Level Crossing Probabilities of the Ornstein - Uhlenbeck Process

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Level Crossing Probabilities of the Ornstein – Uhlenbeck Process

Dr. József Dénes

Institute of Informatics and Mathematics, Faculty of Light Industry, Budapes Tech Department of Science, University of West Hungary

denes.jozsef@nik.bmf.hu

Abstract: The Ornstein Uhlenbeck process is a Gaussian process

X

_twith independent increments and autocorrelation

( )

2

s s t t

X e X E

−

+

=

. First the Laplace transform of the probability density ^P

(

^Xt =^x^X₀ = ^p

)

is computed. Using this, the Laplace transform of

X

_tfirst time reaching a given valuex is derived. It is proved that these results agree with the special case derived earlier by Bellman and Harris (Pacific J. Math. 1, 1951).

1 Definitons

The Ornstein Uhlenbeck process is a stationary Gaussian-Markov process Xtsuch that the joint distribution of X_t₁_,X_t₂KX_tmis a gaussian and is dependent only on the differences

t

_j

− t

_i where

i < j

and the autocorelation function is given by

(

_s _s _t

)

^e ^t

2 X 1

X

E ⋅ ₊ = ⁻ (1.1)

2. EX 1 and 0

EX_t = ²_t = (1.2)

Let

X

be a random vector with normal distribution, then the density of its probability distribution is:

X 2X 1 _T ₁

π e 2

1 ⁻ ^Σ⁻ Σ

(2)

where ⎟⎟

⎠

⎜⎜ ⎞

⎝

=⎛ Y

X X and

Σ

is the correlation matrix:

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

2 1

ρ σ

σ ρ

with ρ₁=EX²,ρ₂ =EY²,σ₁=EXYand Σ =ρ₁ρ₂ −σ². Clearly

σ . ρ ρ

ρ σ

σ ρ

2 2 1

1 2 1

−

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

−

= Σ⁻

Hence the joint probability density

( )

2π ρ ρ σ ^exp ^ρ ^x²

(

^ρ ²^ρ^σ^xy^σ ^ρ

)

^y ^.

y 1 Y , x X

P ₂

2 1

1 2 2 2

2 2

1 ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

− +

− −

= −

=

It follows from here that

( ) ( )

1 ρ 2

x

2 2 1

1 2 2 2

2 2 1

πρ 2 e

σ ρ ρ 2

y ρ xy σ 2 x exp ρ σ ρ ρ π 2

1 x

X y Y P

1

− 2

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

− +

− −

= −

=

. ρ

σ ρ 2ρ

ρ x y σ exp ρ

σ ρ πρ 2

1

1 2 2 1

2

1 2

1 2 2

1 ⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −

− −

=

Applying this to what concerns us, the Ornstein-Uhlembeck process, we can determine the probability density P(X_t =x|X₀ =p).

Clearly

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2 σ e 2 , ρ 1 ρ

-t 2

1= = = so ^t

1 t 2 t

2

1 2 2

1 e

ρ σ . e 1 2

1 4 e 2 1 2 2 1 ρ

) σ ρ ρ (

2 ₋ ₋

−

=

−

⎟=

⎟

⎠

⎞

⎜⎜

⎝

⎛ −

− =

Hence:

( )

(

¹ ^e

)

^.

π ) e p X

| x X (

P 2t

e 1

pe x

0 t

t 2

t2

−

− −

= −

=

−

(1.3) We shall denote this with P(t,p,x)or P(p,x)and call it the fundamental

function. The special cases p=0and x=0are important also:

( )

(

¹ ^e

)

^.

π ) e 0 X

| x X (

P 2t

e 1 0

t

t 2 x2

−

− −

= −

=

−

(1.4)

( )

(

¹ ^e

)

^.

π ) e p X

| 0 X (

P 2t

e 1 0

t

t 2

t e 2 p2

−

− −

= −

=

−

(1.5)

By simple substitution it is easy to prove that (1.3) satisfies the forward equation:

t u x u x

u 2 1 t u

2

2 +

∂ + ∂

∂

= ∂

∂

and the backward equation:

p. p u p

u 2 1 t u

2 2

∂

− ∂

∂

= ∂

∂

This also implies that (1.4) satisfies the forward equation and (1.5) satisfies the backward equation.

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2 The Laplace Transforms of the Fundamental Functions

Since both

( )

(

²^t

)

e 1

pe x

e 1 π e ²^t

t 2

−

− −

−

and

( )

(

²^t

)

e 1

e 1 π e ²^t

x2

−

− −

−

satisfy the _t u_xx u xu_x 2

u =1 + +

forward equation their Laplace transforms must satisfy the

xx U xUx

2

sU=U + + second order ordinary differential equation, that is the equation

0 U ) s 1 ( 2 U x 2

U′′+ ′+ − = _(2.1)

To find the solutions of (2.1) let us consider the confluent hypergeometric equation

0 ay 0 y ) x c ( y

x ′′=+ − ′ − = _(2.2)

The two solutions of this are the:

( ) ( )

!K 2 ) 1 c ( c

x 1 a a

! 1 x c 1 a x

; c , a F

2 1

1 +

+ + +

=

and x¹⁻^c₁F₁

(

a+1−c,2−c;x

)

Kummer functions. Let us consider the following transformation of (2.2) u=y(kx²)where k is an arbitrary nonzero constant.

Clearly:

. y x k 4 y k 2 u

y kx 2 u

) kx ( y u

2 2 2

+ ′′

= ′

′′

= ′

′

=

Hence:

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2 2x k 4

x u u y

kx 2 y u

u y

− ′

′′

′′=

= ′

′

=

Substituting these into (2.2) gives:

0 kx au

2 ) u kx c x (

k 4

x u u kx

2 2

=

′ −

− +

⎟⎠

⎜ ⎞

⎝

⎛ ′

′′−

which in turn, after some simplification becomes:

. 0 kau 4 u kx x 2

1 c

u 2 ⎟ ′− =

⎠

⎜ ⎞

⎝

⎛ − −

′′+

Putting 2

c1 gives: u′′−2kxu′−4kau=0. Let us compare this with (2.1)

0 U ) s 1 ( 2 U x 2

U′′+ ′+ − =

).

s 1 ( 2 ka 4

2 k 2

−

=

−

=

−

Hence we get for kand for ak=−1and . 2

s a 1−

= Therefore the solutions of

(2.1) are ⎟

⎠

⎜ ⎞

⎝

⎛ − −

= ²

1 ; x

2 ,1 2

s F 1

F and ; x .

2 ,3 2 1 s xF

F₂ ²⎟

⎠

⎜ ⎞

⎝

⎛ − −−

=

Now we are in the position to determine the Laplace transform of

( )

(

¹ ^e

)

^.

π e

t 2 e 1

x t 2 2

−

− −

−

Clearly it must be of the form AF₁+xBF₂where Aand Bsome constans. To this end Laplace transform will be evaluated for some special cases. The Laplace

transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

is clearly:

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( )

∫

^∞ − ⁻

− −

−

0

st t 2 e 1

x

. dt e e

1 π e ²^t

2

Writing tinstead of e⁻^t transforms it into a Mellin type integral:

∫ ( )

⁻ ⁻

−

1 0

1 s 2 t 1

x

. dt t t

1 π

e ²

2

Substituing tinstead of tyields

∫

⁻ ⁻

−

1 0

2 1 t s 1

x

. dt t t 1 e π 2

1

2

For x=0this becomes the beta funciton type integral:

. 2

s π 1 2

2 s 2 1 2

,s 2 B 1 π 2 dt 1 t 1 t π 2

1 1 0

2 1 s

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

⎟⎠

⎜ ⎞

⎝ Γ⎛

⎟⎠

⎜ ⎞

⎝ Γ⎛

⎟=

⎠

⎜ ⎞

⎝

= ⎛

∫

−

Hence

. 2

1 2 s

2 s A

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

⎟⎠

⎜ ⎞

⎝ Γ⎛

=

Clearly Ais the Laplace transform of

( )

(

¹ ^e

)

^.

π e

t 2 e 1

x t 2 2

−

− −

−

To determine the value of Blet us consider the xderivative of the Laplace transform, which is:

∫

⁻ ⁻

−

− ₀¹ ² ¹

s

2 3 t 1

x

. dt t ) t 1 ( π

xe

2

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In the present case we cannot take the x→0limit by simply substituing 0

x→ for xbecause

2 3 t x

t π xe

2

−

does not converge uniformly to 0 as x→0.In fact it is a “delta function type function”, its integral being

∫

⁼

− 1 0

2 3 t x

. 1 dt t π xe

2

For it is know from theory of the heat equation that, for an arbitrary continous function f(t)

( )

^t^.

f dr ) r t ( f r e π lim x dr ) r ( f ) r t (

e π

lim x ^t

0

2 3 r x

0 x t

0

2 3 r t

x

0 x

2 2

∫

⁼ ⁻ ⁼

−

→

− −

→

Hence in the present case:

, 1 t

dt t ) t 1 ( π

lim ¹ xe _t ₁

0

2 1 1 s 2 s

2 3 t 1

x

0 x

2

−

=

−

− ⁻ ⁻ ⁻ ₌

−

→

∫

thus B=−1.Therefore the The Laplace transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

is

. x 2; ,3 2 1 s xF x

2; ,1 2

s F 1 2

1 2 s

2 s xF

AF₁ ₂ ² ²⎟

⎠

⎜ ⎞

⎝

⎛ − −

⎟−

⎠

⎜ ⎞

⎝

⎛ − −

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

⎟⎠

⎜ ⎞

⎝ Γ⎛

=

−

(8)

Now we compute the The Laplace transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

. It has been shown

that it statisfies the backward equation

2 pu u

u_t =− _p + ^pp . Therefore its Laplace transform is the solution of the second order linear differential equation

2 pU U

sU=− _p + ^pp that is of the equation

0 sU 2 U p 2

U′′+ ′+ =

Now the solution of u′′−2kxu′−4kau=0 are ⎟

⎠

⎜ ⎞

⎝

⎛ ;kx2

2 ,1 a

F and

⎟⎠

⎜ ⎞

⎝⎛ + ;kx² 2 ,3 2 a 1

xF .

Comparing the two equations we get for k s 2 ka 4

2 k 2

=

that is k=1and . 2

a= s Thus the Laplace transform must be the linear

combination of ⎟

⎠

⎜ ⎞

⎝

= ⎛ ²

1 ;p

2 ,1 2 F s

G and ⎟

⎠

⎜ ⎞

⎝

= ⎛ + ²

2 ;p

2 ,3 2

s pF 1

pG . To find the

conficciens of G₁and pG₂let us inspect the Laplace transform itself.

( )

∫

^∞ − ⁻

− −

−

0

st t 2 e 1

x

. dt e e

1 π e ²^t

2

Writing tinstead of e⁻^t it transforms again into the Mellin type integral:

( ) ( )

∫

⁻ ⁻

−

1 0

1 s 2 t 1

t p

. dt t t

1 π

e ²

2 2

Substituing tinstead of tyields

(9)

∫

⁻ ⁻

−

1 0

2 1 t s 1

t p

. dt t t 1 e π 2

1

2 2

Again putting p=0this becomes:

∫

0¹ ₋ ⁼ 2 s

. A tdt 1

t π 2

1

The coefficient of pG₂can be evaluated the same way as was done for

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

and it is found to be again –1. Thus the Laplace transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

is

. p 2; ,3 2

s 1 1 pF p

2; ,1 2 AF s pG

AG₁ ₂ ² ²⎟

⎠

⎜ ⎞

⎝

⎛ +

−

⎟−

⎠

⎜ ⎞

⎝

= ⎛

−

The above result can be arrived at directly from the Laplace transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

.

To this end let us inspect

( )

∫

^∞ − ⁻

− −

−

0

st t 2 e 1

e p

dt e e

1 π e ²^t

2 t 2

using

. e p

1 p e

1 e

p ₂

t 2 2 t

2 t 2 2

− −

− ⁻ = ⁻

−

(10)

This becomes

( )

∫

^∞ − ⁻

− −

−

0

st t 2 e 1

p

p e dt

e 1 π e e

t 2 2

2 and the integra here is of the same

form as of the Laplace transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

except we have pinstead of x.

Therfore the Laplace transform of

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

is

⎟⎟

⎟

⎠

⎞

⎜⎜

⎜

⎝

⎛

⎟⎠

⎜ ⎞

⎝

⎛ − −

⎟−

⎠

⎜ ⎞

⎝

⎛ − −

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

⎟⎠

⎜ ⎞

⎝ Γ⎛

2 2

p ; p

2 ,3 2 1 s pF p

2; ,1 2

s F 1 2

1 2 s

2 s e ²

Applying Kummer’s formula F(a,c;x)=e^xF(c−a,c;x) we get for the Laplace

transform of

( )

(

¹ ^e ²^t

)

e p

e 1 π e ²^t

t 2 2

−

− −

−

. p 2; ,3 2

s 1 1 pF p 2; ,1 2 F s 2

1 2 s

2 s

2

2 ⎟

⎠

⎜ ⎞

⎝

⎛ − +

⎟−

⎠

⎜ ⎞

⎝

⎛

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

⎟⎠

⎜ ⎞

⎝ Γ⎛

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3 The Laplace Transforms of ( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

We have seen that the

( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

fundamental function satisfies both the forward and backward equations, therefore its Laplace transform must satisfy both of the ordinary differential equations:

0 U ) s 1 ( 2 U p 2

U′′+ ′+ − =

(3.1) .

0 sU 2 U p 2

U′′− ′− = (3.2)

Because of (3.1) must be of the form: HF₁+KxF₂,where Hand Kmust be some linear combinations of G₁and pG₂since it satisfies (3.2) as well. Let us

observe that

( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

is analytic in xfor all values pand texcept when 0

t= and x=p, in the latter case it is undefined. Therefore its Laplace transform

is analytic in the x≤p domain as well. Putting x=0in

( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

gives

( )

(

¹ ^e ²^t

)

e p

e 1 π e ²^t

2 t 2

−

− −

−

and we have seen that its Laplace transform is AG₁−pG₂, so

2

1 pG

AG

H= − (when x≤p). The determination of K is more involved.

Differentiating the fundamental function by xgives:

( )

(

¹ ^e

)

^.

π e pe e 2 e

1 π

e pe 2

2 3 t 2

e 1

e p p t

2 3 t 2

e 1

e p

t ²^t

2 t 2 t 2

2 2 t 2

−

− −

−

− −

−

=

−

(3.3)

Clearly the coefficient K is the Laplace transform of (3.3). To evaluate it let us compute the following convolution integral:

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( )

(

₁ _e

)

^π

(

¹¹^e

)

^.

π e pe

e 2 ₂_t

2 t 3 2

e 1

p p t

t 2 2 2

− −

−

∗ −

−

(3.4)

It has been shown that the Laplace transform of the second factor in (3.4) is A, so the Laplace transform of (3.3) is the Laplace transform of (3.4) divided into A. Next we evaluate (3.4):

( )

( ) ( )

∫

₋ ⁼

−

− −

− ⁻

t

0 2(t r)

2 r 3 2

e 1

p

p r dr

e 1 π

1 e

1 π

e pe e 2

r 2 2 2

putting rfor e⁻^ryields:

( )

( ) ( )

∫

₋ ⁼

−

− − 1

T 2 2

2 2 3

r 1

p

p dr

T r π

1 r

1 π pre e 2

2 2 2

where T=e⁻^r. Substituting rfor rgives:

( )

( ) ( )

( )

. ) 2 1 ( π e t

π 1 t π e pe

rdr π 1 r T 1 π e pe

T dr r π

1 r

1 π e pe

t 2

) 2 1 (

t 2 e p

T 1 3 t

t p p

T 1 0

2 2 3

r T 1

p p

1

T 2 2

2 3 r 1

p p

t 2 2

2 2

2

2 2

2

2 2

−

− −

−

=

−

− − − −

− −

= −

∗

⋅

=

−

=

− −

−

∫

Thus we have for the coefficient

A pG K AG¹− ²

= . Hence the Laplace transform

of

( )

(

¹ ^e

)

) pe x (

e ²^t

2 t

−

− − ₋⁻

is for x≤p:

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( )

A .

) pG AG ( xF AF A

pG xF AG

F ) pG AG

( ₁ ₂ ₁ ₂ ¹ ² ¹+ ² ¹− ²

− = +

−

Next let us consider the case p≤x. If the same computation is repeated but instead of x=0we look at p=0, that is we compute the coefficients of G₁and

pG2. Putting p=0in

( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

gives

( )

(

¹ ^e ²^t

)

x

e 1 π e ²^t

2

−

− −

−

. Its Laplace transform is AF₁−xF₂, carrying through similar computation as was done for the coefficient of xF₂ we get for the coefficient for

A xF pG₂ AF¹− ²

. Thus the

Laplace transform of

( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

when p≤x is:

( )

_.

A

) pG AG ( xF

AF₁+ ₂ ₁− ₂

Hence the Laplace transform of

( )

(

¹ ^e ²^t

)

) pe x (

e 1 π e ²^t

2 t

−

− −

−

is:

( )

⎪⎩

⎪⎨

⎧

− ≤ +

=

⎟⎟

⎟

⎠

⎞

⎜⎜

⎜

⎝

⎛

− ⁻

−

− − ₋⁻

. p x A if

) pG AG ( xF AF

x p A if

) pG AG ( xF AF e

1 π

? e

2 1 2 1

2 1 2 1 t

2 e 1

) pe x (

t 2

2 t

(3.5)

4 Level Crossing Probabilites

Let the random variable FCor FC(x)(first crossing) be the smallest possible value of tsuch that X_t =x given X₀=p. Let φ(t,p,x)be the distribution of

FC, clearly: φ(t,p,x)∗P(t,x,x)=P(t,p,x).

That is:

( )

(

¹ ^e

)

^e^π

(

¹

⁽

^e

⁾ )

^.

π ) e x , p , t (

φ ₂_t

e 1

) pe x (

t 2 e 1

) pe x (

t 2

2 t t

2 2 t

−

− −

−

− −

= −

∗ −

−

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Now the probability that X_tstays below xis: ⎟⎟⎠⁼ ⁻

∫

⎜⎜ ⎞

⎝

⎛

≤

t r 0

dr ) r ( φ 1 X sup

P .

Let us denote the Laplace transform of φby Ψ, then Ψfor 0≤p≤x using (3.5) can be expressed as

( ) ( )

( ) ( ( ) ( ) )

( ) ( )

(

^AG^AG ^x^p ^xG^pG ^x^p

) (

^AF^AF

( )

^x^x ^xF^xF

( )

^x^x

)

ψ

2 1

+ +

+

= + (4.1)

2 1

xG AG

pG AG

+

= + (4.2)

For the special case when p=0, that is when X_treaches level xsubject to the initial condition X₀ =0is

2

1 xG

AG ψ A

= + . (4.3)

For this case Bellman and Harris

[ ]

¹ found the following expression:

. e

2 s 2 1 ψ

0

dy y y² ²^xy ^s ¹

∫

^{∞ −} ⁺ ⁻

⎟⎠

⎜ ⎞

⎝ Γ⎛

= (4.4)

For the case p≥x≥0:

A

) xG AG )(

xF AF (

A

) pG AG )(

xF AF ( ) x , p ( ψ

2 1 2 1

− +

= (4.5)

2 1

xG AG

pG AG

−

= − (4.6)

For the special case p>0, x=0we have:

( )

A

pG 0 AG

, p

ψ ¹− ²

= . (4.7)

Using (4.7) it is not difficult to show that (4.2) holds for p≤xand holds for x

p≥ as well. Formula (4.7) easily invertable, for

(15)

⎟⎠

⎜ ⎞

⎝⎛ + Γ

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

=

⎟⎠

⎜ ⎞

⎝ Γ⎛

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

=

⎟⎠

⎜ ⎞

⎝ Γ⎛

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

=

2 1 s 2 s

2 1 2 s

2 s 2 s

2 1 s s

2 s 2

1 2 s A

1 .

Clearly

⎟⎠

⎜ ⎞

⎝⎛ + Γ

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

2 1 s 2

1 s

is the Laplace transform of

(

²^t

)

t

e 1 π 2 e

−

⋅ − . Hence (4.7) is the Laplace transform of:

( )

( ) ( ) (

¹ ^e

)

^.

e π pe dz 2 π e

1 dt 2 d e

1 π

e e

1 π e dt 2 d

2 3 t 2 e 1

e p t

e 1

pe z t

2 t t

2 e 1

e p

t 2

t 2 2

t 2 t t 2

2 t 2 2

−

− −

∞ −

−

− −

−

=

⋅

=

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

∗ −

⋅ −

−

∫

5 The Equivalence of Bellman-Haris’ and our Result

To show that formulas (4.3) and (4.4) are the same, we have to evaluate the

integral ( )

( )

.

dy e d

! n 1 x e

e e

e _n

y n 0

n n n x

y x x 0

xy 2 y

2 2 2 2

2 ∞ −

=

−

∞ − + −

∑

∫

⁼ ^⋅ ⁼ ⁻

Substituting this into the integral we get:

( )

y dy.

dy e d

! n 1 x e

dy y e e

dy y

e ^s ¹

0

n 0 n

y n n n

x 1 s 0

xy 2 y

2 2 y2

x 2

2 ∞ −

=

∞ −

∞ − + −

∫ ∑ ∫

∫

⁼ ^⋅ ⁻ ⁼ ⁻

(5.1) Let us observe that the integrals on the right hand side are the Mellin transfroms of

the functions _n

y n

dy e d ⁻ ²

. First we compute the Mellin transform of e⁻^y²which is:

2 . s 2 dy 1 y 2 e

dy 1 y

e ² ¹

s 0

y 1

s 0

y² ⎟

⎠

⎜ ⎞

⎝ Γ⎛

=

= ^{∞ −} ⁻

∞ − −

∫

(16)

Let us denote the Mellin transform of a function f by ?or F. It is not difficult to see that:

( ) ( ) ( )

( ) ( )( ) ( )

( )

^f

^{( ) (}

¹ ^s ¹

⁾⁽

^s ²

^{) (}

^s ⁿ

^{) (}

^F^s ⁿ

⁾

^.

2 s F 2 s 1 s f

1 s F 1 s f

n = − n − − − −

Μ

−

′′ = Μ

−

′ = Μ

L K

K

Hence the Mellin transforms of e⁻^y², , dy de⁻^y²

dy , e d

2 2 −^y²

3 K

3

dy e d

y2

−

are

( ) ( ) ( )( ) ( ) ( )( )( ) ( )

( )( )( )( ) ( ) ( )( )( )( )( ) ( )

_, _.

2 5 S 2

1 s 2 s 3 s 4 s 5 ,s 2

4 S 2

1 s 2 s 3 s 4 s

2 , 3 S 2

1 s 2 s 3 , s 2

2 S 2

1 s 2 ,s 2

1 s 2

1 , s 2 s 2 1

− L

− Γ

−

− Γ −

−

Γ −

−

− Γ −

−

− Γ −

− −

⎟⎠

⎜ ⎞

⎝ Γ⎛

Substituing these into (5.1) gives:

( ) ( )( ) ( )

( )( )( ) ( ) ( )( )( )( ) ( )

( )( )( )( )( ) ( )

( )( ) ( )

( )( )( )( ) ( )

⎭⎬ + ⎫ Γ −

−

− + −

⎪⎩

⎪⎨

⎧ ⎟+ − − Γ −

⎠

⎜ ⎞

⎝ Γ⎛

=

⎭⎬ + ⎫ Γ −

−

− + −

Γ −

−

− + −

Γ −

−

− + −

⎪⎩

⎪⎨

⎧ ⎟+ − Γ − + − − Γ −

⎠

⎜ ⎞

⎝ Γ⎛

L L

2 4 s 2

1 s 2 s 3 s 4 s

! 4 x

2 2 s 2

1 s 2 s

! 2 x 2 s 2 e 1

2 5 s 2

1 s 2 s 3 s 4 s 5 s

! 5 x

2 4 s 2

1 s 2 s 3 s 4 s

! 4 x 2

3 s 2

1 s 2 s 3 s

! 3 x

2 2 s 2

1 s 2 s

! 2 x 2

1 s 2

1 s

! 1

x 2 s 2 e 1

4 x 2

5

4 3

x 2

2 2

( ) ( )( )( ) ( )

( )( )( )( )( ) ( )

( ) ( )( )

⎟⎞+ _⎬^⎫

⎜⎛ Γ

−

⎪⎩ +

⎪⎨

⎧ ⎟⎞

⎜⎛ Γ Γ

⎟+

⎜ ⎞ Γ⎛

=

⎭⎬ + ⎫ Γ −

−

− + −

Γ −

−

− + −

⎩⎨

⎧ −

− Γ +

L L

2 1 s s 3 - s

! 2 4 x 2 1 s - s

! 2 2 x 2 s 2 e 1

2 5 s 2

1 s 2 s 3 s 4 s 5 s

! 5 x

2 3 s 2

1 s 2 s 3 s

! 3 x 2

1 s 2

1 s

! 1 e x

4 1 2 0

x 5 x 3

2 2

(17)

( )

( )( )( )

⎪⎪

⎭

⎪⎪⎬

⎫

⎪⎪

⎩

⎪⎪⎨

⎧

+

−

− +

−

⎟ −

⎠

⎜ ⎞

⎝ Γ⎛ + +

⎪⎪

⎭

⎪⎪⎬

⎫

⎪⎪

⎩

⎪⎪⎨

⎧

+

−

− +

−

⎟ −

⎠

⎜ ⎞

⎝ Γ⎛

=

⎭⎬ + ⎫

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

−

− +

⎪⎩

⎪⎨

⎧ ⎟

⎠

⎜ ⎞

⎝ Γ⎛ +

−

⎟+

⎠

⎜ ⎞

⎝ Γ⎛ + +

L L L

2 52 2 s

2 32 1 s

! 2 x 2 32 1 s

! 1 1 x 2 x

1 e s

2 3 2 s 3

2 1 2 s 1

! 2 x 2 12 s 1

! 1 1 x 2 s 2 e 1

2 1 1 s

s 2 s 4 s

! 2 5 x

2 1 2 s

s

! 2 3 x 2

1 2 s

! 1 e x

4 x 2

5 2

3 1 0

x

2 2 2

. x 2; ,3 2

s xF 1 2

1 x s

2; ,1 2 F s 2 s 2 1

x 2; ,3 2 1 s 2 xF

1 e s

x 2,

;1 2

s F 1 2 s 2 e 1

2 2

2 x

2

x² ²

⎟⎠

⎜ ⎞

⎝

⎟ ⎛ +

⎠

⎜ ⎞

⎝ Γ⎛ +

⎟+

⎠

⎜ ⎞

⎝

⎛ −

⎟⎠

⎜ ⎞

⎝ Γ⎛

=

⎟⎠

⎜ ⎞

⎝

⎛ − −

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

⎟+

⎠

⎜ ⎞

⎝

⎛ − −

⎟⎠

⎜ ⎞

⎝ Γ⎛

=

Hence Bellman and Harrises formula becomes:

. x 2; ,3 2

s xF 1 2

1 x s

2; ,1 2 F s 2 s 2 1

2 s 2 1 dy

y e

2 s 2 1

2 0 2

1 s xy 2 y²

⎟⎠

⎜ ⎞

⎝

⎟ ⎛ +

⎠

⎜ ⎞

⎝ Γ⎛ +

⎟+

⎠

⎜ ⎞

⎝

⎟ ⎛

⎠

⎜ ⎞

⎝ Γ⎛

⎟⎠

⎜ ⎞

⎝ Γ⎛

=

⎟⎠

⎜ ⎞

⎝ Γ⎛

∫

^∞ ⁻ ⁺ ⁻

Diving both the numerator and the denominator of the right hand side into

⎟⎠

⎜ ⎞

⎝ Γ⎛ +

2 1 s gives

2

1 xG

AG A

+ and this completes the proof.

Reference

[1] Bellman, R., Harris, T.: Recurence times for the Ehrenfest model. Pacific J.

Math. 1. 179-193 (1951)