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
Two sided exit problems Sample path of a stochastic process X_t X_0 T_1 T_2 T_3 T_4 Time Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.2
Two sided exit problems Sample path of a stochastic process u X_t l T_1 T_2 T_3 T_4 Time Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.2
Two sided exit problems Sample path of a stochastic process u X_t X_0 l T_1 T_2 T_3 T_4 tau Time Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.2
Two sided exit problems Mathematical formulation of problem Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3
Two sided exit problems Mathematical formulation of problem • X stochastic process, X 0 ∈ ( l, u ) • τ = inf { t > 0 | X t / ∈ ( l, u ) } Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3
Two sided exit problems Mathematical formulation of problem • X stochastic process, X 0 ∈ ( l, u ) • τ = inf { t > 0 | X t / ∈ ( l, u ) } • Ultimate goal: Find joint distribution of ( τ, X τ ) Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3
Two sided exit problems Mathematical formulation of problem • X stochastic process, X 0 ∈ ( l, u ) • τ = inf { t > 0 | X t / ∈ ( l, u ) } • Ultimate goal: Find joint distribution of ( τ, X τ ) • Partial solution: Find the Laplace transform E x [exp( − θτ ); A ] where A contains information about X τ • Example: A = ( X τ < l ) Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.3
What kind of process is X ? Compound Poisson dam /shot noise process /generalized Ornstein-Uhlenbeck process • X solves stochastic diff. equation dX t = kX t dt + dL t • L t = � N t n =1 Y n is compound Poisson process Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.4
What kind of process is X ? Compound Poisson dam /shot noise process /generalized Ornstein-Uhlenbeck process • X solves stochastic diff. equation dX t = kX t dt + dL t • L t = � N t n =1 Y n is compound Poisson process • N ∼ Pois ( λ ) with jump times T n • Y n iid ∼ G Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.4
What kind of process is X ? Compound Poisson dam /shot noise process /generalized Ornstein-Uhlenbeck process • X solves stochastic diff. equation dX t = kX t dt + dL t • L t = � N t n =1 Y n is compound Poisson process • N ∼ Pois ( λ ) with jump times T n • Y n iid ∼ G ← Pos. and neg. jumps can occur Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.4
What kind of process is X ? Solution to: dX t = kX t dt + dL t Sample path of a stochastic process X_t X_0 T_1 T_2 T_3 T_4 Time Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.5
Classic solution • A : generator for X �� � A f ( 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
Classic solution • A : generator for X �� � A f ( x ) = kxf ′ ( x ) + λ f ( x + y ) G ( dy ) − f ( x ) • Îto’s formula � t exp( − θt ) f ( X t ) = f ( X 0 )+ ( A f ( X s ) − θf ( X s )) ds + M t 0 Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6
Classic solution • A : generator for X �� � A f ( x ) = kxf ′ ( x ) + λ f ( x + y ) G ( dy ) − f ( x ) • Îto’s formula � t exp( − θt ) f ( X t ) = f ( X 0 )+ ( A f ( X s ) − θf ( X s )) ds + M t 0 • Solve eigenvalue problem: A f = θf Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6
Classic solution • A : generator for X �� � A f ( x ) = kxf ′ ( x ) + λ f ( x + y ) G ( dy ) − f ( x ) • Îto’s formula � t exp( − θt ) f ( X t ) = f ( X 0 )+ ( A f ( X s ) − θf ( X s )) ds + M t 0 • Solve eigenvalue problem: A f = θf • Optional Stopping + dom. convergence ⇒ E x [exp( − θτ ) f ( X τ )] = f ( X 0 ) Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.6
Classic solution Potential problems • How to obtain information about τ from E x [exp( − θτ ) f ( X τ )] ? • 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
Advanced solution • f bounded on ( l, u ) � L x < l • f ( x ) = U x > u • A f ( x ) = θf ( x ) ( x ∈ ( l, u )) Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8
Advanced solution • f bounded on ( l, u ) � L x < l • f ( x ) = U x > u ( x ∈ ( l, u )) ← partial eigenfunction • A f ( x ) = θf ( x ) Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8
Advanced solution • f bounded on ( l, u ) � L x < l • f ( x ) = U x > u ( x ∈ ( l, u )) ← partial eigenfunction • A f ( x ) = θf ( x ) ⇓ ← non standard argument E x [exp( − θτ ) f ( X τ )] = f ( X 0 ) Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8
Advanced solution • f bounded on ( l, u ) � L x < l • f ( x ) = U x > u ( x ∈ ( l, u )) ← partial eigenfunction • A f ( x ) = θf ( x ) ⇓ ← non standard argument E x [exp( − θτ ) f ( X τ )] = f ( X 0 ) Note: f ( X τ ) takes only finitely many diff. values! Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.8
Solving the eigenvalue problem Look for partial eigenfunctions on the form � f ( x ) = exp( − xz ) ψ 0 ( z ) dz ( x ∈ ( l, u )) Γ • Γ : contour in complex plane from γ 1 to γ 2 Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.9
Solving the eigenvalue problem Look for partial eigenfunctions on the form � f ( x ) = exp( − xz ) ψ 0 ( z ) dz ( x ∈ ( l, u )) Γ • Γ : contour in complex plane from γ 1 to γ 2 • Along Γ the “density” ψ 0 must satisfy diff. equation with certain boundary condition Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.9
Solving the eigenvalue problem Look for partial eigenfunctions on the form � f ( x ) = exp( − xz ) ψ 0 ( z ) dz ( x ∈ ( l, u )) Γ • Γ : contour in complex plane from γ 1 to γ 2 • Along Γ the “density” ψ 0 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
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
Simple example: Laplace dist. jumps Laplace distributed jumps G ( dy ) = µ/ 2 exp( − µ | y | ) dy Diff. eq: ψ ′ ψ 0 ( z ) = λ 0 ( z ) µ 2 − z 2 − ( θ/k + 1) 1 µ k z Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.10
Simple example: Laplace dist. jumps Laplace distributed jumps G ( dy ) = µ/ 2 exp( − µ | y | ) dy Diff. eq: ψ ′ ψ 0 ( z ) = λ 0 ( z ) µ 2 − z 2 − ( θ/k + 1) 1 µ k z Local solution: ψ 0 ( z ) = exp( − λ/ (2 k )log( z 2 − µ 2 ) − ( 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
Simple example: Laplace dist. jumps ( k < 0 ) • ψ 0 ( t ) = | t 2 − µ 2 | − λ/ (2 k ) | t | − ( θ/k +1) • � f ( x ) = exp( − xt ) ψ 0 ( t ) dt ( x ∈ [ l, u ]) Γ • µ � L = µ − t exp( − lt ) ψ 0 ( t ) dt Γ • µ � U = µ + t exp( − ut ) ψ 0 ( t ) dt Γ • 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
Extensions / Main result Specification of ψ 0 and integration paths, Γ , when G = pG − + (1 − p ) G + Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.12
Extensions / Main result Specification of ψ 0 and integration paths, Γ , when G = pG − + (1 − p ) G + � G + ( dy ) = α i µ i exp( − µ i y ) ( y > 0) i � G − ( dy ) = β i ν i exp( ν i y ) ( y < 0) i Linear comb. of exponentials Hitting Time Problems for Piecewise Exponential Markov Processes with Two sided Jumps – p.12
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