Ruin problem for integrated stationary Gaussian process Kobelkov S. - PDF document

Ruin problem for integrated stationary Gaussian process Kobelkov S. G. 1 Consider a random process � t 0 X s ds − ct θ , Y t = where X t , t ≥ 0 is a stationary real-valued zero-mean Gaussian process with continuous trajectories and twice differentiable covariation function R ( t ), c > 0, θ > 1 / 2. Such a model arises, for example, in ruin financial problems, telecommunications, and information storage problems [6,7]. Define the ruin probability P ( u ) = P {∃ t ≥ 0 : Y t ≥ u } = P { max t ≥ 0 Y t ≥ u } . � t The random process 0 X s ds is a process with stationary increments. The ruin probability P ( u ) has been studied in a number of papers for various models of processes with stationary increments. In [3,7] an exact asymptotic of P ( u ) as u → ∞ has been found for Y t = B α/ 2 ( t ) − ct , where B α/ 2 is the fractional Brownian motion. The most similar in problem formulation work is the paper [5]. There, an asymptotic of the ruin probability for θ = 1 was found in the following form: P ( u ) = H η G e − Hc 2 /G 2 e − uc/G (1 + o (1)) c 2 as u → ∞ . Here H η is a generalized Pickands constant for the process √ 2) − 1 � t η = c ( G 0 X t dt . But the constant H η has not been calculated, and its dependance on characteristics of the original process X t remains unclear. Application of the Rice’s method allows to obtain the exact asymptotic of the P ( u ) as u → ∞ under some conditions, thus we are able to calculate the value of H η . Intuitively it is clear that for large u the event that the process Y t crosses the level u more than once is rare, so the probability P ( u ) approximately equals the mean number of crossings. The Rice’s method allows to formalize this idea. 1 M.V.Lomonosov Moscow State University, Dept. of Mechanics and Mathemat- ics, Chair of theory of probability, Leninskije Gory, 119992 Moscow, Russia; e-mail: sergey@dodo.inm.ras.ru 1

Let a random variable N u ([0 , T ]) be equal to the number of crossings of the level u by the process Y t on the segment [0 , T ]. In [1,2] it was shown that for a random process with continuously differentiable trajectories and for any segment S the following relation holds 0 ≤ EN u ( S ) − P ( max t ∈ S Y t ≥ u ) ≤ EN u ( S )( N u ( S ) − 1) . Application of this method gives us � ∞ � ∞ Theorem 1 Suppose that G = 0 R ( s ) ds > 0 , H = 0 sR ( s ) ds is finite, � ∞ and u 2 − 2 /θ u 1 /θ sR ( s ) ds → 0 , u → ∞ . Then � R (0) 2 π u − 1+1 /θ (2 θ − 1) 1 / 2 − 1 /θ c − 1 /θ × √ P ( u ) = (1 + τ θ min (2 θ − 1) − 1 )) 2 � � − u 2 − 1 /θ × exp (1 + o (1)) 4 G (2 θ − 1) − 1 /θ c − 1 /θ τ min − 4 Hu − 1 /θ as u → ∞ , where τ min = τ min ( u ) is a point of minimum of the function (1 + τ θ (2 θ − 1) − 1 )) 2 v ( τ ) = 4 G (2 θ − 1) − 1 /θ c − 1 /θ τ − 4 Hu − 1 /θ . It turns out that main part of the probability P ( u ) compose events such that the level crossing occurres for t in some neighborhood of maximum point of variance of the process Y t . To prove this, rewrite the probability P ( u ) in the following form: � t � � 1 P ( u ) = P max 0 X s ds > u . (1 + ct θ /u ) t> 0 � t 1 Then, the variance of the process V t = 0 X s ds can be represented as (1+ ct θ /u ) the sum � ∞ � ∞ 2 Gt R ( s ) ds − 2 sR ( s ) ds 2 Gt − 2 H t t Var V t = (1 + ct θ /u ) 2 − = S 1 ( t )+ R 1 ( t ) . (1) (1 + ct θ /u ) 2 The second term in (1) is negligible due to the assumptions set, and the first term has a unique point of maximum for large enough u . If we denote this 2

point by t max = t max ( u ), then with the help of Piterbarg inequality [1] we can choose a segment I = [ t max − ∆ , t max + ∆] with ∆ = ∆( u ) → 0, such that u 2 � � � � �� P sup t �∈ I Y t > u = o exp − . 2 S 1 ( t max ) Thus, it is sufficient to estimate the values EN u ( I ) and EN u ( I )( N u ( I ) − 1). The first term can be evaluated using the Rice formula [1] � ∞ � E N u ( I ) = 0 | y | p t ( u, y ) dydt, (2) I where p t ( u, y ) is a joint density of the random variables Y t , Y ′ t . Performing change of the variable t = ( u (2 θ − 1) − 1 /c ) 1 /θ τ and applying the generaliza- tion of the Laplace method to the integral (2), we obtain that � R (0) 2 π u − 1+1 /θ (2 θ − 1) 1 / 2 − 1 /θ c − 1 /θ × √ E N u ( I ) = (1 + τ θ min (2 θ − 1) − 1 )) 2 � � − u 2 − 1 /θ × exp (1 + o (1)) , 4 G (2 θ − 1) − 1 /θ c − 1 /θ τ min − 4 Hu − 1 /θ where τ min = τ min ( u ) is defined in the statement of the theorem. The estimation of the EN u ( I )( N u ( I ) − 1) is based on the application of the explicit formula for the second moment � ∞ � ∞ � � EN u ( I )( N u ( I ) − 1) = y 1 y 2 ϕ t,s,t,s ( u, u, y 1 , y 2 ) dy 1 dy 2 ds dt, I I 0 0 where ϕ t,s,t,s ( u, u, y 1 , y 2 ) is a joint density of the variables Y t , Y s , Y ′ t , Y ′ s . Pro- ceeding to conditional densities and applying Taylor formula, we prove that EN u ( I )( N u ( I ) − 1) = o ( EN u ( I )) as u → ∞ . It turns out that the require- ment of twice differentiability of the covariance function R ( t ) is significant in this method. Consideration of the case θ = 1 gives us the asymptotic � R (0) 2 π c − 1 exp {− Hc 2 √ P ( u ) = G 2 } exp {− uc/G } (1 + o (1)) , thus, comparing it with the result of Debicki, we obtain that the Pickands � t √ c � √ constant for the process η ( t ) = 0 X t dt equals R (0) / ( 2 πGc ). G 2 3

Denote the time of ruin τ u = inf { t ≥ 0 : u − Y t ≤ 0 } . The Rice’s method allows to obtain the asymptotic of the conditional distribution of τ u as u → ∞ given the ruin condition max t ≥ 0 Y t ≥ u . It is worth to mention that τ u takes values mostly in some neighborhood of t max . Theorem 2 Let the conditions of the Theorem 1 be fulfilled. Then P ( τ u < f ( x ) | τ u < ∞ ) → Φ( x ) , u → ∞ , where Φ( x ) is a distribution function of the standard normal random variable, and f ( x ) = u 3 / (2 θ ) − 1 √ 2 G (2 θ − 1) − 3 / (2 θ ) c − 3 / (2 θ ) θ − 2 x + ( u (2 θ − 1) − 1 /c ) 1 /θ . Let us start with the estimation of the probability P ( τ u < f ( x )). To prove the result we, as above, restrict consideration to the neighborhood of the point t max : I = [ t max − c − 1 /θ (2 θ − 1) − 1 /θ u 3 / (2 θ ) − 1 ln u, t max +( f ( x ) − t max )]. Substituting t = t ( τ ) = ( u (2 θ − 1) − 1 /c ) 1 /θ τ + t max , we obtain I = ( I 1 + I 2 )(1+ o (1)), where I 1 = ( u (2 θ − 1) − 1 /c ) 1 /θ � 0 σ ( t ( τ )) √ a ( τ ) g ( t ( τ )) exp {− S 3 ( τ ) } dτ, 2 π − u 1 / (2 θ ) − 1 ln u I 2 = ( u (2 θ − 1) − 1 /c ) 1 /θ � u 1 / (2 θ ) − 1 h ( x ) σ ( t ( τ )) (3) √ a ( τ ) g ( t ( τ )) exp {− S 3 ( τ ) } dτ, 2 π 0 u 2 S 3 ( τ ) = 2 S 1 ( t ( τ )) . a ( τ ) , g ( t ), and σ ( t ) are known functions with not more than polynomial growth in u . The first integral in (3) is estimated with the help of the Laplace method, which gives us � R (0) 2 π u − 1+1 /θ (2 θ − 1) 1 / 2 − 1 /θ c − 1 /θ exp {− S 3 (0) } (1 + o (1)) . I 1 = √ 2 In the second integral we can perform the change of variable y 2 = 2( S 3 ( u 1 / (2 θ ) − 1 τ ) − S 3 (0)), thus obtaining � � x R (0) 1 0 e − y 2 2 π u − 1+1 /θ (2 θ − 1) 1 / 2 − 1 /θ c − 1 /θ exp {− S 3 (0) } 2 dy (1+ o (1)) . √ √ I 2 = 2 π 4

The estimation of the EN u ( I )( N u ( I ) − 1) repeats the corresponding part of the proof of the former theorem. To complete the proof, it remains to sum the obtained estimates and to divide it by the probability P (sup t> 0 Y t > u ). References [1] Piterbarg, V. I. Asymptotic Methods in the Theory of Gaussian Processes and Fields , American Mathematical Society, Ser. Translations of Mathemat- ical Monographies, V. 148, Providence. [2] Piterbarg, V. I. (1996) Rice method for Gaussian random fields , Funda- mentalnaya i Prikladnaya Matematika, 2, 187–204. [3] H¨ usler, J., Piterbarg, V. (1999) Extremes of a certain class of Gaussian processes , Stochastic Processes and their Applications, 83, 257–271. [4] Kobelkov, S. (2004) Ruin problem for stationary Gaussian process , Teorija veroyatnostej i ee primenenija, 49, 171–178. [5] Debicki, K. (2002) Ruin probability for Gaussian integrated process , Stochastic Processes and their Applications, 98, 151–174. [6] Norros, I., (1994) A storage model with self-similar input , Queueing Sys- tems, 16, 387–396. [7] Narayan, O. (1998) Exact asymptotic queue length distribution for frac- tional Brownian traffic , Advances in Performance Analysis, 1, 39–63. [8] Chiu, S.N., Yin, C.C. (2003) The time of ruin, the surplus prior to ruin and the deficit at ruin for the classical risk process perturbed by diffusion. Insurance: Mathematics and Economics 33, 59-66. 5