STK 3505/4505: Summary of the course

Transcription

1 November 22, 2016

2 CH 2: Getting started the Monte Carlo Way How to use Monte Carlo methods for estimating quantities ψ related to the distribution of X, based on the simulations X1,..., X m: mean: X = 1 m m i=1 X i standard deviation: s 1 m ( = m 1 i=1 X i X ) 2 ɛ -percentile q ɛ given by P(X q ɛ ) = ɛ: qɛ = X(mɛ), where X(1)... X (m). Inversion method for sampling X with the cdf F (x): 0. Input: parameters of F (x) 1. Draw U U(0, 1) 2. Return X F 1 (U ) % or F 1 (1 U )

3 CH 2: Getting started the Monte Carlo Way The normal distribution and extensions: transformations, such as the log-normal Y = e X, X N(ξ, σ). stochastic volatility: X = ξ + ξ σ Zɛ, with ɛ N(0, 1) and Z > 0 independent of ɛ, for instance Z = 1 G, G Gamma(α). multivariate: bivariate, equicorrelation model. Positive distributions used in insurance: continuous: log-normal, Gamma, Exponential, Weibull, Pareto discrete: Poisson.

4 CH 3: Evaluating risk: A primer Models for general insurance: claim numbers N (policy) and N (portfolio of J policies) during a period of length T, typically N Poisson(µT ) and N Poisson(JµT ). claim sizes Z i > 0, typically iid and independent of N/N. aggregate compensations: X = Z Z N (policy) and X = Z Z N (portfolio). compensation functions: X = H(Z 1 ) H(Z N ), X = H(Z 1 ) H(Z N ), with 0 H(z) z, for instance 0, z < a H(z) = z a, a z < a + b. b, z a + b reinsurance: Zi re = H(Z i ) (contract at policy level) or X re = H(X ) (contract at portfolio level), where 0 H(z) z, for instance as above.

5 CH 3: Evaluating risk: A primer Models for life insurance: sequence of equidistant payments X 1, X 2,... k-step survival probability k p l : probability of surviving till age at least k + l given current age l k p l = p l+k 1 p l+k 2... p l, where p j = 1 p j. expected present value of the payments: V 0 = k=0 d k E(X k ), typically with d = 1/(1 + r). pension scheme with start age l 0, yearly premium π before retirement age l r, yearly payment s from age l r : V 0 = π l r l 0 1 k=0 dk kp l 0 + s k=l r l 0 dk kp l 0. equivalence premium: π given by V 0 = 0. One-time premium: π l0 given by V 0 = 0 when the premium is paid only once, at age l 0.

6 CH 3: Evaluating risk: A primer Protecting financial investments with options: pay-off X = H(R)ν 0 at expiry T, when ν 0 is the investment and R is the return of the asset during T. put-option: H(R) = max(r g R, 0), call option: H(R) = max(r r g, 0), both with guarantee r g. risk-neutral price π: π = e rt E Q (X ), where e rt is the discount over T and Q is the modified, risk-neutral model Black-Scholes: exact formula for π when R = e ξt +σ T ɛ 1, ɛ N(0, 1). Under Q R = e rt 1 2 σ2 T +σ T ɛ 1. π(ν 0) = (e rt (1 + r g )Φ(a) Φ(a σ T ))ν 0 a = log(1 + rg ) rt σ2 T σ. T by simulation: also for other models for R. 1. Simulate X1,..., Xm under Q m 2.π = e rt Xi. m i=1

7 CH 3: Evaluating risk: A primer Net assets in general insurance and the ruin problem: recursion for net assets Y k : Y k = (1 + R k )Y k 1 + Π k O k X k, where Y 0 = ν 0, R k is the return from the financial investments, Π k the total premium, O k the overhead and X k the total pay-off to the portfolio of policies. ruin problem: finding the start capital ν 0ɛ, such that P(min(Y 1,..., Y K ) < 0 Y 0 = ν 0ɛ ) = ɛ.

8 CH 5: Modelling I: Linear dependence Markowitz optimal portfolio selection: R 1,..., R J returns of J assets with means ξ 1,..., ξ J and covariance matrix Σ r risk-free opportunity R = w 0 r + w 1 R R J portfolio return with w 0 + w w J = 1 objective: find w 0, w 1,..., w J so that Var(R) = J J i=1 j=1 w iw j σ ij is minimised when E(R) = r + J j=1 w j(ξ j r) = e g solution: w j = γ w j, j = 1,..., J, w 0 = 1 J j=1 w j, with J j=1 w e jσ ij = ξ i r, i = 1,..., J and γ = g r J j=1 w j (ξ i r).

9 CH 5: Modelling I: Linear dependence Multivariate normal distribution with extensions: simulation: 0. Input: ξ, C 1. Draw η1,..., ηj iid N(0, 1) 2. Return X ξ + Cη, where η = (η 1,..., η J )T and Σ = CC T. transformations: multivariate log-normal model stochastic volatility.

10 CH 5: Modelling I: Linear dependence Random walk: iid standard: Y k = Y k 1 + X k, X k = ξ + σɛ k, ɛ k N(0, 1), k = 1, 2,... geometric: S k = e Y k = (1 + R k )S k 1, log(1 + R k ) = X k, k = 1, 2,... after k periods: Y k = Y 0 + kξ + kση k, S k = S 0 e kξ+ kση k, η k N(0, 1) multivariate simulation. Mean-reversion models: iid AR(1)-model: Y k = ξ + X k, X k = ax k 1 + σɛ k, ɛ k N(0, 1), X 0 = x 0, k = 1, 2,..., with a < 1 for stationarity after k periods: E(X k x 0 ) = a k x 0 1 a sd(x k x 0 ) = 2k 1 a σ 2 k σ 1 a 2 0 k Cov(X k, X k l x 0 ) = a l 1 a2k 1 a σ 2 a l 2 k transformations simulation. σ2 1 a 2

11 CH 6: Modelling II: Conditional and non-linear Conditional models: preservation of normality under conditioning survival probabilities: k p l = P(Y > l + k Y > l), where Y is the age at which a person dies extreme claim sizes Rules of double expectation and variance: E(Y ) = E(E(Y X )) Var(Y ) = Var(E(Y X )) + E(Var(Y X )) Example: N Poisson(JµT ), Z 1,..., Z N iid with E(Z i ) = ξ z, sd(z i ) = σ z E(X N ) = N ξ z, Var(X N ) = N σ 2 z E(X ) = E(E(X N )) = JµT ξ z Var(X ) = Var(E(X N )) + E(Var(X N )) = JµT (σ 2 z + ξ 2 z ).

12 CH 6: Modelling II: Conditional and non-linear Common factor models: X 1,..., X J all depend on a common, stochastic factor ω X 1,..., X J iid, conditionally on ω Risk cannot be diversified away Examples: claim counts with a common, stochastic intensity µ; CAPM Simulation of multivariate models based on conditioning: Draw X 1 from f (x 1) Draw X 2 from f (x 2 X 1 ) Draw X 3 from f (x 3 X 1, X 2 ).. Draw X n from f (x n X 1,..., X n 1 ).