Optimal dividend distribution under a ruin constraint Wolfgang Runggaldier‘s 65th birthday Bressanone/Brixen July 1 7 , 2007 Christian Hipp Institute for Finance, Banking , and Insurance Universität Karlsruhe (TH) (1/3) http://insurance.fbv.uni-karlsruhe.de christian.hipp@kit.edu Christian Hipp Finance, Banking and Insurance
Good to be here! Christian Hipp Finance, Banking and Insurance
Giovanni Andreatta, Università di Padova Alain Bensoussan, University of Texas, Dallas Francesca Biagini, Universität München Michele Bonollo, Banco Popolare di Verona e Novara Carl Chiarella, University of Technology, Sydney Carlo Alberto Clarotti, ENEA, Roma Giovanni Battista Di Masi, Università di Padova Ernst Eberlein, Universität Freiburg Robert J. Elliott, University of Calgary Gino Favero, Università Bocconi, Milano Lorenzo Finesso, ISIB-CNR, Padova Rüdiger Frey, Universität Leipzig Marco Frittelli, Università degli Studi, Milano Martino Grasselli, Università di Padova Stefan Jaschke, Munich Re, Munich Yuri Kabanov, Universitè de Franche-Comtè, Besancon Juerg Kohlas, University of Freiburg Robert Liptser, Tel Aviv University Fabio Mercurio, Banca IMI, Milano Sanjoy Mitter, Massachusetts Institute of Technology Hideo Nagai, University of Osaka Fulvio Ortu, Università Bocconi, Milano Sara Pasquali, IMATI-CNR, Milano Giorgio Picci, Università di Padova Eckhard Platen, University of Technology, Sydney Maurizio Pratelli, Università di Pisa Giorgio Romanin Jacur, Università di Padova Martin Schweizer, ETH, Zurich Dieter Sondermann, University of Bonn Fabio Spizzichino, Università di Roma - La Sapienza Peter J.C. Spreij, University of Amsterdam Michael Taksar, University of Missouri Karl Thomaseth, ISIB-CNR, Padova Marco Tolotti, Università Bocconi, Milano Nizar Touzi, Ecole Polytechnique, Paris Omar Zane, ABN-AMRO, London Good to be here! Christian Hipp Finance, Banking and Insurance
Good to be here! Christian Hipp Finance, Banking and Insurance
1 Summary Dynamic dividend optimization • started with de Finetti (1957, Act. Congress NY) • is connected with one of the major research topics of Wolfgang (discrete approximation of continuous control problems, papers in 1994/5/6/9 and 2001/2) • is a challenge under constraints • is on my desk since 2002 (paper H(2003)) Here and in H(2003): infinite time horizon, path dependent constraint ruin probability . 0-0
Solving a control problem under a constraint • Pareto optimal solution to a two objective problem. • with an extra state variable • this state variable is at the same time a control variable • modification of dynamic equations • stopping times in the continuous case • numerical computation via discrete approximations We consider two objectives: ruin probability and expected discounted dividends. Ruin probability: objective function for policy holders or for supervision. Expected discounted dividends: objective function for the stock holder. 0-1
Optimal dividend distribution – without a ruin constraint – leads to certain ruin which is not acceptable for the policy holders. Minimizing ruin probability leads to no dividend payment which is not acceptable for stock holders. A ruin probability less than one is possible only for a risk process which tends to infinity. Optimal dividend payment with ruin constraint also leads to a reserve tending to infinity, but later. 0-2
Continuous time model The company’s value process modelled by Brownian motion with drift X ( t ) = x + µt + σW ( t ) , t ≥ 0 , µ, σ > 0 , and W ( t ) , t ≥ 0 standard Brownian motion. Maximize the accumulated discounted expected dividends � τ D e − ρt dD ( t )] E [ (1) 0 under the ruin constraint ψ D ( x ) = P { τ D < ∞} ≤ α. (2) τ ruin time of the controlled process ρ positive interest for discounting of future payments. Maximum is taken over all non-decreasing adapted processes D ( t ) , t ≥ 0 . 0-3
The infinite horizon ruin probability without paying dividends, ψ 0 ( x ) = P { X ( t ) > 0 for all t ≥ 0 } = exp( − 2 µ σ 2 x ) . Solve the problem without ruin constraint: via a variational inequality max {− ρu ( x ) + µu ′ ( x ) + 1 2 σ 2 u ′′ ( x ) , 1 − u ′ ( x ) } . 0 = (3) In the range u ′ ( x ) > 1 we have a linear differential equation (LDG) with constant coefficients. The characteristic equation reads 0 = − ρ + µz + 1 2 σ 2 z 2 with solutions z 1 < 0 and z 2 > 0 . The general solution for (LDG) is u ( x ) = C 1 exp( z 1 x ) + C 2 exp( z 2 x ) . Let M be the unique value with u ′ ( M ) = 1 . Using the initial value u (0) = 0 and the smooth paste conditions u ′ ( M ) = 1 , u ′′ ( M ) = 0 , we 0-4
arrive at C 1 = − C 2 = C, where C and M are defined via M = log( z 2 1 ) − log( z 2 2 ) , z 2 − z 1 C = 1 / ( z 1 exp( z 1 M ) − z 2 exp( z 2 M )) . u ( x, 1) = u ( x ) = C (exp( z 1 x ) − exp( z 2 x )) . With ruin constraint (2): � τ D � � e − ρt dD ( t )] : ψ D ( x ) ≤ α u ( x, α ) = sup E [ 0 0-5
Dividends: continuous case Optimal dividend payment: pay all above M , pay nothing below M . Certain ruin, bounded wealth. u ( w ) w M Christian Hipp Finance, Banking and Insurance
Characterization of u ( x, α ): {− ρu ( x, α ) + ( µ − δ ) u x ( x, α ) + 1 2 σ 2 u xx ( x, α ) 0 = sup (4) δ,f + σ 2 fu x,α ( x, α ) + 1 2 σ 2 f 2 u αα ( x, α ) } . or � − ρu ( x, α ) + µu x ( x, α ) + 1 2 σ 2 u xx ( x, α ) 0 = max (5) 2 σ 2 u x,α ( x, α ) 2 − 1 � u αα ( x, α ) , 1 − u x ( x, α ) These equations do not help. 0-6
Discrete approximation (time/state space) For initial surplus x ≥ 0 (integer), discount factor 0 < v < 1 and dividend payment strategy D = ( d 0 , d 1 , ... ) with integers d t ≥ 0 : define X D ( t ) = x + X 1 + ... + X t − d 0 − ... − d t − 1 , t ≥ 0 , P { X t = 1 } = 1 − P { X t = − 1 } = p > 1 / 2 Choose D such that τ D − 1 � = max! v t d t E t =1 under the constraint ψ D ( x ) = P { τ D < ∞} ≤ α τ D ruin time. u ( x, α ) value function. 0-7
Dynamic equation for u ( x ) = u ( x, 1) (without constraint): u ( x ) = d + max v [ pu ( x + 1 − δ ) + (1 − p ) u ( x − 1 − δ )] δ = max [1 + u ( x − 1) , v ( pu ( x + 1) + (1 − p ) u ( x − 1))] Leads to a barrier strategy. Dynamic equation for u ( x, α ) : � � u ( x, α ) = sup δ + v ( pu ( x + 1 − δ, β ) + (1 − p ) u ( x − 1 − δ, β )) (6) δ,β δ ∈ { 0 , 1 , ..., x } , (7) ψ 0 ( x + 1 − δ ) ≤ β ≤ 1 (8) pβ + (1 − p ) β = α (9) ψ 0 ( x − 1 − δ ) ≤ β ≤ 1 (10) 0-8
Alternative: � � � � u ( x, α ) = max sup v ( pu ( x + 1 , β ) + (1 − p ) u ( x − 1 , β )) , 1 + u ( x − 1 , α ) β ✬✩ ψ 0 ( x + 1) ≤ β ≤ 1 pβ + (1 − p ) β = α p x + 1 , β ψ 0 ( x − 1) ≤ β ≤ 1 ✫✪ ✬✩ ✑ ✸ ✑✑✑✑ x, α pβ + (1 − p ) β = α ✫✪ ✬✩ ◗◗◗◗ ◗ s 1 − p x − 1 , β ✫✪ Solves the problem: solution of equation exists, verification argument works, numerical algorithm follows. 0-9
Algorithm: � � � � u n +1 ( x, α ) = max sup v ( pu n ( x + 1 , β ) + (1 − p ) u n ( x − 1 , β )) , 1 + u n ( x − 1 , α ) β Problems: • discretization for the range of β • β not in the grid • truncation of x • slow convergence • needs huge main memory • slow approximation 0-10
Continuous problem: An alternative dynamic equation for u ( x, α ) using stopping times: ✬✩ p ( x ) M, β ✫✪ ✬✩ ✑ ✸ ✑✑✑✑ x, α p ( x ) β + (1 − p ( x )) β = α ✫✪ ✬✩ ◗◗◗◗ ◗ s s, β 1 − p ( x ) ✫✪ t = 0 t = τ s < x < M, τ = inf { t : X D ( t ) / ∈ ( s, M ) } , p ( x ) = P { X τ = M } 0-11
Resulting equation for u ( x, α ) in the region u x ( x, α ) > 1 : � � u ( x, α ) = sup p ( x ) G ( x ) u ( M, β ) + (1 − p ( x )) H ( x ) u ( s, β ) s,M,β G ( x ) = E [exp( − ρτ )1 ( X ( τ )= M ) ] H ( x ) = E [exp( − ρτ )1 ( X ( τ )= s ) ] α = p ( x ) β + (1 − p ( x )) β ψ 0 ( M ) ≤ β ≤ 1 ψ 0 ( s ) ≤ β ≤ 1 p ( x ) solves 0 = µf ′ ( x ) + 1 2 f ′′ ( x ) G ( x ) , H ( x ) solve 0 = − ρf ( x ) + µf ′ ( x ) + 1 2 f ′′ ( x ) 0-12
Algorithm: � � u n +1 ( x, α ) = sup p ( x ) G ( x ) u n ( M, β ) + (1 − p ( x )) H ( x ) u n ( s, β ) s,M,β Too complex for numerical calculation. Experiments with MAPLE or Mathematica show a considerable improve- ment in each iteration; I had to stop after 10 iterations (too many recur- sions). The following numerical results are derived via discrete approximations. 0-13
Approximation: n → ∞ σ/ √ n step size ∆ = µ p = 1 / 2 + 2 σ √ n v = exp( − ρ/n ) In the numerical example we have n = 10 . 000 , µ = σ = 1 , ρ = 0 . 1 . The figures show results after 2.000 iterations. At the end, the improvement in each iteration was still significant: sup | u k +1 ( x, α ) − u k ( x, α ) | = 0 . 0157 . 0-14
Recommend
More recommend
