af(l,r) = frac{1}{2} ( partial^2_{ll} – 2 partial^2_{lr} + partial^2_{rr})f(l,r) + lambda bigg{ int_0^l (f(u,r)-f(l,r))du + int_0^infty (f(l+s,r)-f(l,r))e^{-s} ds bigg} \
+ lambda bigg{ int_0^r (f(l,u)-f(l,r))du + int_0^infty (f(l,r+s)-f(l,r))e^{-s} ds bigg}.
end{multline}
for some $lambda > 0$. This process has the following interpretation. If $(L_0,R_0) = (l,r)$, then
It is straightforward to check that $phi(l,r) = e^{-(l+r)}$ is a stationary density for the process, ie if $L_0$ and $R_0$ are independent exponential random variables with unit mean, then so are $L_t$ and $R_t$ for all $t$. That is,
begin{equation}
int_0^infty int_0^infty Af(l,r) e^{-(l+r)} dl dr = 0.
end{equation}
Aim:
I wish to compute
begin{equation}
mathbf{E}_phi tau = int_0^infty int_0^infty mathbf{E}[tau , |, L_0 = l, R_0 = r] e^{-l} e^{-r} dl dr,
end{equation}
where
begin{equation}
tau = inf{tgeq0: L_t R_t = 0},
end{equation}
that is, the expected time to hit either $L_t = 0$ or $R_t =0$ starting from the stationary distribution.
Thoughts thus far:
In the case that $lambda = 0$, then the problem reduces to the mean first exit time of $(-l,r)$ of a one-dimensional Brownian motion. It is well known that this is given by $lr$.
Although in this setting the boundaries are jumping, since we start from the stationary (exponential distribution), then on average the boundary is a distance 1 to the left and a distance 1 to the right, leading us to guess that $mathbf{E}_phi tau = 1$.
A standard, maybe naive, approach would be to find a solution $g$ the boundary value problem
begin{equation}
begin{cases}
Ag(l,r)=-1, & text{for all } (l,r) in (0,infty)^2, \
g(0,r) = g(l,0) = 0, & l,rin[0,infty);
end{cases}
end{equation}
given which,
begin{equation}
mathbf{E}_phi tau = int_0^infty int_0^infty g(l,r) e^{-(l+r)} dl dr.
end{equation}
However, it is not easy to identify a solution. Taking cues from the classical case, with $lambda = 0$, it is straightforward to compute that with
begin{align}
&h(l,r) = lr,\
&Ah(l,r) = -1 + lambda big{ -frac{1}{2} l^2 r + r – frac{1}{2} l r^2 + l big},
end{align}
so that
begin{equation}
L_t R_t -L_0R_0 + t – lambda int_0^t big{ -frac{1}{2} L_s^2 R_s + R_s – frac{1}{2} L_s R_s^2 + L_s big} ds
end{equation}
is a martingale. Taking $t = tau$ and taking expectations with respect to $mathbf{E}_phi$, a naive application of Dynkin’s formula or the optional stopping theorem would then ‘yield’
begin{equation}
mathbf{E}_phi tau = 1 + lambda mathbf{E}_phi bigg[ int_0^tau big{ -frac{1}{2} L_s^2 R_s + R_s – frac{1}{2} L_s R_s^2 + L_s big} ds bigg].
end{equation}
Specific questions:
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