Trivariate Density Revisited - Columbia University · Trivariate Density Revisited Steven E. Shreve...

Trivariate Density Revisited

Steven E. ShreveCarnegie Mellon University

–Conference in Honor of Ioannis Karatzas

Columbia UniversityJune 4–8, 2012

June 1, 2012

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Outline

Rank-based diffusion

Bang-bang control

Trivariate density

Personal reflections

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Rank-based diffusion

E. R. Fernholz, T. Ichiba, I. Karatzas & V. Prokaj, Planardiffusions with rank-based characteristics and perturbed Tanakaequation, Probability Theory and Related Fields, to appear.

dX1 = −h dt + ρ dB1

dX2 = g dt + σ dB2

dX1 = g dt + σ dB1

dX2 = −h dt + ρ dB2

ρ2 + σ2 = 1g + h > 0

B1, B2 independent Br. motions

3 / 94

Remarks

Karatzas, et. al. examine:

I Weak and strong existence and uniqueness in law;

I Properties of X1 ∨ X2 and X1 ∧ X2;

I The reversed dynamics of (X1,X2);

I Transition probabilities (X1,X2).

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Remarks

Karatzas, et. al. examine:

I Weak and strong existence and uniqueness in law;

I Properties of X1 ∨ X2 and X1 ∧ X2;

I The reversed dynamics of (X1,X2);

I Transition probabilities (X1,X2).

5 / 94

The difference processDefine

Y (t) = X1(t)− X2(t).

dY = lI{Y≤0}(g + h) dt − lI{Y>0}(g + h) dt

+lI{Y≤0}σ dB1 − lI{Y≤0}ρ dB2 + lI{Y>0}ρ dB1 − lI{Y>0}σ dB2

= −λ sgn(Y (t)

)dt + dW ,

λ = g + h > 0,

dW = σ(lI{Y≤0} dB1 − lI{Y>0} dB2

)︸︷︷︸dW1

+ρ(lI{Y>0} dB1 − lI{Y≤0} dB2

)︸︷︷︸dW2

6 / 94

Y (t) = X1(t)− X2(t).

= −λ sgn(Y (t)

)dt + dW ,

λ = g + h > 0,

)︸︷︷︸dW1

+ρ(lI{Y>0} dB1 − lI{Y≤0} dB2

)︸︷︷︸dW2

7 / 94

Y (t) = X1(t)− X2(t).

= −λ sgn(Y (t)

)dt + dW ,

λ = g + h > 0,

)︸︷︷︸dW1

+ρ(lI{Y>0} dB1 − lI{Y≤0} dB2

)︸︷︷︸dW2

8 / 94

Y (t) = X1(t)− X2(t).

= −λ sgn(Y (t)

)dt + dW ,

λ = g + h > 0,

)︸︷︷︸dW1

+ρ(lI{Y>0} dB1 − lI{Y≤0} dB2

)︸︷︷︸dW2

9 / 94

Y (t) = X1(t)− X2(t).

= −λ sgn(Y (t)

)dt + dW ,

λ = g + h > 0,

)︸︷︷︸dW1

+ρ(lI{Y>0} dB1 − lI{Y≤0} dB2

)︸︷︷︸dW2

10 / 94

The sum processDefine

Z (t) = X1(t) + X2(t).

dZ = (g − h) dt + lI{Y≤0}σ dB1 + lI{Y≤0}ρ dB2

+lI{Y>0}ρ dB1 + lI{Y>0}σ dB2

= ν dt + dV ,

ν = g − h,

dV = σ(lI{Y≤0} dB1 + lI{Y>0} dB2

)︸︷︷︸dV1

+ρ(lI{Y>0} dB1 + lI{Y≤0} dB2

)︸︷︷︸dV2

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Z (t) = X1(t) + X2(t).

= ν dt + dV ,

ν = g − h,

)︸︷︷︸dV1

+ρ(lI{Y>0} dB1 + lI{Y≤0} dB2

)︸︷︷︸dV2

12 / 94

Z (t) = X1(t) + X2(t).

= ν dt + dV ,

ν = g − h,

)︸︷︷︸dV1

+ρ(lI{Y>0} dB1 + lI{Y≤0} dB2

)︸︷︷︸dV2

13 / 94

Z (t) = X1(t) + X2(t).

= ν dt + dV ,

ν = g − h,

)︸︷︷︸dV1

+ρ(lI{Y>0} dB1 + lI{Y≤0} dB2

)︸︷︷︸dV2

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Z (t) = X1(t) + X2(t).

= ν dt + dV ,

ν = g − h,

)︸︷︷︸dV1

+ρ(lI{Y>0} dB1 + lI{Y≤0} dB2

)︸︷︷︸dV2

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Summary

X1(t) + X2(t) = Z (t)

= X1(0) + X2(0) + νt + V (t),

X1(t)− X2(t) = Y (t)

= X1(0) + X2(0)− λ∫ t

0sgn(Y (s)

)ds + W (t),

where V and W are correlated Brownian motions.

I To determine transition probabilities for (X1,X2), it suffices todetermine transition probabilites for (Z ,Y ).

I (Z ,W ) is a Gaussian process.I Y is determined by W by

dY (t) = −λ sgn(Y (t)

)dt + dW (t).

I It suffices to determine the distribution of (W (t),Y (t)).

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Summary

X1(t) + X2(t) = Z (t)

= X1(0) + X2(0) + νt + V (t),

X1(t)− X2(t) = Y (t)

= X1(0) + X2(0)− λ∫ t

0sgn(Y (s)

)ds + W (t),

)dt + dW (t).

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Summary

X1(t) + X2(t) = Z (t)

= X1(0) + X2(0) + νt + V (t),

X1(t)− X2(t) = Y (t)

= X1(0) + X2(0)− λ∫ t

0sgn(Y (s)

)ds + W (t),

)dt + dW (t).

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Summary

X1(t) + X2(t) = Z (t)

= X1(0) + X2(0) + νt + V (t),

X1(t)− X2(t) = Y (t)

= X1(0) + X2(0)− λ∫ t

0sgn(Y (s)

)ds + W (t),

)dt + dW (t).

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Bang-bang control

Minimize

E∫ ∞

0e−tX 2

Subject to

Xt = x +

0us ds + Wt , a ≤ ut ≤ b, t ≥ 0.

Benes, Shepp & Witsenhausen (1980): Solution is ut = f (Xt),where

f (x) =

{b, if x < δ,a, if x ≥ δ,

and δ = (√

b2 + 2 + b)−1 − (√

a2 + 2− a)−1.What is the transition density for the controlled process X ?(Benes, et. al. computed its Laplace transform.)

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Bang-bang control

Minimize

E∫ ∞

0e−tX 2

Subject to

Xt = x +

0us ds + Wt , a ≤ ut ≤ b, t ≥ 0.

f (x) =

and δ = (√

b2 + 2 + b)−1 − (√

a2 + 2− a)−1.

What is the transition density for the controlled process X ?(Benes, et. al. computed its Laplace transform.)

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Bang-bang control

Minimize

E∫ ∞

0e−tX 2

Subject to

Xt = x +

0us ds + Wt , a ≤ ut ≤ b, t ≥ 0.

f (x) =

and δ = (√

b2 + 2 + b)−1 − (√

a2 + 2− a)−1.What is the transition density for the controlled process X ?(Benes, et. al. computed its Laplace transform.)

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Transition density by GirsanovCompute the transition density for X , where

Xt = x +

0f (Xs) ds + Wt .

Start with a probability measure PX under which X is a Brownianmotion. Define W by

Wt = Xt − x −∫ t

0f (Xs) ds.

Change to a probability measure P under which W is a Brownianmotion. Compute the transition density for X under P.

P{Xt ∈ B} = EX

[lI{Xt∈B}

∣∣∣∣Ft

[∫ t

0f (Xs) dXs −

0f 2(Xs) ds

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Xt = x +

0f (Xs) ds + Wt .

Wt = Xt − x −∫ t

0f (Xs) ds.

P{Xt ∈ B} = EX

[lI{Xt∈B}

∣∣∣∣Ft

[∫ t

0f (Xs) dXs −

0f 2(Xs) ds

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Xt = x +

0f (Xs) ds + Wt .

Wt = Xt − x −∫ t

0f (Xs) ds.

P{Xt ∈ B} = EX

[lI{Xt∈B}

∣∣∣∣Ft

[∫ t

0f (Xs) dXs −

0f 2(Xs) ds

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Xt = x +

0f (Xs) ds + Wt .

Wt = Xt − x −∫ t

0f (Xs) ds.

P{Xt ∈ B} = EX

[lI{Xt∈B}

∣∣∣∣Ft

[∫ t

0f (Xs) dXs −

0f 2(Xs) ds

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Transition density by Girsanov (continued)To compute P{Xt ∈ B}, we need the joint distribution of

0f (Xs) dXs ,

0f 2(Xs) ds.

Assume without loss of generality that δ = 0. Define

F (x) =

0f (ξ) dξ =

{bx , if x ≤ 0,ax , if x ≥ 0.

Tanaka’s formula implies

F (Xt) =

0f (Xs) dXs +

2(a− b)LX

so ∫ t

0f (Xs) dXs = F (Xt)−

2(a− b)LX

We need the joint distribution of

Xt , LXt ,

0f 2(Xs) ds.

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0f (Xs) dXs ,

0f 2(Xs) ds.

F (x) =

0f (ξ) dξ =

{bx , if x ≤ 0,ax , if x ≥ 0.

F (Xt) =

0f (Xs) dXs +

2(a− b)LX

so ∫ t

2(a− b)LX

Xt , LXt ,

0f 2(Xs) ds.

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0f (Xs) dXs ,

0f 2(Xs) ds.

F (x) =

0f (ξ) dξ =

{bx , if x ≤ 0,ax , if x ≥ 0.

F (Xt) =

0f (Xs) dXs +

2(a− b)LX

so ∫ t

2(a− b)LX

Xt , LXt ,

0f 2(Xs) ds.

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Transition density by Girsanov (continued)

Define

Γ+(t) =

0lI(0,∞)

(X (s)

Γ−(t) =

0lI(−∞,0)

(X (s)

)ds = t − Γ+(t).

Then∫ t

0f 2(Xs) ds = b2Γ−(t) + a2Γ+(t) = b2t + (a2 − b2)Γ+(t).

Xt , LXt , Γ+(t) =

0lI(0,∞)(Xs) ds,

where X is a Brownian motion.

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Define

Γ+(t) =

0lI(0,∞)

(X (s)

Γ−(t) =

0lI(−∞,0)

(X (s)

)ds = t − Γ+(t).

Then∫ t

Xt , LXt , Γ+(t) =

0lI(0,∞)(Xs) ds,

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Define

Γ+(t) =

0lI(0,∞)

(X (s)

Γ−(t) =

0lI(−∞,0)

(X (s)

)ds = t − Γ+(t).

Then∫ t

Xt , LXt , Γ+(t) =

0lI(0,∞)(Xs) ds,

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Trivariate density

Theorem (Karatzas & Shreve (1984))

Let W be a Brownian motion, let L be its local time at zero, andlet

Γ+(t) ,∫ t

0lI(0,∞)(Ws) ds

denote the occupation time of the right half-line. Then for a ≤ 0,b ≥ 0, and 0 < t < T ,

P{WT ∈ da, LT ∈ db, Γ+(T ) ∈ dt}

=b(b − a)

t3(T − t)3exp

[−b2

2t− (b − a)2

2(T − t)

]da db dt.

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RemarkBecause the occupation time of the left half-line is

Γ−(T ) ,∫ T

0lI(−∞,0)(Wt) dt = T − Γ+(T ),

we also know P{WT ∈ da, LT ∈ db, Γ−(T ) ∈ dt} for a ≤ 0, b ≥ 0,and 0 < t < T . Applying this to −W , we obtain a formula for

P{WT ∈ da, LT ∈ db, Γ+(T ) ∈ dt}, a ≥ 0, b ≥ 0, 0 < t < T .

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Classic trivariate density

Theorem (Levy (1948))

Let W be a Brownian motion, let MT = max0≤t≤T Wt , and let θTbe the (almost surely unique) time when W attains its maximumon [0,T ]. Then for a ∈ R, b ≥ max{a, 0}, and 0 < t < T ,

P{WT ∈ da,MT ∈ db, θT ∈ dt}

=b(b − a)

t3(T − t)3exp

[−b2

2t− (b − a)2

2(T − t)

]da db dt.

The elementary proof uses the reflection principle and the Markovproperty.

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Positive part of Brownian motion

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

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Γ+(T ) t0

W+(t) , WΓ−1+ (t)

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Γ+(T ) t0

W+(t) , WΓ−1+ (t)

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Full decomposition

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

W−(t) , −WΓ−1− (t)

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Full decomposition

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

W−(t) , −WΓ−1− (t)

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Full decomposition

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

W−(t) , −WΓ−1− (t)

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Full decomposition

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

W−(t) , −WΓ−1− (t)

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Local time

Local time at zero of W :

0δ0(Ws) ds = lim

ε↓0

0lI(−ε,ε)(Ws) ds

= limε↓0

0lI(0,ε)(Ws) ds = lim

ε↓0

0lI(−ε,0)(Ws) ds.

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Tanaka’s formulaTanaka formula:

max{Wt , 0} =

0lI(0,∞)(Ws) dWs +

0δ0(Ws) ds

= −Bt + Lt ,

Bt = −∫ t

0lI(0,∞)(Ws) dWs , 〈B〉t = Γ+(t).

Then B+(t) , BΓ−1+ (t) is a Brownian motion.

Time-changed Tanaka formula:

W+(t) = −B+(t) + L+(t),

where L+(t) = LΓ−1+ (t).

Conclusion: W+ is a reflected Brownian motion. It is the Brownianmotion −B+ plus the nondecreasing process L+ that grows onlywhen W+ is at zero.

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max{Wt , 0} =

0lI(0,∞)(Ws) dWs +

0δ0(Ws) ds

= −Bt + Lt ,

Bt = −∫ t

0lI(0,∞)(Ws) dWs ,

〈B〉t = Γ+(t).

W+(t) = −B+(t) + L+(t),

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max{Wt , 0} =

0lI(0,∞)(Ws) dWs +

0δ0(Ws) ds

= −Bt + Lt ,

Bt = −∫ t

0lI(0,∞)(Ws) dWs , 〈B〉t = Γ+(t).

W+(t) = −B+(t) + L+(t),

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max{Wt , 0} =

0lI(0,∞)(Ws) dWs +

0δ0(Ws) ds

= −Bt + Lt ,

Bt = −∫ t

0lI(0,∞)(Ws) dWs , 〈B〉t = Γ+(t).

W+(t) = −B+(t) + L+(t),

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max{Wt , 0} =

0lI(0,∞)(Ws) dWs +

0δ0(Ws) ds

= −Bt + Lt ,

Bt = −∫ t

0lI(0,∞)(Ws) dWs , 〈B〉t = Γ+(t).

W+(t) = −B+(t) + L+(t),

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max{Wt , 0} =

0lI(0,∞)(Ws) dWs +

0δ0(Ws) ds

= −Bt + Lt ,

Bt = −∫ t

0lI(0,∞)(Ws) dWs , 〈B〉t = Γ+(t).

W+(t) = −B+(t) + L+(t),

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Skorohod representation

W+(t) = −B+(t) + L+(t).

Skorohod representation:The nondecreasing process added to −B+ that grows only whenW+ = 0 is

L+(t) = max0≤s≤t

B+(s).

In particular,

B+(t) = −W+(t) + L+(t) = −W+(t) + max0≤s≤t

B+(s).

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W+ and B+

B+(t) = −W+(t) + L+(t) = −W+(t) + max0≤s≤t

B+(s).

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

B+ L+(T ) = max0≤t≤Γ+(t) B+(t)

51 / 94

W+ and B+

B+(t) = −W+(t) + L+(t) = −W+(t) + max0≤s≤t

B+(s).

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

B+ L+(T ) = max0≤t≤Γ+(t) B+(t)

52 / 94

W+ and B+

B+(t) = −W+(t) + L+(t) = −W+(t) + max0≤s≤t

B+(s).

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

L+(T ) = max0≤t≤Γ+(t) B+(t)

53 / 94

W+ and B+

B+(t) = −W+(t) + L+(t) = −W+(t) + max0≤s≤t

B+(s).

Γ+(T ) t0

W+(t) , WΓ−1+ (t)

B+ L+(T ) = max0≤t≤Γ+(t) B+(t)

54 / 94

W+ and B+. W− and B−.B+(t) = −W+(t) + L+(t) = −W+(t) + max

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

−W− run backwards

B+ B− run backwards

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0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

56 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

57 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

58 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

59 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

B− run backwards

60 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

61 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

62 / 94

0≤s≤tB+(s)

B−(t) = −W−(t) + L−(t) = −W−(t) + max0≤s≤t

B−(s).

Γ+(T ) tT0

W− run backwards

Local time L+(T ) = LT time has become the maximum.Γ+(T ) has become the time of the maximum.

63 / 94

Personal reflections

64 / 94

In the beginning....S. Shreve, Reflected Brownian motion in the “bang-bang” controlof Browian drift, SIAM J. Control Optimization 19, 469–478,(1981).

Acknowledgment in the paper

The author wishes to acknowledge the aid of V. E.Benes, who found an error in some preliminary work onthis subject and suggested the applicability of Tanaka’sformula. He also wishes to thank the referee for pointingout the uniqueness of the transition densitycorresponding to the weak solution in Section 5.

I Work on stochastic control (monotone follower, boundedvariation follower, finite-fuel, ....)

I Work on optimal investment, consumption and duality withJohn Lehoczky, Suresh Sethi, Gan-Lin Xu, Jaksa Cvitanic ....

65 / 94

66 / 94

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68 / 94

....and then THE BOOK,

69 / 94

....and then THE BOOK,

70 / 94

and mathematical finance took off.

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and mathematical finance took off.

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Ten things I learned from Ioannis Karatzas

I Greek culture10. Macedonia is a province in Greece.

9. All Greeks plan to return home some day, until the opportunityto do so actually arises.

I Spelling8. Lemmata (From the Greek ληµµατα; not commonly used in

West Virginia dialect.)7. Coordinate (Diaeresis: [Ancient Greek] The separate

pronunciation of two vowels in a diphthong.)I Scholarship

6. Appreciate the work of others.5. Revise, revise, revise.

I Advising4. Choose excellent students.3. Teach students to appreciate the work of others and to revise,

revise, revise.I Administration

2. Avoid it.

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Ten things I learned from Ioannis KaratzasI Greek culture

10. Macedonia is a province in Greece.9. All Greeks plan to return home some day, until the opportunity

to do so actually arises.I Spelling

8. Lemmata (From the Greek ληµµατα; not commonly used inWest Virginia dialect.)

7. Coordinate (Diaeresis: [Ancient Greek] The separatepronunciation of two vowels in a diphthong.)

I Scholarship6. Appreciate the work of others.5. Revise, revise, revise.

2. Avoid it.

74 / 94

10. Macedonia is a province in Greece.

9. All Greeks plan to return home some day, until the opportunityto do so actually arises.

2. Avoid it.

75 / 94

10. Macedonia is a province in Greece.9. All Greeks plan to return home some day,

until the opportunityto do so actually arises.

2. Avoid it.

76 / 94

to do so actually arises.

2. Avoid it.

77 / 94

2. Avoid it.

78 / 94

8. Lemmata

(From the Greek ληµµατα; not commonly used inWest Virginia dialect.)

2. Avoid it.

79 / 94

2. Avoid it.

80 / 94

7. Coordinate

(Diaeresis: [Ancient Greek] The separatepronunciation of two vowels in a diphthong.)

2. Avoid it.

81 / 94

2. Avoid it.

82 / 94

I Scholarship

2. Avoid it.

83 / 94

I Scholarship6. Appreciate the work of others.

5. Revise, revise, revise.I Advising

4. Choose excellent students.3. Teach students to appreciate the work of others and to revise,

2. Avoid it.

84 / 94

2. Avoid it.

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I Advising

4. Choose excellent students.3. Teach students to appreciate the work of others and to revise,

2. Avoid it.

86 / 94

I Advising4. Choose excellent students.

3. Teach students to appreciate the work of others and to revise,revise, revise.

I Administration2. Avoid it.

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revise, revise.

I Administration2. Avoid it.

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2. Avoid it.

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2. Avoid it.

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Number one

1. A professional colleague who is also a friend is more preciousthan Euros/Drachmae.

Thank you, Yannis,

for all you have done

I for your students,

I for your colleagues,

I for science,

I and for me personally.

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Number one

Thank you, Yannis,

I for science,

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Number one

Thank you, Yannis,

I for science,

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Number one

Thank you, Yannis,

I for science,

94 / 94

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