Discounted compound renewal sums with a stochastic force of interest Ghislain Léveillé, Franck Adékambi Université Laval, Québec, Canada 45 th Actuarial Research Conference Vancouver, Canada July 26-28, 2010. 1
Abstract Recursive moments, moments generating functions, distributions functions and risk measures have been found for the compound renewal sums with discounted claims, for a constant force of real interest. In this talk we present several results on the (joint) moments, on the (joint) moments generating functions and on regression aspects of these discounted renewal sums, in a context that may involve a stochastic force of real interest. Examples will be given for the counting Poisson process and for the Ho-Lee-Merton interest rate model. Keywords : Compound Poisson process; discounted aggregate claims; force of interest; Itô process; joint moments; renewal process; stochastic interest rate. 2
Our risk model { ( ) , t ≥ 0 } and { ( ) , t ≥ 0 } (i) The claims counting processes N t N d t form respectively an ordinary and a delayed renewal process, { } : k ∈ = 1, 2, 3, ... and for { } , • the positive claim occurrence times are given by Tk , k ∈ • the positive claim inter-arrival times are given by τ k = T k − T k − 1 , T 0 = 0 . with { } are (ii) The corresponding deflated claim severities Xk , k ∈ such that { } are i.i.d. . • Xk , k ∈ { } are mutually independent. • Xk , τ k ; k ∈ • The m.g.f. of X 1 exists in a neighourhood Ω ⊂ containing zero. 3
(iii) The aggregate discounted value at time 0 of the inflated claims recorded over the period 0, t ⎦ are given respectively, for ⎡ ⎤ ⎣ the ordinary and the delayed renewal case, by ( ) ( ) N d t N t ( ) = ( ) X k ( ) = ( ) X k ∑ ∑ , , Z t D T k Z d t D T k k = 1 k = 1 where ⎧ ⎫ T k ⎪ ⎪ ( ) = Z d t ( ) = 0 if N t ( ) = N d t ( ) = 0 , ( ) = exp − δ x ( ) dx ∫ , Z t D T k ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ 0 ( ) is the force of real interest, which can be a deterministic and δ x function or a random variable. ( ) for the risk process generated by an Remark : We will note Z o t embedded ordinary renewal process. 4
A reminder Moments of compound renewal sums with discounted claims have been considered for the first time by Léveillé and Garrido (2001), for a positive constant force of real interest. Using essentially renewal arguments, these recursives formulas have been obtained : • for the ordinary renewal case t ⎛ ⎞ n − 1 n ( ) ( ) ( ) ( ) = E N t ( ) E Z n t ⎦ e − n δ v E Z k t − v ∑ ∫ ⎡ ⎤ ⎟ E X n − k ⎡ ⎤ ⎡ ⎤ ⎦ = ⎡ ⎤ ⎦ dm v m t , ⎣ ⎦ . ⎜ ⎣ ⎣ ⎣ ⎝ ⎠ k k = 0 0 • for the delayed renewal case t ⎛ ⎞ n − 1 n ( ) ( ) ( ) ( ) = E N d t ( ) n t ∑ k t − v ∫ ⎡ ⎤ ⎟ E X n − k ⎡ ⎦ e − n δ v E Z o ⎤ ⎡ ⎤ ⎦ = ⎡ ⎤ ⎦ . E Z d ⎦ dm d v m d t ⎣ ⎜ , ⎣ ⎣ ⎣ ⎝ ⎠ k k = 0 0 5
Recursive joint moments for a constant δ We need first a lemma in order to get these recursive joint moments. Lemma 1 : Consider an ordinary or a delayed renewal counting t > 0 , h > 0 , process, such as defined previously. Then, for any δ ≥ 0 ( ) ∈Ω × Ω , the joint m.g.f. of our risk process satisfies and u , v respectively the following integral equations: (1) For the ordinary renewal case : t + h ( ) M Z t + h − x ( ) dF ( ) = F ( ) + ( ) ∫ M X ve −δ x ) ve −δ x τ 1 t + h M Z t ) u , v τ 1 x ( ) , Z t + h ( ( t t ( ) M Z t − x ( ) dF ( ) e −δ x ( ) ∫ ) ue −δ x , ve −δ x + M X u + v τ 1 x . ( ) , Z t + h − x ( 0 6
(2) For the delayed renewal case : t + h ( ) M Z o t + h − x ( ) dF ( ) = F ( ) + ( ) ∫ M X ve −δ x ) ve −δ x τ 1 t + h M Z d t ) u , v τ 1 x ( ) , Z d t + h ( ( t t ( ) M Z o t − x ( ) dF ( ) e −δ x ( ) . ∫ + M X u + v ) ue −δ x , ve −δ x τ 1 x ( ) , Z o t + h − x ( 0 ( ) = 1 − F ( ) . where F τ 1 t τ 1 t ( ) , N t + h ( ) , Proof of (1): We condition first on N t T 1 ,..., T N t + h ) , ( which yields ( ) ( ) ⎡ ⎤ N t + h N t ( ) ( ) ( ) = E ∏ ∏ ( ) e −δ T k u + v M X ve −δ T k ⎥ , M Z t ) u , v M X ⎢ ( ) , Z t + h ( ⎣ ⎦ ( ) + 1 k = 1 k = N t and thereafter we condition on τ 1 to get the result. 7
Theorem 1 : According to the hypotheses of lemma 1, the joint n , m ∈ , moments of our risk process are given respectively, for by : (1) For the ordinary renewal case : ( ) ⎛ ⎞ ⎛ ⎞ n + m min k , n n m ∑ ∑ ⎡ ⎤ ⎡ ⎤ E Z n ( t ) Z m ( t + h ) ⎦ = k E X 1 ⎣ ⎣ ⎦ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ k − i ⎠ i [ ] + k = 1 i = k − m t ∫ ( ) ( ) δ u E Z n − i ( t − u ) Z m − k − i ( ) ( t + h − u ) ⎡ ⎤ × e − n + m ⎣ ⎦ dm u . 0 (2) For the delayed renewal case : ( ) n + m ⎛ ⎞ ⎛ ⎞ min k , n n m ∑ ∑ ⎡ m ( t + h ) ⎤ ⎦ = ⎡ ⎤ n ( t ) Z d k E Z d E X 1 ⎣ ⎣ ⎦ ⎜ ⎟ ⎜ ⎟ k − i ⎝ ⎠ ⎝ ⎠ i [ ] + k = 1 i = k − m t ∫ ( ) ( ) δ u E Z o ( ) ( t + h − u ) × e − n + m ⎡ n − i ( t − u ) Z o m − k − i ⎤ ⎣ ⎦ dm d u . 0 8
Proof : The preceding equations follow directly by taking the appropriate partial derivatives of the integral equations of lemma 1, a number of times with respect to u and with respect ( ) = 0,0 ( ) to v , then each time evaluating these expressions at u , v and thereafter using induction. h = 0 in the equations of theorem 1, then Remark 1 : (1) If we set we retrieve the preceding recursive expressions for the moments. (2) For n = m = 1 , the joint moments of the risk process of theorem 1 can be written for the ordinary renewal case as follows t { } [ ] e − 2 δ u E Z t − u ( ) Z t + h ( ) ( ) ( ) ( ) ∫ ⎦ = E X 1 ⎦ + E Z t + h − u ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ E Z t dm u ⎣ ⎣ ⎣ ⎦ 0 t ( ) ∫ ⎡ ⎤ ⎦ e − 2 δ u dm u + E X 1 2 ⎣ 0 9
which is equivalent to t + h − u t [ ] ( ) Z t + h ( ) ( ) ) dm v ( ) dm u ( ) ⎦ = E Z 2 t ⎦ + E 2 X 1 ∫ ∫ ( ⎡ ⎤ e −δ v + 2 u ⎡ ⎤ E Z t ⎣ ⎣ , t − u 0 and similarly for the delayed renewal case t { } [ ] e − 2 δ u E Z o t − u ( ) Z d t + h ( ) ( ) ( ) ( ) ∫ ⎦ = E X 1 ⎦ + E Z o t + h − u ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ E Z d t dm d u ⎣ ⎣ ⎣ ⎦ 0 t ( ) ∫ + E X 1 ⎡ ⎤ ⎦ e − 2 δ u dm d u 2 ⎣ 0 which yields t + h − u t [ ] ( ) Z d t + h ( ) ( ) ) dm o v ( ) dm d u ( ) 2 t ⎦ + E 2 X 1 ∫ ∫ ( ⎡ ⎤ e −δ v + 2 u ⎡ ⎦ = E Z d ⎤ E Z d t ⎣ ⎣ . t − u 0 10
Example 1 : Consider a constant force of real interest δ > 0 and a counting Poisson process with parameter λ > 0 . Then formula (1) of theorem 1 yields [ ] ⎟ + λ 2 E 2 X 1 ⎛ 1 − e − 2 δ t ⎞ ( ) . ( ) 1 − e ( ) Z t + h ( ) ( ) ⎡ ⎤ 1 − e −δ t −δ t + h ⎡ ⎦ = λ E X 1 ⎤ 2 E Z t ⎣ ⎣ ⎦ ⎜ ⎝ 2 δ ⎠ δ 2 Thus ⎛ 1 − e − 2 δ t ⎞ ( ) Z t + h ( ) ⎡ ⎤ ⎦ = λ E X 1 ⎡ ⎤ 2 Cov Z t ⎣ ⎟ , ⎣ ⎦ ⎜ ⎝ 2 δ ⎠ ( ) which is independent of h (and then equal to ⎦ ) and is ⎡ ⎤ V Z t ⎣ almost constant for large t . 11
( ) is the correlation coefficient between ( ) Furthermore, if ρ t , h Z t ( ) , then and Z t + h ⎛ ⎞ 1 − e − 2 δ t ⎡ ⎤ λ E X 1 2 ⎣ ⎦ ⎜ ⎟ 2 δ ⎝ ⎠ ( ) = ρ t , h ( ) 1 2 1 2 ⎛ ⎞ ⎡ − 2 δ t + h ⎤ ⎡ ⎛ 1 − e − 2 δ t ⎞ ⎤ 1 − e λ E X 1 ⎡ ⎤ λ E X 1 ⎡ ⎤ 2 2 ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎜ ⎟ ⎣ ⎦ ⎜ ⎟ 2 δ 2 δ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ 1 2 ⎡ 1 − e − 2 δ t ⎤ = . ⎢ ⎥ ( ) − 2 δ t + h 1 − e ⎣ ⎦ 1 2 when ( ) → 1 − e − 2 δ t ( ) is almost 0 for a ⎡ ⎤ So ρ t , h h → ∞ , and ρ t , h ⎣ ⎦ small t and a large h … as normally expected. 12
For our discounted compound Poisson process, if the correlation is ‘‘strong enough’’ on the period t , t + h ⎦ then we can eventually ⎡ ⎤ ⎣ ( ) from a use a linear predictor to estimate the value of Z t + h ( ) . known value of Z t Hence assume that the equation of the linear predictor is given by ( ) 1 2 ) V Z t + h ⎧ ⎡ ⎤ ⎫ ⎣ ⎦ ( ) = E Z t + h ( ) { ( ) − E Z t ( ) } , ( ⎦ + ρ t , h ⎡ ⎤ ⎡ ⎤ L t , h Z t ⎨ ⎬ ⎣ ⎣ ⎦ ( ) ⎡ ⎤ V Z t ⎩ ⎣ ⎦ ⎭ then, for our example, we get ⎧ λ ⎫ ( ) = Z t ( ) + e − δ t ( ) ( ) + e − δ t E Z h ( ) ⎦ 1 − e − δ h = Z t ⎦ . ⎡ ⎤ ⎡ ⎤ L t , h δ E X 1 ⎨ ⎬ ⎣ ⎣ ⎩ ⎭ 13
Joint moments for a stochastic δ ( x ) If we now consider a stochastic force of real interest, the preceding method (that use renewal arguments) does not work anymore and, which more is, it is not possible to get recursive formulas for the joint moments. So we need a more general method that will help us to find explicit formulas for the joint moments of our risk process for a stochastic discount rate. This method will be based essentially on the following lemma that gives the conditional joint distribution of the claims arrival times knowing the number of claims, for any renewal process. This lemma generalizes the well-known similar formulas obtained for the Poisson process by using the order statistic property. 14
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