Maximizing the Mean Exit Time of a Brownian Motion from an Interval

.


Introduction
Consider the one-dimensional controlled standard Brownian motion process {X t ,t ≥ 0} defined by the stochastic differential equation dX t b 0 X t k u X t dt dB t , 1.1 where u is the control variable, b 0 > 0, k ∈{0, 1,...} and {B t ,t≥ 0} is a standard Brownian motion.Assume that X 0 x ∈ −d, d and define the first passage time Our aim is to find the control u * that minimizes the expected value of the cost function where q 0 and λ are positive constants.

International Journal of Stochastic Analysis
In the case when k 0, Lefebvre and Whittle 1 were able to find the optimal control u * by making use of a theorem in Whittle 2, page 289 that enables us to express the value function in terms of a mathematical expectation for the uncontrolled Brownian motion {B t ,t ≥ 0} obtained by setting u ≡ 0i n 1.1 .Moreover, Lefebvre 3 has also obtained the value of u * when k 0 if the cost function J in 1.3 is replaced by Although we cannot appeal to the theorem in Whittle 2 in that case, the author was able to express the function F x in terms of a mathematical expectation for an uncontrolled geometric Brownian motion.
In Section 2, we will find u * when k 1.The problem cannot then be reduced to the computation of a mathematical expectation for an uncontrolled diffusion process.Therefore, we will instead find the optimal control by considering the appropriate dynamic programming equation.Moreover, if the instantaneous reward λ given for survival in the interval −d, d is too large, then the value function F x becomes infinite.We will determine the maximal value that λ can take in Section 3.

Optimal Control
The value function F x satisfies the following dynamic programming equation: It follows that the optimal control is given by Substituting this value into 2.1 , we find that we must solve the nonlinear ordinary differential equation The boundary conditions are Making use of a mathematical software program, we find that the solution of 2.3 can be expressed as where J ν and Y ν are Bessel functions and c 1 is a constant that must be chosen so that F d 0. Unfortunately, it seems very difficult to evaluate the integral explicitly.Notice however that actually we do not need to find F x ,b u to n l yF ′ x to determine the optimal value of the control variable u.
We will prove the following proposition.
Proposition 2.1.The control u * x that minimizes the expected value of the cost function Proof.We deduce from 2.6 that

2.8
Moreover, from the formula see Abramowitz and Stegun 4, page 358 which is valid for ν / − 1, −2,..., we find that the function F ′ x may be rewritten as

2.10
Now, because the optimizer is trying to maximize the time spent by X t in the interval −d, d , taking the quadratic control costs into account, we can assert, by symmetry, that u * x should be equal to zero when x 0. One can check that it is indeed the case for any value of the constant c 1 .Furthermore, the function F x must have a minimum that is, a maximum in absolute value at x 0, so that F ′ 0 0 as well.
With the help of the formula see Abramowitz and Stegun 5, page 360 Hence, we deduce that the constant c 1 must be equal to 1, so that

2.13
Formula 2.7 for the optimal control then follows at once from 2.2 .

Maximal Value of λ
Because the optimizer wants X t to remain in the interval −d, d as long as possible and because u X t is multiplied by b 0 X t with b 0 > 0 in 1.1 , we can state that the optimal control u * x should always be negative when x / 0. However, if we plot u * against x for particular values of the constants λ, b 0 , q 0 ,andd, we find that it is sometimes positive.This is due to the fact that the formula in Proposition 2.1 is actually only valid for λ less than a critical value λ crit .Thisλ crit depends on the other parameters.Conversely, if we fix the value of λ, then we can find the largest value that d can take.One way to determine λ crit is to find out for what value of λ the value function becomes infinite.However, because we were not able to obtain an explicit expression for F x without an integral sign , we must proceed differently.
Another way that can be used to obtain the value of λ crit is to determine the smallest value of x positive for which the denominator in 2.13 vanishes. Let Using a mathematical software program, we find that f x 0at approximately x 1.455.Hence, we deduce that we must have δd ≃ 1.455.

3.2
We can now state the following proposition.

3.3
To conclude, one gives the value of d max when λ is fixed.
Corollary 3.2.For fixed values of b 0 , q 0 ,andλ, the maximal value that d can take is 3.4