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Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps Anders Tolver Jensen, Postdoc tolver@math.ku.dk Department of Applied Mathematics and Statistics, University of Copenhagen - Joint work with Martin Jacobsen

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SLIDE 1

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps

Anders Tolver Jensen, Postdoc

tolver@math.ku.dk

Department of Applied Mathematics and Statistics, University of Copenhagen

  • Joint work with Martin Jacobsen

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.1

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SLIDE 2

Two sided exit problems

Sample path of a stochastic process

Time X_t X_0 T_1 T_2 T_3 T_4

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.2

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SLIDE 3

Two sided exit problems

Sample path of a stochastic process

Time X_t l u T_1 T_2 T_3 T_4

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.2

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SLIDE 4

Two sided exit problems

Sample path of a stochastic process

Time X_t T_1 T_2 T_3 T_4 tau l X_0 u

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.2

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SLIDE 5

Two sided exit problems

Mathematical formulation of problem

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3

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SLIDE 6

Two sided exit problems

Mathematical formulation of problem

  • X stochastic process, X0 ∈ (l, u)
  • τ = inf{t > 0|Xt /

∈ (l, u)}

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3

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SLIDE 7

Two sided exit problems

Mathematical formulation of problem

  • X stochastic process, X0 ∈ (l, u)
  • τ = inf{t > 0|Xt /

∈ (l, u)}

  • Ultimate goal: Find joint distribution of (τ, Xτ)

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3

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SLIDE 8

Two sided exit problems

Mathematical formulation of problem

  • X stochastic process, X0 ∈ (l, u)
  • τ = inf{t > 0|Xt /

∈ (l, u)}

  • Ultimate goal: Find joint distribution of (τ, Xτ)
  • Partial solution: Find the Laplace transform

Ex[exp(−θτ); A]

where A contains information about Xτ

  • Example: A = (Xτ < l)

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3

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SLIDE 9

What kind of process is X?

Compound Poisson dam /shot noise process /generalized Ornstein-Uhlenbeck process

  • X solves stochastic diff. equation

dXt = kXtdt + dLt

  • Lt = Nt

n=1 Yn is compound Poisson process

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.4

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SLIDE 10

What kind of process is X?

Compound Poisson dam /shot noise process /generalized Ornstein-Uhlenbeck process

  • X solves stochastic diff. equation

dXt = kXtdt + dLt

  • Lt = Nt

n=1 Yn is compound Poisson process

  • N ∼ Pois(λ) with jump times Tn
  • Yn iid ∼ G

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.4

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SLIDE 11

What kind of process is X?

Compound Poisson dam /shot noise process /generalized Ornstein-Uhlenbeck process

  • X solves stochastic diff. equation

dXt = kXtdt + dLt

  • Lt = Nt

n=1 Yn is compound Poisson process

  • N ∼ Pois(λ) with jump times Tn
  • Yn iid ∼ G ← Pos. and neg. jumps can occur

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.4

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SLIDE 12

What kind of process is X?

Solution to:

dXt = kXtdt + dLt

Sample path of a stochastic process

Time X_t X_0 T_1 T_2 T_3 T_4

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.5

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Classic solution

  • A : generator for X

Af(x) = kxf′(x) + λ

  • f(x + y)G(dy) − f(x)
  • Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6
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Classic solution

  • A : generator for X

Af(x) = kxf′(x) + λ

  • f(x + y)G(dy) − f(x)
  • Îto’s formula

exp(−θt)f(Xt) = f(X0)+ t (Af(Xs) − θf(Xs)) ds+Mt

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6

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SLIDE 15

Classic solution

  • A : generator for X

Af(x) = kxf′(x) + λ

  • f(x + y)G(dy) − f(x)
  • Îto’s formula

exp(−θt)f(Xt) = f(X0)+ t (Af(Xs) − θf(Xs)) ds+Mt

  • Solve eigenvalue problem: Af = θf

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6

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SLIDE 16

Classic solution

  • A : generator for X

Af(x) = kxf′(x) + λ

  • f(x + y)G(dy) − f(x)
  • Îto’s formula

exp(−θt)f(Xt) = f(X0)+ t (Af(Xs) − θf(Xs)) ds+Mt

  • Solve eigenvalue problem: Af = θf
  • Optional Stopping+dom. convergence⇒

Ex[exp(−θτ)f(Xτ)] = f(X0)

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6

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SLIDE 17

Classic solution

Potential problems

  • How to obtain information about τ from

Ex[exp(−θτ)f(Xτ)]?

  • Formal proofs straightforward if f is bounded
  • Bounded eigenfunctions rarely exist

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.7

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Advanced solution

  • f bounded on (l, u)
  • f(x) =
  • L

x < l U x > u

  • Af(x) = θf(x)

(x ∈ (l, u))

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8

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Advanced solution

  • f bounded on (l, u)
  • f(x) =
  • L

x < l U x > u

  • Af(x) = θf(x)

(x ∈ (l, u)) ← partial eigenfunction

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8

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SLIDE 20

Advanced solution

  • f bounded on (l, u)
  • f(x) =
  • L

x < l U x > u

  • Af(x) = θf(x)

(x ∈ (l, u)) ← partial eigenfunction ⇓ ← non standard argument Ex[exp(−θτ)f(Xτ)] = f(X0)

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8

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SLIDE 21

Advanced solution

  • f bounded on (l, u)
  • f(x) =
  • L

x < l U x > u

  • Af(x) = θf(x)

(x ∈ (l, u)) ← partial eigenfunction ⇓ ← non standard argument Ex[exp(−θτ)f(Xτ)] = f(X0)

Note: f(Xτ) takes only finitely many diff. values!

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8

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SLIDE 22

Solving the eigenvalue problem

Look for partial eigenfunctions on the form

f(x) =

  • Γ

exp(−xz)ψ0(z)dz (x ∈ (l, u))

  • Γ : contour in complex plane from γ1 to γ2

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.9

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SLIDE 23

Solving the eigenvalue problem

Look for partial eigenfunctions on the form

f(x) =

  • Γ

exp(−xz)ψ0(z)dz (x ∈ (l, u))

  • Γ : contour in complex plane from γ1 to γ2
  • Along Γ the “density” ψ0 must satisfy diff. equation

with certain boundary condition

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.9

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Solving the eigenvalue problem

Look for partial eigenfunctions on the form

f(x) =

  • Γ

exp(−xz)ψ0(z)dz (x ∈ (l, u))

  • Γ : contour in complex plane from γ1 to γ2
  • Along Γ the “density” ψ0 must satisfy diff. equation

with certain boundary condition

  • Constants L and U can be adjusted in appropriate

ways

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.9

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SLIDE 25

Simple example: Laplace dist. jumps

Laplace distributed jumps

G(dy) = µ/2 exp(−µ|y|)dy

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.10

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SLIDE 26

Simple example: Laplace dist. jumps

Laplace distributed jumps

G(dy) = µ/2 exp(−µ|y|)dy

  • Diff. eq:

ψ′

0(z)

ψ0(z) = λ k µ µ2 − z2 − (θ/k + 1) 1 z

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.10

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SLIDE 27

Simple example: Laplace dist. jumps

Laplace distributed jumps

G(dy) = µ/2 exp(−µ|y|)dy

  • Diff. eq:

ψ′

0(z)

ψ0(z) = λ k µ µ2 − z2 − (θ/k + 1) 1 z

Local solution:

ψ0(z) = exp(−λ/(2k)log(z2 − µ2) − (θ/k + 1)log(z))

Complex log and contour Γ chosen with care!

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.10

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Simple example: Laplace dist. jumps (k < 0)

  • ψ0(t) = |t2 − µ2|−λ/(2k)|t|−(θ/k+1)
  • f(x) =
  • Γ

exp(−xt)ψ0(t)dt(x ∈ [l, u])

  • L =
  • Γ

µ µ − t exp(−lt)ψ0(t)dt

  • U =
  • Γ

µ µ + t exp(−ut)ψ0(t)dt

  • Γ one of the intervals (−µ, 0), (0, µ), or (µ, ∞).

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.11

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Extensions / Main result

Specification of ψ0 and integration paths, Γ, when

G = pG− + (1 − p)G+

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.12

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Extensions / Main result

Specification of ψ0 and integration paths, Γ, when

G = pG− + (1 − p)G+ G+(dy) =

  • i

αiµi exp(−µiy) (y > 0) G−(dy) =

  • i

βiνi exp(νiy) (y < 0)

Linear comb. of exponentials

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.12

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SLIDE 31

Extensions / Main result

Specification of ψ0 and integration paths, Γ, when

G = pG− + (1 − p)G+ G+(dy) =

  • i

αiµi exp(−µiy) (y > 0) G−(dy) =

  • i

βiνi exp(νiy) (y < 0)

Linear comb. of exponentials

Applicable for numerical purposes!

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.12

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Example: Different types of ruin

Laplace distributed jumps, negative drift k = −1

5 10 15 20 25 0.0 0.2 0.4 0.6 0.8 1.0

Downcrossings of l=5 caused by a jump when mu=0.1

X_0 / Initial value p_j / Frequency

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.13

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Example: Different types of ruin

Laplace distributed jumps, negative drift k = −1

5 10 15 0.0 0.2 0.4 0.6 0.8 1.0

Downcrossings of l caused by a jump when mu=0.1 and X_0=15

l / Level to be crossed p_j / Frequency

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.13

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SLIDE 34

Example: Different types of ruin

Laplace distributed jumps, negative drift k = −1

0.00 0.05 0.10 0.15 0.20 0.0 0.2 0.4 0.6 0.8 1.0

Downcrossings of l=5 caused by a jump when X_0=15

mu / Scale parameter for jumps p_j / Frequency

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.13

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SLIDE 35

Example: A nice result

  • Jump distribution

G = pG− + (1 − p)G+

with G− exponential and G+ arbitrary

  • pλ/k = 1
  • τ = inf{t > 0|Xt < l}

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.14

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SLIDE 36

Example: A nice result

  • Jump distribution

G = pG− + (1 − p)G+

with G− exponential and G+ arbitrary

  • pλ/k = 1
  • τ = inf{t > 0|Xt < l} ← exit caused by neg. jump!

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.14

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Example: A nice result

  • Jump distribution

G = pG− + (1 − p)G+

with G− exponential and G+ arbitrary

  • pλ/k = 1
  • τ = inf{t > 0|Xt < l} ← exit caused by neg. jump!

⇓ τ follows defective exponential distribution!

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.14

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Example: A nice result

  • Jump distribution

G = pG− + (1 − p)G+

with G− exponential and G+ arbitrary

  • pλ/k = 1
  • τ = inf{t > 0|Xt < l} ← exit caused by neg. jump!

⇓ τ follows defective exponential distribution!

Use Γ small circle around µ+calc. of residues

Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.14