1 Brownian Motion Hitting Probabilities for
General Two-Sided Square-Root Boundaries
Doncho S. Donchev
St. Kliment Ohridsky University
Department of Mathematics and Informatics
5, James Bourchier Str.
1164 Sofia, Bulgaria
(e-mail:[email protected])
√
√
Abstract. Let Bt be a Brownian motion, g(t) = a t + c, f (t) = b t + c, t ≥ 0,
a < b, c > 0, T > 0, and τ be the first hitting time of Bt either in f (t) or in g(t).
We study the hitting probabilities v(t, x) = Pt,x (τ ≤ T, Bτ = f (τ )) for 0 < t < T
and g(t) < x < f (t), where Pt,x is a probability such that Pt,x (Bt = x) = 1. We
give general description of v(t, x) and find explicit series expansion for it in case of
some special boundaries .
Keywords: Brownian motion, hitting time, square-root boundary.
1.1
Introduction
Let Bt be a Brownian motion, f (t) and g(t), f (t) > g(t), be two deterministic
functions and τ = inf(t > 0 : Bt = f (t) or Bt = g(t)) be the first time when
Bt hits either f (t) or g(t). [Skorohod, 1964] has shown that for any fixed
T > 0, the function
u(t, x) = Pt,x (τ > T ), 0 < t < T, g(t) < x < f (t),
(1.1)
is a solution to the mixed problem
∂u 1 ∂ 2 u
+
= 0,
∂t
2 ∂x2
u(t, g(t)) = u(t, f (t)) = 0, u(T, x) = 1.
In (1.1), Pt,x is a measure such that Pt,x (Bt = x) = 1. Similarly, the functions
v(t, x) = Pt,x (Bτ = f (τ ), τ < T ),
v(t, x) = Pt,x (Bτ = g(τ ), τ < T ),
solve the problems
∂v 1 ∂ 2 v
+
= 0,
∂t
2 ∂x2
v(t, f (t)) = 1, v(t, g(t)) = v(T, x) = 0
(1.2)
2
Doncho S. Donchev
and
∂v 1 ∂ 2 v
+
= 0,
∂t
2 ∂x2
v(t, g(t)) = 1, v(t, f (t)) = v(T, x) = 0
(1.3)
respectively.
In this note, we study problems (1.2) and (1.3) and find conditions under
which they admit closed solutions. Evidently, it suffices to consider only the
first of them and, for a sake of brevity, we set v = v. Introducing a new
unknown function v1 (t, x) by the formula
x − g(t)
v(t, x) = v1 t,
,
(1.4)
f (t) − g(t)
we reduce (1.2) to the problem
(f (t) − g (t))x + g (t) ∂v1
1
∂ 2 v1
∂v1
−
·
+
·
= 0,
2
∂t
f (t) − g(t)
∂x
2(f (t) − g(t))
∂x2
v1 (t, 1) = 1, v1 (t, 0) = v1 (T, x) = 0.
(1.5)
The coefficients of the parabolic equation in (1.5) usually depend on both t
and x. However, for some special boundaries they become time-independent.
A trivial example is the case of two parallel lines g(t) = at + b, f (t) =
at + c. An infinite horizon version of this problem has been studied
√ by
t + c,
[Anderson,
1960].
Here,
we
consider
square-root
boundaries
g(t)
=
a
√
f (t) = b t + c, b > a. It is shown that in this case (1.5) also reduces to
a problem with time independent coefficients to which the Laplace transformation method applies. This is done in Section 2, where we state our
main result. In Section 3 we consider some corollaries about transition densities of the first hitting time of Brownian motion and reflecting Brownian
motion in one-sided square-root boundaries. They agree with the results of
[Novikov, 1971] and [Novikov et al., 1999].
1.2
The main result
Consider the matrix function
A(p, a, y) =
D−p (−a) D−p (a)
D−p (−y) D−p (y)
,
where a < y < b, Re(p) > 0 and
2
Dp (y) =
2−p/2 e−y
Γ (−p)
/4
∞
0
is the parabolic cylinder function.
2
e−t
√
− 2ty −p−1
t
dt, Re(p) < 0
1
Brownian Motion Hitting Probabilities
√
√
Theorem 1 Let a t + c < x < b t + c and 0 < t < T . Then
x
1 T +c
v(t, x) := Pt,x (Bτ = f (τ ), τ < T ) = V √
, ln
,
t+c
t+c 2
3
(1.6)
where V (y, t) = L−1 V (y, p) is the inverse Laplace transform of the function
V (y, p) =
1 (y2 −b2 )/4 det A(p, a, y)
e
,
p
det A(p, a, b)
(1.7)
and det A is a determinant of the matrix A.
√
√
Proof Setting in (1.5) g(t) = a t + c, f (t) = b t + c, b > a, we obtain the
equation
1
1
∂ 2 v1
∂v1
a
∂v1
−
+
·
= 0,
(1.8)
x+
·
∂t
2(t + c)
b−a
∂x
2(b − a)2 (t + c) ∂x2
with the same boundary and final conditions as in (1.5). Making use of time
1 T +c
and the substitution
change t → ln
2
t+c
1 T +c
v1 (t, x) = v2
ln
, x
(1.9)
2
t+c
in (1.8), we reduce (1.5) to the following problem with time-independent
coefficients
∂v2
a
1
∂ 2 v2
∂v2
−
− x+
+
·
= 0,
·
∂t
b−a
∂x
(b − a)2 ∂x2
(1.10)
v2 (t, 1) = 1, v2 (t, 0) = v2 (0, x) = 0,
1 T +c
.
0 < x < 1, 0 < t < ln
2
c
Denote by V2 (x, p) the Laplace transform w.r.t. t of v2 (t, x). In view of
(1.10), the function V2 satisfies the problem
1
a
V
−
x
+
V2 − pV2 = 0,
(b − a)2 2
b−a
(1.11)
1
V2 (0) = 0, V2 (1) = .
p
Applying substitutions
a
= x̂, V1 (x̂, p) = V2 (x, p),
b−a
(b − a)x̂ = y, V (y, p) = V1 (x̂, p),
x+
(1.12)
(1.13)
we reduce (1.11) to the problem
V − yV − pV = 0,
(1.14)
V (a) = 0, V (b) = 1/p.
(1.15)
4
Doncho S. Donchev
The general solution of Equation (1.14) is well-known and can be found
in Kamke’s reference book on ODE, [Kamke, 1959, Part 3, Section 2, case
(10) of Equation 2.273]. It can be represented in terms of Witheker’s functions Mk,m (y), Kummer’s confluent hypergeometric functions 1 F1 (k, m, y) or
parabolic cylinder functions Dk (y) as follows
2
1 M 1 − p , 1 y 2 /2 + C
2 M 1 − p ,− 1 y 2 /2
V (y, p) = y −1/2 ey /4 C
4
2 4
4
2
4
2
2
1 y1 F1 1 + p , 3 , y
2 1 F1 p , 1 , y
=C
+C
(1.16)
2
2 2
2 2 2
= ey
2
/4
(C1 (D−p (−y) − D−p (y)) + C2 (D−p (−y) + D−p (y))) .
These representations of the function V (y, p) coincide in view of the formulas
Mk,m (y) = y 1/2+m e−y/2 1 F1 (1/2 + m − k, 2m + 1, y),
(1.17)
2k−1
1 y/2 1
D−2k (− 2y) + D−2k ( 2y) ,(1.18)
= √ Γ k+
e
1 F1 k, , y
2
π
2
2k ey/2
1 3
√
,
y
=
D1−2k (− 2y) − D1−2k ( 2y) . (1.19)
F
Γ
k
−
k,
1 1
2
2
4 2πy
Making use of the third representation in (1.16) and taking into consideration
the boundary conditions (1.15) we determine both constants C1 and C2 :
2
C1 = −
e−b /4 D−p (−a) + D−p (a)
·
,
2p
det A(p, a, b)
C2 = −
e−b /4 D−p (−a) − D−p (a)
·
.
2p
det A(p, a, b)
2
Substituting them in (1.16) we get (1.7). Finally, (1.6) follows from (1.9),
(1.12) and (1.13).
1.3
Corollaries
In this section we consider some special square-root boundaries for which the
function v(t, x) in (1.6) can be calculated explicitly.
1.3.1
The case b = −a > 0
In this case, in view of (1.18) and (1.19), the denominator in (1.7) is equal to
2
2
2
D−p
(b)−D−p
(−b)
π25/2−p e−b /2 b
=−
1 F1
Γ (p/2)Γ ((1 + p)/2)
p 1 b2
, ,
2 2 2
1 F1
1 + p 3 b2
, ,
2
2 2
.
1
Brownian Motion Hitting Probabilities
5
Representing the parabolic cylinder functions in the numerator in (1.7) by
means of Kummer’s functions we get that it equals
2
−
where
2
π23/2−p e−(b +y )/4
(K1 (p, y, b) + K2 (p, y, b)) ,
Γ (p/2)Γ ((1 + p)/2)
p 1 b2
1 + p 3 y2
, ,
,
,
F
,
1 1
2 2 2
2
2 2
p 1 y2
1 + p 3 b2
, ,
,
,
F
.
K2 (p, y, b) = b 1 F1
1 1
2 2 2
2
2 2
K1 (p, y, b) = y1 F1
Thus, we obtain
V (y, p) = (W1 (y, p) + W2 (y, p))/2,
where
1 F1
1+p 3 y 2
2 , 2, 2
(1.20)
y
·
,
(1.21)
3 b2
bp 1 F1 1+p
2 , 2, 2
p 1 y2
F
,
,
1
1
2 2 2
1
W2 (y, p) = ·
.
(1.22)
p 1 F1 p2 , 12 , b22
For any fixed b both functions 1 F1 (1 + p)/2, 3/2, b2/2 and 1 F1 p/2, 1/2, b2/2
have simple zeros on the negative half-line. Applying a standard analytical
technique we get
W1 (y, p) =
Ṽ (y, t) = L−1 V (y, p) = (Ṽ1 (y, t) + Ṽ2 (y, t))/2,
where
2
√ 3 y
(2qn −1)t
∞
erfi y/ 2
2y 1 F1 qn , 2 , 2 e
, (1.23)
√ +
Ṽ1 (y, t) =
∂
3 b2
b n=1 (2qn − 1) ∂q
erfi b/ 2
1 F1 qn , 2 , 2
2
∞ 1 F1 pn , 1 , y
e2pn t
2 2
Ṽ2 (y, t) = 1 +
(1.24)
2,
p ∂ F pn , 12 , b2
n=1 n ∂p 1 1
pn and qn , n = 1, 2, ..., being zeros of 1 F1 p, 1/2, b2/2 and 1 F1 q, 3/2, b2 /2 ,
z 2
respectively, and erfi(z) = √2π 0 ez dz. In (1.23) we make use of the formula
√
√
√
2 z 1 F1 (1/2, 3/2, z) = πerfi( z),
see [Prudnikov et al., 1986, p. 580, formula 11].
Let us observe that V (−y) is a solution to Equation (1.14) provided V (y) is.
It follows that the function
V (y, p) + V (−y, p) = W2 (y, p)
6
Doncho S. Donchev
√
√
is a Laplace transform of the probability of hitting either b t + c or −b t + c
till time T . The last probability
is the same as the probability of hitting the
√
one-sided boundary b t + c by the reflecting Brownian motion. Setting in
(1.24)
t = 12 ln T +c
c , y = 0, we find the probability of hitting the boundary
√
b t + c till time T by a reflecting Brownian motion Btr such that B0r = 0 a.s.
It is
−p
∞
1
n
1 F1 pn , 2 , 0 c
(T + c)pn .
1+
1 b2
∂
p
F
,
,
p
n 2 2
n=1 n ∂p 1 1
Differentiating the√last series w.r.t. T we find the density of the first hitting
time τ1 of Btr in b t + c:
−p
∞
1
n
1 F1 pn , 2 , 0 c
(T + c)pn −1 .
P (τ1 ∈ dT ) = dT
∂
1 b2
F pn , 2 , 2
n=1 ∂p 1 1
1.3.2
(1.25)
The case a → −∞
In this case
lim
a→−∞
D−p (−y)
det A(p, a, y)
=
,
det A(p, a, b)
D−p (−b)
since D−p (a) grows faster than D−p (−a) as a → −∞. Consequently,
2
Ψ ( p2 , 12 , z2 )|z=−y
1 2 2 D−p (−y)
=
,
V (y, p) = e(y −b )/4
2
p
D−p (−b)
pΨ ( p2 , 12 , z2 )|z=−b
(1.26)
where
√
√
p 1 z2
p 1 z2
1 + p 3 z2
π
2π
Ψ
, ,
,
,
z
,
,
F
F
=
−
1 1
1 1
2 2 2
Γ ((1 + p)/2)
2 2 2
Γ (p/2)
2
2 2
is the Trikomi’s confluent hypergeometric function whose relationship with
the parabolic cylinder function is given by the formula
p 1 z2
−p/2 −z 2 /4
, ,
D−p (z) = 2
e
Ψ
.
2 2 2
Inverting the Laplace transform in (1.26) we get
Ṽ (y, t) = 1 +
∞
2
Ψ (νn , 12 , z2 )|z=−y
2
n=1
∂
νn ∂ν
Ψ (νn , 12 , z2 )|z=−b
exp(2νn t),
2
where νn , n = 1, 2, ... are the negative zeros of the function Ψ (ν, 12 , z2 )|z=−b
√
√
for a fixed b. Setting in the last formula t = 12 ln T +c
c , y = −a/ c, −a < b c,
1
Brownian Motion Hitting Probabilities
7
√
we obtain the probability of hitting the one-sided boundary b t + c till time
T by a Brownian motion Bt such that B0 = −a, a.s. It is equal to
2
1+
∞
c−νn Ψ (νn , 12 , z2 )|z=a/√c
2
n=1
∂
νn ∂ν
Ψ (νn , 12 , z2 )|z=−b
(T + c)νn
√
and is the same as the probability of hitting the boundary a + b t + c by
a Brownian motion starting from zero. Differentiating the last series w.r.t.
T we√find the density of the first hitting time τ2 of Bt , B0 = 0 a.s., in
a + b t + c:
2
P (τ2 ∈ dT ) = dT
∞
c−νn Ψ (νn , 12 , z2 )|z=a/√c
n=1
∂
1 z2
∂ν Ψ (νn , 2 , 2 )|z=−b
(T + c)νn −1 .
(1.27)
Remark Formulas (1.25) and (1.27) coincide with formulas (15) and (17)
in [Novikov et al., 1999] where the corresponding results are formulated in
terms of the functions
2
2ν
ey /4
Ψ (−ν/2, 1/2, z 2/2)|z=−y =
D2ν (−y),
Γ (−2ν)
Γ (−2ν)
SH(ν, y) = (HK(ν, y) + HK(ν, −y))/2.
HK(ν, y) =
1.3.3
The case a = 0
Since D−p (0) = 0, formula (1.19) implies that
V (p, y) =
1 (y2 −b2 )/4 D−p (−y) − D−p (y)
e
= W1 (y, p),
p
D−p (−b) − D−p (b)
where W1 (y, p) is given by (1.21). The inverse Laplace transform of W1 (y, p)
1
is the function
√ Ṽ (y, t) defined by (1.23). Setting in (1.23) t =√2 ln(T +
c)/c, y = x/ c, we find the probability of hitting the boundary b t + c till
time T by a Brownian motion Bt , B0 = x, b > x > 0, without reaching the
level x = 0. It is
√
3 x2
1/2−qn
∞
q
c
F
,
,
1
1
n
2 2c
erfi(x/ 2c)
2x
√
+ √
(T + c)qn −1/2 .
∂
3 b2
b
c
erfi(b/ 2)
q
(2q
−
1)
F
,
,
n
1
n
1
∂q
2 2
n=1
Evidently, the first term in the last formula is the probability of reaching the
square-root boundary before the level x = 0.
References
[Anderson, 1960]T. Anderson. A modification of the sequential probability ratio
test to reduce the sample size. Ann. Math. Statist., 31:165–197, 1960.
8
Doncho S. Donchev
[Kamke, 1959]E. Kamke. Differentialgleichungen, Leipzig, 1959.
[Novikov, 1971]A. Novikov. On stopping times for a Wiener process. Theory Prob.
Appl., 16:458–465, 1971.
[Novikov et al., 1999]A. Novikov, –, and et al. Approximations of boundary crossing
probabilities for a Brownian motion. J. Appl.Prob.,36:1019–1030, 1999.
[Prudnikov et al., 1986]A. Prudnikov, –, and et al. Integrals and Series, Volume 3.
Nauka, Moscow, 1986.
[Skorohod, 1964]A. Skorohod. Random Processes with Independent Increments,
Nauka, Moscow, 1964.