an empirical based approach for optimal reinsurance
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An Empirical-based Approach for Optimal Reinsurance C HENGGUO W ENG Department of Statistics & Actuarial Science University of Waterloo This is a joint work with Ken Seng Tan August 31, 2009 C. Weng (c2weng@uwaterloo.ca) p.1/23


  1. An Empirical-based Approach for Optimal Reinsurance C HENGGUO W ENG Department of Statistics & Actuarial Science University of Waterloo This is a joint work with Ken Seng Tan August 31, 2009 C. Weng (c2weng@uwaterloo.ca) – p.1/23

  2. ✞ ☎ Outline of Today’s Presentation ✝ ✆ Background Motivation Empirical-based Approach Conclusion C. Weng (c2weng@uwaterloo.ca) – p.2/23

  3. ✞ ☎ Effect of Reinsurance ✝ ✆ Premium π 0 Policyholders Insurance Company (Insureds) (Insurer or Cedent) Loss X C. Weng (c2weng@uwaterloo.ca) – p.3/23

  4. ✞ ☎ Effect of Reinsurance ✝ ✆ Premium π 0 Policyholders Insurance Company (Insureds) (Insurer or Cedent) Loss X f ( X ) : Ceded Claims e.g. stop loss: f ( X ) = ( X − d ) + , d ≥ 0 e.g. quota-share: Reinsurance Company (Reinsurer) f ( X ) = cX, 0 ≤ c ≤ 1 C. Weng (c2weng@uwaterloo.ca) – p.4/23

  5. ✞ ☎ Effect of Reinsurance ✝ ✆ Premium π 0 Policyholders Insurance Company (Insureds) (Insurer or Cedent) Loss X f ( X ) : Ceded Claims Π( f ) : Reinsurance Premium Π( f ) e.g. stop loss: e.g. Expectation premium principle: f ( X ) = ( X − d ) + , d ≥ 0 Π( f ) = (1 + θ )E[ f ( X )] e.g. quota-share: Reinsurance Company (Reinsurer) f ( X ) = cX, 0 ≤ c ≤ 1 C. Weng (c2weng@uwaterloo.ca) – p.5/23

  6. ✞ ☎ Effect of Reinsurance ✝ ✆ Premium π 0 Policyholders Insurance Company (Insureds) (Insurer or Cedent) Loss X f ( X ) : Ceded Claims Π( f ) : Reinsurance Premium Π( f ) e.g. stop loss: e.g. Expectation premium principle: f ( X ) = ( X − d ) + , d ≥ 0 Π( f ) = (1 + θ )E[ f ( X )] e.g. quota-share: Reinsurance Company (Reinsurer) f ( X ) = cX, 0 ≤ c ≤ 1 • Insurer’s retained risk: R f ( X ) = X − f ( X ) • Insurer’s total risk: T f ( X ) = R f ( X ) + Π( f ) = X − f ( X ) + Π( f ) • tradeoff between the amount of loss retained and the reinsurance premium payable to a reinsurer C. Weng (c2weng@uwaterloo.ca) – p.6/23

  7. ✞ ☎ Risk Measure Minimization Reinsurance Models ✝ ✆ A plausible optimal reinsurance model: � � min f ρ ( T f ( X )) = ρ X − f ( X ) + Π( f ( X )) s.t. Π( f ( X )) ≤ π 0 ≤ f ( x ) ≤ x for all x ≥ 0 where ρ is a chosen risk measure, such as variance, VaR and CTE. C. Weng (c2weng@uwaterloo.ca) – p.7/23

  8. ✞ ☎ Risk Measure Minimization Reinsurance Models ✝ ✆ A plausible optimal reinsurance model: � � min f ρ ( T f ( X )) = ρ X − f ( X ) + Π( f ( X )) s.t. Π( f ( X )) ≤ π 0 ≤ f ( x ) ≤ x for all x ≥ 0 where ρ is a chosen risk measure, such as variance, VaR and CTE. Complexity of solving the above risk measure minimization models: if the form of f is specified: technically tractable. stop-loss f ( x ) = ( x − d ) + quota-share f ( x ) = cx if a general f is considered: infinite dimensional problem, very challenging to obtain the explicit solutions. relies on the premium principle and the risk measure ρ . C. Weng (c2weng@uwaterloo.ca) – p.8/23

  9. ✞ ☎ Empirical Approach: Motivation ✝ ✆ The general reinsurance model can be formulated as:   min f ρ ( X, f )   s.t. 0 ≤ f ( x ) ≤ x for all x ≥ 0 ,    Π( f ( X )) ≤ π. Often difficult to solve (due to infinite dimension and nonlinearity) In practice, the distribution of the underlying risk X is estimated from the observed data { x 1 , · · · , x N } . Empirical-based reinsurance model : exploits the observed data directly C. Weng (c2weng@uwaterloo.ca) – p.9/23

  10. ✞ ☎ Empirical-based Model: Formulation ✝ ✆ Collect samples x := ( x 1 , x 2 , · · · , x N ) corresponding to the underlying risk X Introduce decision vector f := ( f 1 , f 2 , · · · , f N ) where each f i represents the reinsurance indemnification for the loss amount x i , i = 1 , 2 , · · · , N . Define empirical-based estimates as: � ρ ( X, f ) → ρ ( x , f ) 0 ≤ f ( x ) ≤ x, for all x ≥ 0 → 0 ≤ f i ≤ x i , i = 1 , 2 , · · · , N, � Π( f ) ≤ π → Π( f ) ≤ π C. Weng (c2weng@uwaterloo.ca) – p.10/23

  11. ✞ ☎ Empirical-based Model: Formulation (Cont’d) ✝ ✆ General reinsurance model   min f ρ ( x , f )   s.t. 0 ≤ f ( x ) ≤ x, for all x ≥ 0 ,    Π( f ) ≤ π. Empirical reinsurance model:  �  min f ρ ( x , f )   s.t. 0 ≤ f i ≤ x i , i = 1 , 2 , · · · , N,    � Π( f ) ≤ π. Many empirical reinsurance model can be cast as Second-Order Conic (SOC) programming: A wide class of optimization problems Efficient softwares are available for solving SOC programming: e.g., CVX (Grant and Boyd, 2008) C. Weng (c2weng@uwaterloo.ca) – p.11/23

  12. ✞ ☎ Empirical Model: Variance Minimization ✝ ✆ Consider the following variance minimization model:   min f Var ( T f ) = Var ( X − f ( X ) + Π( f ))  s.t. 0 ≤ f ( x ) ≤ x, Π( f ) ≤ π. Empirical version of the goal function: N � �� �� 2 , 1 � x − ¯ Var ( T f ) = x i − f i ) − (¯ f N − 1 i =1 x denotes the mean of x , and ¯ where ¯ f denotes the mean of f . Empirical version of the constraints: 0 ≤ f i ≤ x i , i = 1 , 2 , · · · , N, and � Π( f ) ≤ π. Empirical variance minimization model:  �� �� 2 � N x − ¯  min f ∈ R N x i − f i ) − (¯ f   i =1 s.t. 0 ≤ f i ≤ x i , i = 1 , 2 , · · · , N.    � Π( f ) ≤ π. C. Weng (c2weng@uwaterloo.ca) – p.12/23

  13. ✞ ☎ Reinsurance Premium Budget Constraint ✝ ✆ P1. Expectation principle: Π( f ) = (1 + θ ) E [ f ] with θ > 0 . � ⇒ (1 + θ ) ¯ Π( f ) ≤ π ⇐ f ≤ π, � P2. Standard deviation principle: Π( f ) = E [ f ] + β Var [ f ] , where β > 0 . � N � 1 / 2 � � � 2 β � ⇒ ¯ f i − ¯ √ Π( f ) ≤ π ⇐ f + ≤ π. f N − 1 i =1 � The empirical variance minimization model can be cast as a SOC programming problem for as many as ten principles. � We analyzed the empirical solutions by some numerical examples (under expectation and std premium principles). C. Weng (c2weng@uwaterloo.ca) – p.13/23

  14. ✞ ☎ Empirical Model: CTE minimization ✝ ✆ Theoretical CTE minimization model:  � �  min f CTE α ( T f ) ≡ CTE α X − f ( X ) + Π[ f ( X )]  s.t. 0 ≤ f ( x ) ≤ x, Π[ f ( X )] ≤ π, Technical model:  �� � + � G α ( ξ, f ) ≡ ξ + 1   min ( ξ,f ) X − f ( X ) + Π( f ( X )) − ξ . E α   s.t. 0 ≤ f ( x ) ≤ x, Π( f ( X )) ≤ π. We developed the following fact: ( ξ ∗ , f ∗ ) ∈ arg min ( ξ,f ) G α ( ξ, f ) if and only if f ∗ ∈ arg min f CTE α ( T f ) , ξ ∗ ∈ arg min ξ G α ( ξ, f ∗ ) . We proved that stop-loss is the optimal solution given that Π is the expectation principle. C. Weng (c2weng@uwaterloo.ca) – p.14/23

  15. ✞ ☎ Empirical Model: CTE minimization (Cont’d) ✝ ✆ Empirical model:  �� � + � N  �  1  � x i − f i + � min ( ξ, f ) G α ( ξ, f ) ≡ ξ + Π( f ) − ξ , αN i =1    � s.t. 0 ≤ f i ≤ x i , i = 1 , 2 , · · · , N, Π( f ) ≤ π. Empirical CTE minimization model:  N �  1   min ( ξ, f , z ) ξ + z i ,   αN    i =1 � 0 ≤ f i ≤ x i , Π( f ) ≤ π, s.t.    z i ≥ 0 , z i ≥ �  Π( f ) − f i − ξ + x i ,     i = 1 , 2 , · · · , N. The above empirical model can be cast as Second-order conic programming for as many as ten reinsurance premium principles. C. Weng (c2weng@uwaterloo.ca) – p.15/23

  16. ✞ ☎ Example: CTE minimization ✝ ✆ Use samples from an exponential loss distribution with mean 1000 Reinsurance premium principle: expectation principle with loading factor θ = 0 . 2 standard deviation principle with loading factor β = 0 . 2 Consider different levels of the reinsurance premium budget The solutions f ∗ are illustrated by their scatter plots against sample x C. Weng (c2weng@uwaterloo.ca) – p.16/23

  17. ✞ ☎ Soln’s: CTE min & Expectation Principle (1/2) ✝ ✆ 6000 6000 5000 5000 4000 4000 3000 3000 2000 2000 1000 1000 0 0 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 1) π = 80 2) π = 200 6000 6000 5000 5000 4000 4000 3000 3000 2000 2000 1000 1000 0 0 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 3) π = 400 4) π = 600 C. Weng (c2weng@uwaterloo.ca) – p.17/23

  18. ✞ ☎ Soln’s: CTE min & Expectation Principle (2/2) ✝ ✆ 6000 6000 5000 5000 4000 4000 3000 3000 2000 2000 1000 1000 0 0 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 5) π = 800 6) π = 1000 6000 6000 5000 5000 4000 4000 3000 3000 2000 2000 1000 1000 0 0 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 7) π = 1500 8) π = 2000 C. Weng (c2weng@uwaterloo.ca) – p.18/23

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