Regime-Switching Interest Rate Models With Randomized Regimes · Regime-Switching Interest Rate...

Regime-Switching Interest Rate Models WithRandomized Regimes

James G. Bridgeman, FSA

University of Connecticut

Actuarial Science Seminar Jan. 29, 2008

Bridgeman (University of Connecticut) Random RegimesActuarial Science Seminar Jan. 29, 2008 1

Introduction

Work in progress on cash �ow testing interest rate models

Empirical Work In Valuation Actuary Practice (1990�s):

Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings

Theoretical Work (2006):

Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1

More Recent Results (2007):

Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1

This year? Numerical examples and extensions

Introduction

Work in progress on cash �ow testing interest rate modelsEmpirical Work In Valuation Actuary Practice (1990�s):

Introduction

Unconstrained lognormal models have too much tail

Mean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings

Introduction

Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulder

Randomizing the reversion target �xes itTrial and error calibration2001 Valuation Actuary Symposium Proceedings

Introduction

Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes it

Trial and error calibration2001 Valuation Actuary Symposium Proceedings

Introduction

Unconstrained lognormal models have too much tailMean-reverting ones have too little shoulderRandomizing the reversion target �xes itTrial and error calibration

2001 Valuation Actuary Symposium Proceedings

Introduction

Theoretical Work (2006):Closed form calibration for the mean-reverting lognormal

A surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1

Introduction

Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormal

Couldn�t get closed form calibration with randomized targetsARCH 2007.1

Introduction

Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targets

ARCH 2007.1

Introduction

Theoretical Work (2006):Closed form calibration for the mean-reverting lognormalA surprising drift formula for the mean-reverting lognormalCouldn�t get closed form calibration with randomized targetsARCH 2007.1

Introduction

More Recent Results (2007):Asymptotic closed form calibration with randomized targets

Interesting probability results/techniquesARCH 2008.1

Introduction

More Recent Results (2007):Asymptotic closed form calibration with randomized targetsInteresting probability results/techniques

ARCH 2008.1

Introduction

More Recent Results (2007):Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1

Introduction

More Recent Results (2007):Asymptotic closed form calibration with randomized targetsInteresting probability results/techniquesARCH 2008.1

This year? Numerical examples and extensionsBridgeman (University of Connecticut) Random Regimes

Actuarial Science Seminar Jan. 29, 2008 2/ 36

Example: 55 Years of the 10-year Treasury Rate

10 YEAR TREASURY RATE 19532007 (monthly data)

The Distribution of those Interest Rates

FREQUENCY OF 10 YEAR RATES

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

DATA LOGNORMAL

Lognormal 4th Moment Is Just Too High (6th too)

FREQUENCY OF 10 YEAR RATES

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

DATA LOGNORMAL

55 Years of Changes in the 10 Year Treasury Rate

MONTHLY LOGCHANGE IN 10 YEAR RATE

What is the Distribution of Those Changes?

FREQUENCY OF MONTHLY LOGCHANGE IN 10 YEAR RATES

DATA GAUSSIAN

For Rate Changes, Lognormal 4th Moment Too Low

FREQUENCY OF MONTHLY LOGCHANGE IN 10 YEAR RATE

DATA GAUSSIAN

The Fix: Randomize the Reversion Target

50 YEAR SAMPLE PATH (A DANGEROUS ONE)

0.1PATH TARGET

Lognormal Models

Unconstrained:

d ln (r t )=Dtdt + σpdtNt

d ln(rt ) = Dtdt + σdWt

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i[ln(T0)� ln(rt�dt )]

+ (1� F )dt Dtdt + (1� F )dt σpdtNt

actuarial folklore (circa 1970)d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)

With Randomized Reversion Target

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

1[j,j+1)(t) ln(Tj )� ln(rt�dt )#

+ (1� F )dt Dtdt + (1� F )dt σpdtNt , where 1[j,j+1) (t) is

the indicator for t to be in a random interval�tj, tj+1

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

actuarial folklore (circa 1970)

d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

Lognormal Models

Unconstrained:

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

�Bridgeman (University of Connecticut) Random Regimes

Lognormal Models

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

�d ln(rt ) = � ln (1� F )

∑j=0

1[j ,j+1)(t) ln(Tj )� ln(rt )#dt

+Dtdt + σdWt

Lognormal Models

Mean-reverting:

d ln (r t )=h1� (1� F )dt

actuarial folklore (circa 1970)

d ln(rt ) = f� ln (1� F ) [ln(T0)� ln(rt )] +Dtg dt + σdWtBlack-Karasinski (1991)

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

�d ln(rt ) = � ln (1� F )

∑j=0

+Dtdt + σdWt

Lognormal Models

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

�d ln(rt ) = � ln (1� F )

∑j=0

+Dtdt + σdWt

Lognormal Models

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

�d ln(rt ) = � ln (1� F )

∑j=0

+Dtdt + σdWt

Lognormal Models

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

d ln(rt ) = � ln (1� F )"

∑j=0

+Dtdt + σdWt

Lognormal Models

Mean-reverting:

d ln (r t )=h1� (1� F )dt

i " ∞

∑j=0

�d ln(rt ) = � ln (1� F )

∑j=0

+Dtdt + σdWt

Drift Compensation and Calibration: plain mean-reversion

It would be intuitive to have:

E [rt ] = r(1�F )t0 T

[1�(1�F )t ]0

To �nd out what drift Dt will ensure it, you can integrate d ln(rt ) :

ln(rt ) = ln(r0) (1� F )tdt dt + σ

∑s=1

Nt�(s�1)dt (1� F )sdt

+ ln(T0)h1� (1� F )dt

∑s=1(1� F )(s�1)dt (= notice geom. series

∑s=1

Dt�(s�1)dt (1� F )sdt which simpli�es to:

ln(rt ) = ln(r0) (1� F )t + σpdt

∑s=1

+ ln(T0)�1� (1� F )t

�+ dt

∑s=1

Dt�(s�1)dt (1� F )sdt , which is

Gaussian.

It would be intuitive to have:

E [rt ] = r(1�F )t0 T

[1�(1�F )t ]0

To �nd out what drift Dt will ensure it, you can integrate d ln(rt ) :

ln(rt ) = ln(r0) (1� F )tdt dt + σ

∑s=1

+ ln(T0)h1� (1� F )dt

∑s=1(1� F )(s�1)dt (= notice geom. series

∑s=1

Dt�(s�1)dt (1� F )sdt which simpli�es to:

∑s=1

+ ln(T0)�1� (1� F )t

�+ dt

∑s=1

Dt�(s�1)dt (1� F )sdt , which is

Gaussian.Bridgeman (University of Connecticut) Random Regimes

Since ln(rt ) is Gaussian, E [rt ] = eµ+ 12 σ2where the µ and σ2 are

some mess determined by the constants in the expression for ln(rt ).

If you work that mess out and set it equal to r (1�F )t

0 T[1�(1�F )t ]0 ,

and require that it be true for all t, you can arrive at what the driftcompensation function Dt must be to deliver the intuitive E [rt ] :

Dt = � 12σ2(1�F )dt

1+(1�F )dth1+ (1� F )2t�dt

Dt = � 14σ2h1+ (1� F )2t

iin the continuous case

There is a similar closed form for the variance of rt based onE�r2t�= e2µ+ 1

2 (2σ)2which can help calibrate the model to historical

variance F = 1�n1� σ2obsdt

ln(Vobs+T 2)�ln(T 2)

o 12dt

Practical work with the randomized reversion target model all butrequires you to know similar closed forms for drift compensation andvariance, but now when you integrate no geometric series appears.

0 T[1�(1�F )t ]0 ,

Dt = � 12σ2(1�F )dt

1+(1�F )dth1+ (1� F )2t�dt

Dt = � 14σ2h1+ (1� F )2t

o 12dt

0 T[1�(1�F )t ]0 ,

Dt = � 12σ2(1�F )dt

1+(1�F )dth1+ (1� F )2t�dt

Dt = � 14σ2h1+ (1� F )2t

o 12dt

0 T[1�(1�F )t ]0 ,

Dt = � 12σ2(1�F )dt

1+(1�F )dth1+ (1� F )2t�dt

Dt = � 14σ2h1+ (1� F )2t

o 12dt

0 T[1�(1�F )t ]0 ,

Dt = � 12σ2(1�F )dt

1+(1�F )dth1+ (1� F )2t�dt

Dt = � 14σ2h1+ (1� F )2t

o 12dt

0 T[1�(1�F )t ]0 ,

Dt = � 12σ2(1�F )dt

1+(1�F )dth1+ (1� F )2t�dt

Dt = � 14σ2h1+ (1� F )2t

o 12dt

Drift Compensation & Calibration: random mean-reversion

∑s=1

+h1� (1� F )dt

∑s=1

∑j=01[j,j+1)(sdt) ln(Tj ) (1� F )t�sdt (ugly

∑s=1

Dt�(s�1)dt (1� F )sdt

∑s=1

+h1� (1� F )dt

∑s=1

Dt�(s�1)dt (1� F )sdt , but switch the order of summation:

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

i(=after telescoping

∑s=1

Dt�(s�1)dt (1� F )sdt , so rt 6= lognormal, = log-log-gamma?

∑s=1

+h1� (1� F )dt

∑s=1

Dt�(s�1)dt (1� F )sdt , but switch the order of summation:

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

i(=after telescoping

∑s=1

Dt�(s�1)dt (1� F )sdt , so rt 6= lognormal, = log-log-gamma?Bridgeman (University of Connecticut) Random Regimes

Condition on the Times When Regimes Switch

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Conditioning on the tj random variables and using a lognormal model(reasonable) for the random targets Tj (so each ln(Tj ) is Gaussian)we again have a (messy) Gaussian for the conditional ln(rt ). Canthat help in calculating an unconditioned E [rt ] and variance?

The answer is "Yes" ... up to an approximate expansion.

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Edgeworth Expansion for the Unconditioned Moments

We expect the tails of ln(rt ) to be supressed in favor of the shoulders.That suggests that E [rt ], and higher moments as well, might beapproximated e¢ ciently by an Edgeworth expansion for ln(rt ). Itworks out to be surprisingly simple:

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

�oHere σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

)where (2n)? = (2n� 1) (2n� 3) � � � (1)Conditional Gaussian ensures that odd higher moments vanish.

The problem now is to calculate µ, σ2, and the µ2j

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

Here σ2 and µ4 stand for central moments of ln(rt ) and µ is itsmean.

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

)where (2n)? = (2n� 1) (2n� 3) � � � (1)

Conditional Gaussian ensures that odd higher moments vanish.

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

Eh(rt )

li� e lµ+ 1

2 (lσ)2n1+ l4

�µ4 � 3σ4

� e lµ+ 12 (lσ)

2n1+ l4

�µ4 � 3σ4

� �1� 3

4! (lσ)2�+ l6

�µ6 � 15σ6

e lµ+12 (lσ)

(1+ limN!∞

N∑j=2

l2j(2j)!

hµ2j � (2j)?σ2j

i N�j∑n=0

(�1)n(2n)?(2n)! (lσ)2n

Expected Value Easy

Remember,

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Dt�(s�1)dt (1� F )sdt and condition on the tj

So µ = E[ ln(rt )] is given by

ln(r0) (1� F )t + ln(T0)n

Eh(1� F )(t�t1)+

i� (1� F )t

o+µTE

∑j=1

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]

Expected Value Easy

Remember,

∑s=1

+ ln(T0)h(1� F )(t�t1)+ � (1� F )t

∑j=1ln(Tj )

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Dt�(s�1)dt (1� F )sdt and condition on the tj

ln(r0) (1� F )t + ln(T0)n

Eh(1� F )(t�t1)+

i� (1� F )t

o+µTE

∑j=1

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Dt�(s�1)dt (1� F )sdt where µT = E [ln(Tj )]Bridgeman (University of Connecticut) Random Regimes

Expected Value Easy

ln(r0) (1� F )t + ln(T0)n

Eh(1� F )(t�t1)+

i� (1� F )t

o+µTE

∑j=1

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

Telescoping,= ln(r0) (1� F )t + ln(T0)

nEh(1� F )(t�t1)+

i� (1� F )t

n1�E

h(1� F )(t�t1)+

∑s=1

Eh(1� F )(t�t1)+

iturns out to be a Laplace transform that we can

calculuate (later).

Expected Value Easy

ln(r0) (1� F )t + ln(T0)n

Eh(1� F )(t�t1)+

i� (1� F )t

o+µTE

∑j=1

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

nEh(1� F )(t�t1)+

i� (1� F )t

n1�E

h(1� F )(t�t1)+

∑s=1

Eh(1� F )(t�t1)+

calculuate (later).

Expected Value Easy

ln(r0) (1� F )t + ln(T0)n

Eh(1� F )(t�t1)+

i� (1� F )t

o+µTE

∑j=1

h(1� F )(t�tj+1)+ � (1� F )(t�tj )+

∑s=1

nEh(1� F )(t�t1)+

i� (1� F )t

n1�E

h(1� F )(t�t1)+

∑s=1

Eh(1� F )(t�t1)+

calculuate (later).

Higher Moments Hard

Remembering that the even central moments of std normal are(2n)? = (2n� 1) (2n� 3) � � � (1), the even central moments ofln(rt ) are E

hfln(rt )�E [ln(rt )]g2n

i= (2n)?E

248<:σ2dt

∑s=1(1� F )2sdt + σ2T

∑j=1e2j

9=;n35

= (2n)?E

"(σ2dt (1� F )2dt 1�(1�F )

1�(1�F )2dt+ σ2T

∑j=1e2j

σ2T is the common variance of the ln(Tj ) Gaussians

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

ofor each j

The fgn part can be expanded binomially, but that still leaves termslike...

Higher Moments Hard

i= (2n)?E

248<:σ2dt

∑s=1(1� F )2sdt + σ2T

∑j=1e2j

9=;n35

= (2n)?E

"(σ2dt (1� F )2dt 1�(1�F )

1�(1�F )2dt+ σ2T

∑j=1e2j

)n#σ2T is the common variance of the ln(Tj ) Gaussians

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

ofor each j

Higher Moments Hard

i= (2n)?E

248<:σ2dt

∑s=1(1� F )2sdt + σ2T

∑j=1e2j

9=;n35

= (2n)?E

"(σ2dt (1� F )2dt 1�(1�F )

1�(1�F )2dt+ σ2T

∑j=1e2j

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

ofor each j

Higher Moments Hard

i= (2n)?E

248<:σ2dt

∑s=1(1� F )2sdt + σ2T

∑j=1e2j

9=;n35

= (2n)?E

"(σ2dt (1� F )2dt 1�(1�F )

1�(1�F )2dt+ σ2T

∑j=1e2j

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

ofor each j

Still Need To Evaluate Terms Like

∑j=1e2j

!m#where the

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

ofail to be independent and

are each complicated in their own right.

But they do have a uniform correlation property

Lemma: Ehe2a1j1 � � � e2akjk

i= ρa1,...,akE

he2a1j1

i� � �E

he2akjk

iindependent of fj1, ..., jkg for distinct fj1, ..., jkgρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,akHow does that help?

∑j=1e2j

!m#where the

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

i= ρa1,...,akE

he2a1j1

i� � �E

he2akjk

∑j=1e2j

!m#where the

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

i= ρa1,...,akE

he2a1j1

i� � �E

he2akjk

iindependent of fj1, ..., jkg for distinct fj1, ..., jkg

ρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,akHow does that help?

∑j=1e2j

!m#where the

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

i= ρa1,...,akE

he2a1j1

i� � �E

he2akjk

iindependent of fj1, ..., jkg for distinct fj1, ..., jkgρa1,...,ak can be computed using Laplace transforms and there�s even arecursive relationship ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak

How does that help?

∑j=1e2j

!m#where the

ej =n(1� F )(t�tj+1)+ � (1� F )(t�tj )+

i= ρa1,...,akE

he2a1j1

i� � �E

he2akjk

For Example

24 ∞

∑j=1e2j

!235 = E

∑j=1e4j +

∑j=1e2j

∑i=1e2i

!� e2j

24 ∞

∑j=1e2j

!235 = E

∑j=1e4j

∑j=1e2j

#!2� E

∑j=1e2j E

he2ji#9=; using monotone

convergence to run expectations across ∞ sums

It gets complicated fast

For Example

24 ∞

∑j=1e2j

!235 = E

∑j=1e4j +

∑j=1e2j

∑i=1e2i

!� e2j

24 ∞

∑j=1e2j

!235 = E

∑j=1e4j

∑j=1e2j

#!2� E

∑j=1e2j E

For Example

24 ∞

∑j=1e2j

!235 = E

∑j=1e4j +

∑j=1e2j

∑i=1e2i

!� e2j

24 ∞

∑j=1e2j

!235 = E

∑j=1e4j

∑j=1e2j

#!2� E

∑j=1e2j E

For m=3

24 ∞

∑j=1e2j

!335=E

26666666664

∑j=1e6j + 3

∑j=1e4j

∑i=1e2i

!� e2j

∑j=1e2j

8>>>><>>>>:

∑i=1e2i

∑k=1

!� e2i � e2j

�e2j

∑k=1

!� e2j

#+ e4j

9>>>>=>>>>;

37777777775=E

∑j=1e6j

#+ 3ρ2,1E

∑j=1e4j

∑j=1e2j

��3ρ2,1 � ρ1,1,1

∑j=1e4j E

he2ji#+ρ1,1,1

∑j=1e2j

�3E"

∑j=1e2j

∑j=1e2j E

he2ji#

∑j=1e2j�

Ehe2ji�2#)

Complicated, but each piece is simpler

Now all you need to be able to evaluate are terms like

∑j=1e2nj

n∏k=1

�Ehe2kji�nk#

, where ∑nk=1 knk � m� n

In fact, we will develop a calculation that includes the odd powers

too, E

∑j=1enj

n∏k=1

�Ehekji�nk#

Some notation: to save ink later let ν(x) stand for xnn∏k=1

E�xk�nk so

our expression abbreviates to E

∑j=1

ν (ej )

∑j=1e2nj

n∏k=1

�Ehe2kji�nk#

too, E

∑j=1enj

n∏k=1

�Ehekji�nk#

E�xk�nk so

∑j=1

ν (ej )

∑j=1e2nj

n∏k=1

�Ehe2kji�nk#

too, E

∑j=1enj

n∏k=1

�Ehekji�nk#

E�xk�nk so

∑j=1

ν (ej )

The Set-Up

Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law d

De�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =

P[d�x ]E[d]

De�ne�d1 = d1 ^ (�d0 + t)��d0, so we begin at a random point in the�rst i.i.d. interval

Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ

The Set-Up

Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)

Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =

P[d�x ]E[d]

The Set-Up

Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of d

The density f�d (x) =P[d�x ]

The Set-Up

Let d1,d2, ...,dj , ...be i.i.d inter-arrival intervals with common law dDe�ne�d0 by the relationships 0 ��d0 � d1 and�d0 ' (d1 ��d0)Let�d stand for the common law of�d0 and (d1 ��d0), the equilibriumdistribution of dThe density f�d (x) =

P[d�x ]E[d]

The Set-Up

P[d�x ]E[d]

The Set-Up

P[d�x ]E[d]

Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ dj

Let J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ

The Set-Up

P[d�x ]E[d]

Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")

De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ

The Set-Up

P[d�x ]E[d]

Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < J

Set�dJ = t � tJ�1 and�dJ+1 = tJ � tSo t =�d1 + d2 + ...+ dJ�1 +�dJ

The Set-Up

P[d�x ]E[d]

Set t0 = 0, t1 =�d1, ... , tj =�d1 + d2 + ...+ djLet J=min fj : tj � tg (a "stopping regime")De�ne random indicators f1j<Jgj�1 by 1j<J = 0 for j � J and1j<J = 1 for j < JSet�dJ = t � tJ�1 and�dJ+1 = tJ � t

So t =�d1 + d2 + ...+ dJ�1 +�dJ

The Set-Up

P[d�x ]E[d]

The Result

∑j=1

ν (ej )

Ehν�(1� F )�d^t

�i�P [�d �t] ν

�(1� F )t

�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]

+Ehν�1� (1� F )�d^t

�i�P [�d �t] ν

�1� (1� F )t

Where K = 1�E

��Ehν�(1� F )d

�i�J�2j J > 1

�= 1� (1� G )t E[(1�G )�

�d^t ]�P[�d�t ](1�G )�t

E[(1�G )�d^t ]�P[�d�t ](1�G )t

And G is de�ned by (1� G ) = expn�L�1d (E

hν�(1� F )d

�i)o,

Ld being the Laplace transformMeaningEh(1� G )d

nL�1d (E

hν�(1� F )d

�io= E

hν�(1� F )d

The Result

∑j=1

ν (ej )

�i�P [�d �t] ν

�(1� F )t

�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]

+Ehν�1� (1� F )�d^t

�i�P [�d �t] ν

�1� (1� F )t

�Where K = 1�E

��Ehν�(1� F )d

�i�J�2j J > 1

�= 1� (1� G )t E[(1�G )�

�d^t ]�P[�d�t ](1�G )�t

E[(1�G )�d^t ]�P[�d�t ](1�G )t

hν�(1� F )d

�i)o,

nL�1d (E

hν�(1� F )d

�io= E

hν�(1� F )d

The Result

∑j=1

ν (ej )

�i�P [�d �t] ν

�(1� F )t

�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]

+Ehν�1� (1� F )�d^t

�i�P [�d �t] ν

�1� (1� F )t

�Where K = 1�E

��Ehν�(1� F )d

�i�J�2j J > 1

�= 1� (1� G )t E[(1�G )�

�d^t ]�P[�d�t ](1�G )�t

E[(1�G )�d^t ]�P[�d�t ](1�G )t

hν�(1� F )d

�i)o,

Ld being the Laplace transform

MeaningEh(1� G )d

nL�1d (E

hν�(1� F )d

�io= E

hν�(1� F )d

The Result

∑j=1

ν (ej )

�i�P [�d �t] ν

�(1� F )t

�o E[ν(1�(1�F )d)]1�E[ν((1�F )d)]

+Ehν�1� (1� F )�d^t

�i�P [�d �t] ν

�1� (1� F )t

�Where K = 1�E

��Ehν�(1� F )d

�i�J�2j J > 1

�= 1� (1� G )t E[(1�G )�

�d^t ]�P[�d�t ](1�G )�t

E[(1�G )�d^t ]�P[�d�t ](1�G )t

hν�(1� F )d

�i)o,

nL�1d (E

hν�(1� F )d

�io= E

hν�(1� F )d

�iBridgeman (University of Connecticut) Random Regimes

Asymptotically

limt!∞ E

∑j=1

ν (ej )

E[ν((1�F )�d)]E[ν(1�(1�F )d)]

1�E[ν((1�F )d)]+E

hν�1� (1� F )�d

Everything Is Closed Form

Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ t

E [xv] = Lv [� ln (x)], where Lv is the Laplace transform of vIf, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then

Ld(x) = (1+ βx)�α, L�1d (y) = 1β

�y�

1α � 1

L�d (x) = 1αβx

h1� (1+ βx)�α

i, L�d^t (x) =

1αβx

n1� e�xt

h1� Γ

�α; tβ

�i� (1+ βx)�α Γ

�α; (

1+βx )tβ

�o+e�xt

n1� Γ

�α+ 1; tβ

�� t

h1� Γ

�α; tβ

�io,

P [�d � t] = 1� Γ�

α+ 1; tβ

�� t

h1� Γ

�α; tβ

Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ tE [xv] = Lv [� ln (x)], where Lv is the Laplace transform of v

If, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then

Ld(x) = (1+ βx)�α, L�1d (y) = 1β

�y�

1α � 1

L�d (x) = 1αβx

h1� (1+ βx)�α

i, L�d^t (x) =

1αβx

n1� e�xt

h1� Γ

�α; tβ

�i� (1+ βx)�α Γ

�α; (

1+βx )tβ

�o+e�xt

n1� Γ

�α+ 1; tβ

�� t

h1� Γ

�α; tβ

�io,

P [�d � t] = 1� Γ�

α+ 1; tβ

�� t

h1� Γ

�α; tβ

Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ tE [xv] = Lv [� ln (x)], where Lv is the Laplace transform of vIf, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then

Ld(x) = (1+ βx)�α, L�1d (y) = 1β

�y�

1α � 1

L�d (x) = 1αβx

h1� (1+ βx)�α

i, L�d^t (x) =

1αβx

n1� e�xt

h1� Γ

�α; tβ

�i� (1+ βx)�α Γ

�α; (

1+βx )tβ

�o+e�xt

n1� Γ

�α+ 1; tβ

�� t

h1� Γ

�α; tβ

�io,

P [�d � t] = 1� Γ�

α+ 1; tβ

�� t

h1� Γ

�α; tβ

Everything is now of the form P [�d � t] and E [xv] for v one of therandom variables d, �d and�d^ tE [xv] = Lv [� ln (x)], where Lv is the Laplace transform of vIf, for example, we take the interarrival distribution d forregime-switches to be gamma(α, β), then

Ld(x) = (1+ βx)�α, L�1d (y) = 1β

�y�

1α � 1

L�d (x) = 1αβx

h1� (1+ βx)�α

i, L�d^t (x) =

1αβx

n1� e�xt

h1� Γ

�α; tβ

�i� (1+ βx)�α Γ

�α; (

1+βx )tβ

�o+e�xt

n1� Γ

�α+ 1; tβ

�� t

h1� Γ

�α; tβ

�io,

P [�d � t] = 1� Γ�

α+ 1; tβ

�� t

h1� Γ

�α; tβ

Even Those Uniform Correlation Coe¢ cients

ρa,b =1D

nEh(1� F )2(a�b)�d^t

i�P [�d � t] (1� F )2(a�b)t

where D =n

Eh(1� F )2a�d^t

i�P [�d � t] (1� F )2at

Eh(1� F )�2b�d^t

i�P [�d �t] (1� F )�2bt

oand ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak recursively

ρa,b =1D

i�P [�d � t] (1� F )2(a�b)t

owhere D =

nEh(1� F )2a�d^t

i�P [�d � t] (1� F )2at

Eh(1� F )�2b�d^t

i�P [�d �t] (1� F )�2bt

and ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak recursively

ρa,b =1D

i�P [�d � t] (1� F )2(a�b)t

owhere D =

nEh(1� F )2a�d^t

i�P [�d � t] (1� F )2at

Eh(1� F )�2b�d^t

i�P [�d �t] (1� F )�2bt

oand ρa1,...,ak = ρa1,a2+...+ak ρa2,...,ak recursively

Where Does the Edgeworth Come From?

De�ne the Fourier Transform f̂ (t) =Z ∞

�∞e�itx f (x) dx

Let W have mean 0 and variance 1 and let φ be std normal density

Write cfW (t) = hcfW (t) � 1bφ(t)�i bφ (t) and Taylor expand the bracket

cfW (t) =(

∑n=0

hcfW (t) � 1bφ(t)�i(n)

t=0tn) bφ (t)

So fW (w) =∞

∑n=0

t=0i�nφ(n) (w) and use Leibniz�s

rulehcfW (t) � 1bφ(t)�i(n)

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

cfW (t) =(

∑n=0

t=0tn) bφ (t)

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

cfW (t) =(

∑n=0

t=0tn) bφ (t)

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

cfW (t) =(

∑n=0

t=0tn) bφ (t)

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

cfW (t) =(

∑n=0

t=0tn) bφ (t)

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

cfW (t) =(

∑n=0

t=0tn) bφ (t)

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

0 =hbφ (t) � 1bφ(t)

�i(n)t=0

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

bφ(j) (0) � 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)

�i(n)t=0

∑j=1

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0hcfW (t) � 1bφ(t)

�i(n)t=0

∑j=3

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0

because W is mean 0 variance 1

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

0 =hbφ (t) � 1bφ(t)

�i(n)t=0

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

�i(n)t=0

∑j=1

n!j !(n�j)!

�i(n)t=0

∑j=3

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

0 =hbφ (t) � 1bφ(t)

�i(n)t=0

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

bφ(j) (0) � 1bφ(t)�(n�j)t=0

∑j=1

n!j !(n�j)!

�i(n)t=0

∑j=3

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

0 =hbφ (t) � 1bφ(t)

�i(n)t=0

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

�i(n)t=0

∑j=1

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0

∑j=3

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0

So fW (w) =∞

∑n=0

t=0=�

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

cfW(j) (0) � 1bφ(t)�(n�j)t=0

0 =hbφ (t) � 1bφ(t)

�i(n)t=0

1bφ(t)�(n)t=0

∑j=1

n!j !(n�j)!

�i(n)t=0

∑j=1

n!j !(n�j)!

�i(n)t=0

∑j=3

n!j !(n�j)!

�cfW(j) (0)� bφ(j) (0)�� 1bφ(t)�(n�j)t=0

But derivatives of Fourier transforms evaluated at 0 are just momentsso

∑j=3

n!(n�j)?j !(n�j)! i

�j �E �Wj�� j?

�and

fW (w) = φ (w) +∞

∑n=0

∑j=3

n!(n�j)?j !(n�j)! i

�n�j �E �Wj�� j?

�φ(n) (w)

φ (w) + limN!∞

∑j=3

�E�Wj�� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+j φ(2n+j) (w)

φ(2n+j) (w) =

24n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

2n+j�k w2n+j�2k

35 φ (w)

But derivatives of Fourier transforms evaluated at 0 are just momentssohcfW (t) � 1bφ(t)

�i(n)t=0

∑j=3

n!(n�j)?j !(n�j)! i

�j �E �Wj�� j?

�and

fW (w) = φ (w) +∞

∑n=0

∑j=3

n!(n�j)?j !(n�j)! i

�n�j �E �Wj�� j?

�φ(n) (w)

φ (w) + limN!∞

∑j=3

�E�Wj�� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+j φ(2n+j) (w)

φ(2n+j) (w) =

24n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

2n+j�k w2n+j�2k

35 φ (w)

�i(n)t=0

∑j=3

n!(n�j)?j !(n�j)! i

�j �E �Wj�� j?

�and

fW (w) = φ (w) +∞

∑n=0

∑j=3

n!(n�j)?j !(n�j)! i

�n�j �E �Wj�� j?

�φ(n) (w)

φ (w) + limN!∞

∑j=3

�E�Wj�� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+j φ(2n+j) (w)

φ(2n+j) (w) =

24n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

2n+j�k w2n+j�2k

35 φ (w)

�i(n)t=0

∑j=3

n!(n�j)?j !(n�j)! i

�j �E �Wj�� j?

�and

fW (w) = φ (w) +∞

∑n=0

∑j=3

n!(n�j)?j !(n�j)! i

�n�j �E �Wj�� j?

�φ(n) (w)

φ (w) + limN!∞

∑j=3

�E�Wj�� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+j φ(2n+j) (w)

φ(2n+j) (w) =

24n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

2n+j�k w2n+j�2k

35 φ (w)

�i(n)t=0

∑j=3

n!(n�j)?j !(n�j)! i

�j �E �Wj�� j?

�and

fW (w) = φ (w) +∞

∑n=0

∑j=3

n!(n�j)?j !(n�j)! i

�n�j �E �Wj�� j?

�φ(n) (w)

φ (w) + limN!∞

∑j=3

�E�Wj�� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+j φ(2n+j) (w)

φ(2n+j) (w) =

24n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

2n+j�k w2n+j�2k

35 φ (w)

Finally, if Y = σW+ µ a change of variables gives the Edgeworthexpansion

fY (y) = 1σ φ�y�µ

�+ limN!∞

∑j=3

�µjσj� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

k�y�µ

�2n+j�2k1σ φ�y�µ

�where µj is the j-th central moment of YFor Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)

�iaround a di¤erent point than 0

For something even more �exible, use a di¤erent function than φ; trylogistic, gamma, inverse gamma or inverse logistic

�+ limN!∞

∑j=3

�µjσj� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

k�y�µ

where µj is the j-th central moment of YFor Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)

�+ limN!∞

∑j=3

�µjσj� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

k�y�µ

�where µj is the j-th central moment of Y

For Esscher (aka Saddlepoint) Expansion, Taylor expandhcfW (t) � 1bφ(t)�iaround a di¤erent point than 0

�+ limN!∞

∑j=3

�µjσj� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

k�y�µ

�+ limN!∞

∑j=3

�µjσj� j?

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n+b j2 c∑k=0

(2n+j)!(2k )?(2n+j�2k )!(2k )! (�1)

k�y�µ

How Do You Get The Moments?

E�rlt�= E

hel ln(rt )

�elY�=

∞Z�∞

e ly fY (y) dy

Substitute the Edgeworth expression, complete the square to integratejust as if you were integrating for the lognormal, and expand thebinomials that occur when you change variables and you wind up with

E�rlt�=

e lµ+12 (lσ)

8<:1+ limN!∞

∑j=3

l jj !

�µj � j?σj

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n (lσ)2n �

�n+b j2 c

∑m=0

(2n+j)!(2n+j�2m)! (lσ)

�2mm

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

9=;Remarkably,

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

k = 0 when m > 0

E�rlt�= E

hel ln(rt )

�elY�=

∞Z�∞

e ly fY (y) dy

E�rlt�=

e lµ+12 (lσ)

8<:1+ limN!∞

∑j=3

l jj !

�µj � j?σj

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n (lσ)2n �

�n+b j2 c

∑m=0

(2n+j)!(2n+j�2m)! (lσ)

�2mm

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

9=;Remarkably,

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

k = 0 when m > 0

E�rlt�= E

hel ln(rt )

�elY�=

∞Z�∞

e ly fY (y) dy

E�rlt�=

e lµ+12 (lσ)

8<:1+ limN!∞

∑j=3

l jj !

�µj � j?σj

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n (lσ)2n �

�n+b j2 c

∑m=0

(2n+j)!(2n+j�2m)! (lσ)

�2mm

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

Remarkably,m

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

k = 0 when m > 0

E�rlt�= E

hel ln(rt )

�elY�=

∞Z�∞

e ly fY (y) dy

E�rlt�=

e lµ+12 (lσ)

8<:1+ limN!∞

∑j=3

l jj !

�µj � j?σj

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n (lσ)2n �

�n+b j2 c

∑m=0

(2n+j)!(2n+j�2m)! (lσ)

�2mm

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

9=;Remarkably,

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

k = 0 when m > 0

Why? 0 =hbφ (t) � 1bφ(t)

�i(2m)t=0

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

Finally, E�rlt�=

e lµ+12 (lσ)

8<:1+ limN!∞

∑j=3

l jj !

�µj � j?σj

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n (lσ)2n

Why? 0 =hbφ (t) � 1bφ(t)

�i(2m)t=0

∑k=0

(2k )?(2(m�k ))?(2k )!(2(m�k ))! (�1)

Finally, E�rlt�=

e lµ+12 (lσ)

8<:1+ limN!∞

∑j=3

l jj !

�µj � j?σj

� b N�j2 c∑n=0

(2n)?(2n)! (�1)

n (lσ)2n

What About the Main Results?

Proofs were inspired by techniques in Decoupling: From Dependenceto Independence, by de la Peña and Giné

Essential lemmata are:

As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ tConditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,

��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJgThis is enough independence to get a geometric series inside the main

expectation E

∑j=1

ν (ej )

#and to pull apart the two sides of the

correlation expectation for ρa,b , leaving a common term involving�d1and�dJ which can be evaluated by writing�d1 = t� (�dJ + dJ�1 + ...+ d2)

expectation E

∑j=1

ν (ej )

As joint distributions fJ,�dJg ' fJ,�d1g and�dJ '�d1 '�d^ t

Conditional on J = j 0 > 1 the following are each independentsets:fJ,�d1g,

expectation E

∑j=1

ν (ej )

��d1,d2, ...,dj 0�1, �d2, ...,dj 0�1,�dj 0 and fJ,�dJg

This is enough independence to get a geometric series inside the main

expectation E

∑j=1

ν (ej )

expectation E

∑j=1

ν (ej )

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