12
Applied probability meets Bessel,
Hermite, Kummer, Tricomi,
Wiener & Hopf
(and also Ornstein & Uhlenbeck)
Onno Boxma
EURANDOM & Eindhoven University of Technology
Based on collaborations with Hansjoerg Albrecher, Shaul Bar-Lev,
Rim Essifi, Guido Janssen, Richard Kuijstermans, Britt Mathijsen and
David Perry.
June 4, 2015
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1. Introduction
An insurance risk model
capital
u
t
Sample path of capital as a function of time.
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A queueing/inventory model
work
t
inventory
Sample path of work/inventory as a function of time.
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A blood bank model
Sample path of net amount of blood available as a function of time.
Common aspects of the three models:
• two-sided process
• state-dependent rates (here ξb v + αb , ξd v + αd )
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A blood bank model
Sample path of the net amount of blood present if ξb = ξd = 0.
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A blood bank model
Xb(t)
t
Xd(t)
Sample path of the net amount of blood present if αb = αd = 0.
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Overview of the talk
1. Introduction
2. The blood bank model
2.1 Model description
2.2 Exact analysis of total inventory and of total demand
2.3 A special case: ξb = ξd = 0
2.4 Scaling limits
2.5 Numerical results
3. The queueing/inventory model
4. The insurance risk model
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2. The blood bank model
Blood bank operation (Israeli blood bank, director Eilat Shanir):
Donated blood is tested at the Central Blood Bank (CBB).
The CBB delivers to hospitals with Local Blood Banks (LBB).
This is intended, a.o., for planned surgeries.
If a hospital has an emergency surgery, it may cancel the order from CBB:
demand impatience.
Blood components are perishable:
red blood cells can be kept for 35-42 days, platelets for 5 days.
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2.1 Model description
Consider one type of blood.
Blood amounts arrival process: Pois(λb )
Amounts: i.i.d. B1 , B2 , · · · ∼ Fb (·)
Blood is discarded (perishes) at rate ξb x + αb if total amount = x
Blood demands arrival process: Pois(λd )
Amounts: i.i.d. D1 , D2 , · · · ∼ Fd (·)
Demand is discarded at rate ξd x + αd if total demand = x
(demand impatience)
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Consider the following key performance measures:
f (·): density of total demand
g(·): density of total amount of blood in inventory
The steady-state distributions/densities exist if ξb , ξd > 0.
Else we need ‘drift to the origin’ conditions:
λdED < αd + λbEB ,
λbEB < αb + λdED.
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2.2 Exact analysis
Xb(t)
------------------------------------------ v
t
Xd(t)
First we consider the amount of blood in inventory. We equate the rate at
which some positive blood level v is upcrossed and downcrossed, respectively. LCT leads to the following integral equation: for v > 0,
Z v
Z ∞
g(y)F̄b(v − y)dy + λb
λb
Z0
= λd
∞
f (y)F̄b(v + y)dy
0
g(y)F̄d(y − v)dy + (ξbv + αb)g(v).
(1)
v
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Next we consider the amount of demand:
Z v
f (y)F̄d(v − y)dy + λd
λd
Z 0∞
= λb
Z
∞
g(y)F̄d(v + y)dy
0
f (y)F̄b(y − v)dy + (ξdv + αd)f (v).
(2)
v
It should be noted that these two, coupled, equations are symmetric:
swap f and g , and the b and d parameters.
Flavor of convolutions, and flavor of Wiener-Hopf integral equations.
Can be handled with LST, for Fb (·) and fd (·) having a rational LST (Phasetype).
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Special case: D ∼ exp(µd ), B ∼ exp(µb ). Take αd = αb = 0.
The integral equations become:
Z v
Z ∞
g(y)e−µb(v−y)dy + λbe−µbv
λb
Z0 ∞
= λd
f (y)e−µby dy
0
g(y)e−µd(y−v)dy + ξbvg(v), v > 0.
v
Z
v
f (y)e
λd
Z 0∞
= λb
−µd (v−y)
−µd v
Z
dy + λde
∞
g(y)e−µdy dy
0
f (y)e−µb(y−v)dy + ξdvf (v), v > 0.
v
A direct approach is possible, without LST.
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Take the equation for f (·).
Z v
λd
f (y)e
Z 0∞
= λb
−µd (v−y)
−µd v
Z
∞
g(y)e−µdy dy
dy + λde
0
f (y)e−µb(y−v)dy + ξdvf (v), v > 0.
(3)
v
Differentiate (3) w.r.t. v :
λdf (v) − µd[λd
v
Z
= −λbf (v) + λbµb
−µd (v−y)
f (y)e
Z0
∞
Z
dy + λd
∞
g(y)e−µd(v+y)dy]
0
f (y)e−µb(y−v)dy + ξdf (v) + ξdvf 0(v).
(4)
v
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Using (3) once more, now to replace the term between square brackets in
(4), we get:
ξdvf 0(v) = (λd + λb − ξd)f (v)
− µd(λb
− µbλb
Z
Z
∞
f (y)e−µb(y−v)dy + ξdvf (v))
v
∞
f (y)e−µb(y−v)dy,
(5)
v
and once more differentiating w.r.t. v then gives:
ξdvf 00(v) + ξdf 0(v) − (λd + λb − ξd − µdξdv)f 0(v)
= −µdξdf (v) + (µb + µd)λbf (v) − µb(µb + µd)λb
Z
(6)
∞
f (y)e−µb(y−v)dy.
v
The integral that appears in (6) can be eliminated by using (5), yielding a
second order homogeneous differential equation:
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ξdvf 00(v) + (2ξd − λd − λb + µdξdv − µbξdv)f 0(v)
+ (µdξd − µbξd − µdλb + µbλd − µbµdξdv)f (v) = 0.
Additional equation:
Z
∞
f (v)dv = πd.
0
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After two simple transformations f (v) = eβv h(v) and h(x) = w((µb +µd )x)
one arrives at Kummer’s equation (cf. Slater (1960)):
xw00(x) + (b − x)w0(x) − aw(x) = 0.
It has two linearly independent solutions, called Kummer’s hypergeometric
function and Tricomi’s hypergeometric function:
w(x) = c1M (a, b, x) + c2U (a, b, x),
where
M (a, b, x) = 1F1(a, b, x) =
∞
X
(a)n
n=0
U (a, b, x) =
(b)nn!
xn ,
Γ(b − 1) 1−b
Γ(b − 1)
M (a, b, x) +
x M (1 + a − b, 2 − b, x),
Γ(1 + a − b)
Γ(a)
with (a)n := a(a + 1) . . . (a + n − 1).
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c1 = 0 because otherwise f (v) → ∞ for v → ∞.
Use that
1 F1 (a, b, x) ∼
Γ(b) x a−b
e x , x → ∞.
Γ(a)
(7)
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Using
R∞
0
f (x)dx = πd (= P(total demand is positive)) we get:
Proposition
Γ 1+
f (v) = πd
Γ
λb
ξd
λb +λb
ξd
× U 1−
λd
,2
ξd
e−µdv
2 F1
−
1−
λd
, 1, 1
ξd
λb +λd
, (µb
ξd
with corresponding Laplace transform
+
λb
, − µµdb
ξd
+ µd)v ,
(8)
λd
, 1, 1
ξd
λb s−µb
,
ξd s+µd
ξd
ξd
+
µd 2F1 1 −
.
φ(s) = πd
µd + s 2F1 1 − λd , 1, 1 + λb , − µb
(9)
µd
Exchange parameters to get a similar expression for the density and LT of
the amount of blood present.
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Use the continuity condition (following from level crossing at 0):
Z ∞
Z ∞
f (y)e−µby dy = λd
λb
0
g(y)e−µdy dy,
(10)
0
i.e., λb φ(µb ) = λd γ(µd ), and πb + πd = 1, to get the missing constants πb
and πd .
The asymptotic behavior of Tricomi’s hypergeometric function gives
f (v) ∼ Ce−µdv v λd/ξd−1, v → ∞.
(11)
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Remark 1. If λb = 0, the formulas simplify a lot (and the model becomes
irrelevant for the blood bank). It becomes a one-sided shot noise model, as
studied by Keilson & Mermin and Bekker, Borst, B., Kella.
d
and f (v) = πd µd e−µd v .
Remark 2. If λd = ξd , then φ(s) = πd µdµ+s
Is there coincidence in chance ...?
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Remark 3. In the symmetric case λb = λd = λ, µb = µd = µ, ξb = ξd = ξ ,
one has πb = πd and, with η := λ/ξ :
f (v) =
1
µ
Γ(1 + η)
η−
2 K1
√
(2µv)
(µv) ,
−η
2 Γ(2η)2F1 2η, 1, 1 + η, 21 2 π
2
(12)
where Kα (·) is the modified Bessel function of the second kind.
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Corollary.
The mean amount of demand (blood) present, given it is positive, equals


E[Xd |Xd > 0] =
1  λd λb λb
1
 ,
− −
ξd µd µb µb 2F1 1 − λd , 1, 1 + λb , − µb
ξd
ξd
µd

E[Xb |Xb > 0] =

1  λb λd λd
1
 .
−
−
ξb µb µd µd 2F1 1 − λd , 1, 1 + λb , − µb
ξd
ξd
µd
Accordingly, the expected net amount of blood present equals
λb λd
πb π d
λbλd 1
1
EQ =
−
+
+
−
.
µb
µd
ξb
ξd
C̄
ξd
ξb
Notice that, if ξb = ξd = ξ , then
EQ = (
λb λd 1
m
− ) =: .
µb µd ξ
ξ
(13)
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Expected mean amount of blood, demand, and net blood present.
. ξ ∈ [0, 1].
λb = µd = 1.2, λd = µb = 1, so m = 11
30
Expected mean amount of blood, demand, and net blood present.
ξb = ξd = 0.5, λb = µd = 1.2θ, λd = µb = θ, so m = 11
. θ ∈ [0, 1].
30
Also note that we Rhave the probability that a demand is immediately fully
∞
satisfied; it equals 0 g(u)(1 − e−µd u )du = πb − γ(µd ).
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2.3 A special case: ξb = ξd = 0.
Stability condition:
λdED < αd + λbEB,
λbEB < αb + λdED.
Now
αdf 00(v)+(−λd −λb +µdαd −µbαd)f 0(v)+(−µdλb +µbλd −µbµdαd)f (v) = 0.
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Repeat:
αdf 00(v)+(−λd −λb +µdαd −µbαd)f 0(v)+(−µdλb +µbλd −µbµdαd)f (v) = 0.
Hence f (·) is a mixture of two exponentials: f (v) = R+ e−x+ v + R− e−x− v ,
with x+ and x− the positive and negative root of the equation
αdx2 − (µdαd − µbαd − λd − λb)x + (−µdλb + µbλd − µbµdαd) = 0.
Requirement: R− = 0.
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Related model (Boucherie, B.):
M/M/1 queue with work removal/negative customers.
In both models, one can also handle the case of blood/demand requirements with rational LST, using a Wiener-Hopf factorization approach.
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2.4 Scaling limits
Take αb = αd = 0; ξb = ξd = ξ ; general demand/blood distributions.
Consider the net amount of blood Q(t) := Xb (t) − Xd (t).
Start from the integral representation
Q(t) = Q(0) − ξb
Z
t
+
Z
Q (s) ds + ξd
0
t
−
Q (s) ds +
0
Nb (t)
Nd (t)
X
X
i=1
Bi −
Di.
i=1
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Fluid limit.
In the n-th process Qn (t), we have arrival rates nλb , nλd . Let
Xn(t) =
Nb (nt)
Nd (nt)
X
X
Bi −
i=1
and
X̄n(t) :=
i=1
Xn(t)
,
n
Then
Q̄n(t) = Q̄n(0) − ξ
Di,
Q̄n(t) :=
Z
Qn(t)
.
n
t
Q̄n(s) ds + X̄n(t),
0
or, with m := λb EB − λd ED and the centralized Ȳn (t) = X̄n (t) − mt:
Z t
Q̄n(t) = Q̄n(0) − ξ
Q̄n(s) −
0
m
ξ
ds + Ȳn(t).
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Use Thm 4.1 of Pang, Talreja, Whitt (2007).
Let D[0, ∞) be the space of all one-dimensional real functions on [0, ∞),
endowed with the J1 Skorohod topology.
The integral representation
Z t
x(t) = y(t) +
u(x(s))ds,
0
with u(0) = u0 , has a unique solution given that u is a Lipschitz function.
Proposition.
Let E[B], E[D] < ∞ and Q̄n (0) = Qn (0)/n → q(0) ∈ R, as n → ∞.
Then for n → ∞,
Q̄n ⇒ q,
(14)
where ⇒ denotes convergence in distribution and
q(t) =
m
m −ξt
+ q(0) −
e .
ξ
ξ
(15)
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Diffusion limit.
Again use
Z t
m
Q̄n(s) −
Q̄n(t) = Q̄n(0) − ξ
ds + Ȳn(t).
ξ
0
√
Subtract q(t) on both sides, and multiply by n:
√
√
n Q̄n(t) − q(t) = n Q̄n(0) − q(0)
Z t
√
√
− ξ
n Q̄n(s) − q(s) ds + n Ȳn(t).
0
Let Q̂n ≡
√
n Q̄n − q and Ŷn ≡
√
Q̂n(t) = Q̂n(0) − ξ
n Ȳn, then this reduces to
Z
t
Q̂n(s) ds + Ŷn(t).
0
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Use the FCLT for renewal-reward processes to prove:
Lemma.
Let EB, ED, E[B 2 ], E[D2 ] < ∞. Then Ŷn ⇒ σW as n → ∞, where
σ 2 := λbE[B 2] + λdE[D2] and W is a standard Brownian motion.
Use the CMT to prove:
Proposition.
Let EB, ED, E[B 2 ], E[D2 ] < ∞. If Q̂n (0) → Q̂(0), then Q̂n ⇒ Q̂ as
n → ∞, where Q̂ satisfies the integral equation
Z t
Q̂(s) ds + σW (t).
(16)
Q̂(t) = Q̂(0) − ξ
0
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So Q̂ is an Ornstein-Uhlenbeck diffusion process with infinitesimal mean
−ξx and infinitesimal variance σ 2 := λbE[B 2] + λdE[D2].
Intuition: the original blood process has mean-reverting behavior, with
a drift towards that equilibrium that is proportional to the distance – just
like in OU.
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2.5 Numerical results
Fix µb = 5, µd = 1, ξb , ξd ; arrival rates nλb = n, nλd = 1.2n.
Take, successively, n = 1, 5, 25.
Density function of the net amount of blood present (black) and the
Gaussian limiting density function (dashed) for ξb = ξd = 0.5.
The Gaussian density corresponds to the stationary density of the corresponding OU-process.
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Fix µb = 5, µd = 1, ξb , ξd ; arrival rates nλb = n, nλd = 1.2n.
Take, successively, n = 1, 5, 25.
Density function of the net amount of blood present (black) and the
Gaussian limiting density function (dashed) for ξb = 0.6, ξd = 0.1.
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A queueing/inventory model
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A queueing/inventory model
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3. The queueing/inventory model
Model description: M/G/1, arrival rate λ, service time B ∼ B(·);
offered load ρ = λEB < 1.
The server keeps on working even when there is no service requirement left;
he builds up inventory.
At rate ω(x) when the inventory level is x, all the inventory is removed.
work
t
inventory
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Consider
the densities
v+(x) of required work and v−(x) of inventory;
R∞
R∞
0 v+ (x)dx + 0 v− (x)dx = 1.
Level crossing technique: For x > 0,
Z x
Z
v+(x) = λ
P(B > x − y)v+ (y)dy + λ
0
∞
P(B > x + y)v− (y)dy,
0
Z
v−(x) = λ
x
∞
P(B > y − x)v− (y)dy +
Z
∞
ω(y)v−(y)dy.
x
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First we restrict ourselves to B ∼ exp(µ). Then, for x > 0,
Z x
Z ∞
v+(x) = λe−µx
eµy v+(y)dy + λe−µx
e−µy v−(y)dy,
0
µx
0
Z
v−(x) = λe
∞
−µy
e
∞
Z
v−(y)dy +
ω(y)v−(y)dy.
x
x
Let z+ (x) := eµx v+ (x), z− (x) := e−µx v− (x). Then
Z x
Z ∞
z+(x) = λ
z+(y)dy + λ
0
Z
z−(x) = λ
∞
z−(y)dy + e
x
z−(y)dy,
0
−µx
Z
∞
ω(y)eµy z−(y)dy.
x
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Z
z+(x) = λ
x
Z
z+(y)dy + λ
0
∞
z−(y)dy,
0
immediately yields z+ (x) = Cλeλx , hence v+ (x) = Cλe−(µ−λ)x , x > 0.
Differentiate the other equation, viz.,
Z ∞
Z ∞
z−(y)dy + e−µx
z−(x) = λ
x
ω(y)eµy z−(y)dy,
x
w.r.t. x, which gives
z−0 (x)
= −λz−(x) − ω(x)z−(x) − µ[z−(x) − λ
Z
∞
z−(y)dy],
x
and after another differentiation:
z−00 (x) + (λ + µ + ω(x))z−0 (x)
+ (λµ + ω 0(x))z−(x) = 0, x > 0.
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Case 1: ω(x) ≡ ω .
z−(x) = C1es1x + C2es2x, x > 0,
and v− (x) = eµx z− (x). Here µ + s1 < 0 < µ + s2 , so C2 = 0.
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Case 2: ω(x) = ax, a > 0.
z−00 (x) + (λ + µ + ω(x))z−0 (x)
+ (λµ + ω 0(x))z−(x) = 0, x > 0,
now reduces to
z−00 (x) + (λ + µ + ax)z−0 (x) + (λµ + a)z−(x) = 0, x > 0.
This is a Hermite differential equation. Solution:
λ + µ + ax
√
)
2a
λµ 1 (λ + µ + ax)2
+ c4 1F1(− , ,
)],
2a 2
2a
2
z−(x) = e−(λ+µ)x−ax /2[c3H λµa (
where Hn (x) is a Hermite polynomial. They were originally only defined for
n integer.
The asymptotic behavior of the confluent hypergeometric function implies
that c4 = 0.
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Case 3: ω(x) = ebx , b > 0.
Again a Kummer differential equation!
Remark.
It makes sense that v+ (x) corresponds to an M/M/1 queue in all these
cases. For general service times, v+ (x) would correspond to an M/G/1
queue with a different first service time.
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A brief look at that case of general service times: β(s) := E[e−sB ], for
ω(x) ≡ ω .
Let φ+ (s) and φ− (s) be the Laplace transforms of v+ (x) and v− (x).
We get
[s − λ(1 − β(s))]φ+(s)
= [ω − s + λ(1 − β(s))]φ−(−s) − ωφ−(0), Re s = 0.
N (s)
Assume that B has a rational LST, i.e., β(s) = D(s) , Re s ≥ 0;
D(s) is a polynomial of degree n, N (s) is a polynomial of lower degree.
Solution via Wiener-Hopf factorization.
Multiply by D(s):
[sD(s) − λ(D(s) − N (s))]φ+(s)
= [(ω − s)D(s) + λ(N (s) − D(s))]φ−(−s) − ωD(s)φ−(0), Re s = 0.
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[sD(s) − λ(D(s) − N (s))]φ+(s)
(17)
= [(ω − s)D(s) + λ(N (s) − D(s))]φ−(−s) − ωD(s)φ−(0), Re s = 0.
LHS: analytic for Re s > 0, continuous for Re s ≥ 0.
RHS: analytic for Re s < 0, continuous for Re s ≤ 0.
Both sides are O(|s|n+1 ) for |s| → ∞.
Liouville’s theorem now states that both sides equal the same polynomial of
degree n + 1.
The coefficients of that polynomial are found by (i) observing that sD(s) −
λ(D(s) − N (s)) has n zeros in the LHP, and that s = 0 is another zero;
(ii) behavior for s → ∞.
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4. An insurance risk model
Albrecher and Lautscham (2013) determine the bankruptcy probabiliity for
exponential claim sizes.
Extension to claim sizes with a rational LST is possible, via Wiener-Hopf
factorization.
capital
u
t
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