Non-Life Insurance: Mathematics and Statistics

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1 ETH Zürich, D-MATH HS 08 Prof. Dr. Mario V. Wüthrich Coordinator Andrea Gabrielli Non-Life Insurance: Mathematics and Statistics Solution sheet 9 Solution 9. Utility Indifference Price (a) Suppose that there exist two utility indifference prices π = π (u, S, c 0 ) and π = π (u, S, c 0 ) with π π. By definition of a utility indifference price, we have Eu(c 0 + π S)] = u(c 0 ) = Eu(c 0 + π S)]. () Without loss of generality, we assume that π < π. Then, we have c 0 + π S < c 0 + π S a.s., which implies u(c 0 + π S) < u(c 0 + π S) since u is a utility function and, thus, strictly increasing by definition. Finally, by taking the expectation, we get Eu(c 0 + π S)] < Eu(c 0 + π S)], which is a contradiction to (). We conclude that if the utility indifference price π exists, then it is unique. Moreover, being a risk-averse utility function, u is strictly concave by definition. Hence, we can apply Jensen s inequality to get a.s., u(c 0 ) = Eu(c 0 + π S)] < u(ec 0 + π S]) = u(c 0 + π ES]). Note that we used that S is non-deterministic and, thus, Jensen s inequality is strict. Since u is strictly increasing, this implies π ES] > 0, i.e. π > ES]. (b) Note that E Y () ] = γ c = = 000 and that ] E Y () = = 00. Since S and S both have a compound Poisson distribution, Proposition. of the lecture notes gives and We conclude that ES ] = λ v E ES ] = λ v E Y () Y () ] = = ] = = ES] = ES + S ] = ES ] + ES ] = (c) The utility indifference price π = π(u, S, c 0 ) is defined through the equation u(c 0 ) = Eu(c 0 + π S)]. Updated: November 0, 08 / 8

2 HS 08 Solution sheet 9 Using that in this exercise the utility function u is given by for all x R, with α =.5 0 6, we get u(x) = exp { αx}, α u(c 0 ) = Eu(c 0 + π S)] α exp { αc 0} = E ] α exp { α(c 0 + π S)} exp { αc 0 } = E exp { α(c 0 + π S)}] exp {απ} = E exp {αs}] π = log E exp {αs}]. α Note that we can write S = S + S and use the independence of S and S to get π = α log E exp {α(s + S )}] = α log (E exp {αs }] E exp {αs }]) = α (log E exp {αs }] + log E exp {αs }]) = α log M S (α) + log M S (α)], where M S and M S denote the moment generating functions of S and S, respectively. Moreover, since S and S both have a compound Poisson distribution, Proposition. of the lecture notes gives where M Y () π = α and M Y () (λ v M Y () ] (α) + λ v M Y () ]) (α), denote the moment generating functions of Y () and Y (), respectively. Using that Y () Γ(γ = 0, c = 0.0) and that Y () expo(0.005), we get and ( ) γ ( c 0.0 M () Y (α) = = c α M () Y (α) = α = In particular, since α < c and α < 0.005, both M () Y (α) and M () Y (α) and thus also M S (α) and M S (α) exist. Inserting all the numerical values, we find the utility indifference price π = 3 06 = Note that we have ( 000 π ES] ES] ) 0 ( ) ] = = % Thus, the loading π ES] is given by approximately 0.46% of the pure risk premium. ] ) Updated: November 0, 08 / 8

3 HS 08 Solution sheet 9 (d) The moment generating function M X of X N (µ, ) with µ R and > 0 is given by M X (r) = exp {rµ + r }, for all r R. Hence, if we assume Gaussian distributions for S and S, then we get π = α log M S (α) + log M S (α)] = ) (αes ] + α α Var(S ) + αes ] + α Var(S ) = ES ] + ES ] + α Var(S ) + Var(S )] = ES] + α Var(S), where in the last equation we used that S and S are independent. We see that in this case the utility indifference price is given according to a variance loading principle. Since here we assume Gaussian distributions for S and S with the same corresponding first two moments as in the compound Poisson case in part (c), in order to calculate Var(S ) and Var(S ), we again assume that S and S have compound Poisson distributions. Note that ( ) ] E Y () γ(γ + ) 0 = c = 0.0 = , and that E ( ) ] Y () = = Then, Proposition. of the lecture notes gives ( ) ] Var(S ) = λ v E Y () = = and Var(S ) = λ v E ( ) ] Y () = = , 0 which leads to Var(S) = Var(S + S ) = Var(S ) + Var(S ) = We conclude that the utility indifference price is given by Note that we have π = ES] + α Var(S) = = π ES] ES] = = %. Thus, as in part (c), the loading π ES] is given by approximately 0.46% of the pure risk premium. The reason why we get the same results in (c) and (d) is the Central Limit Theorem. In particular, neither the gamma distribution nor the exponential distribution are heavy-tailed distributions. Moreover, the skewness ς S of S and also the skewness ς S of S are rather small (ς S and ς S 0.067). Thus, λ v = λ v = 000 are large enough for the normal approximations to be valid approximations for the compound Poisson distributions. Updated: November 0, 08 3 / 8

4 HS 08 Solution sheet 9 Solution 9. Value-at-Risk and Expected Shortfall (a) Since S LN(µ, ) with µ = 0 and = 0.05, we have } ES] = exp {µ Let z denote the VaR of S ES] at security level q = 99.5%. Then, since the distribution function of a lognormal distribution is continuous and strictly increasing, z is defined via the equation PS ES] z] = q. By writing Φ for the distribution function of a standard Gaussian distribution, we can calculate z as follows PS ES] z] = q PS z + ES]] = q ] log S µ log(z + ES]) µ P = q ] log(z + ES]) µ Φ = q log(z + ES]) = µ + Φ ( q) For q = 99.5%, we have Φ ( q).576. Thus, we get In particular, π CoC is then given by Note that we have z = exp { µ + Φ ( q) } ES] ( z = exp{µ} exp { Φ ( q) } { }) exp. z π CoC = ES] + r CoC z π CoC ES] ES] = %. Thus, the loading π CoC ES] is given by approximately.64% of the pure risk premium. (b) For all u (0, ), let VaR u and ES u denote the VaR risk measure and the expected shortfall risk measure, respectively, at security level u. Note that actually in part (a) we found that and that by a similar computation we get for all u (0, ). In particular, we have VaR u (S ES]) = exp { µ + Φ (u) } ES], VaR u (S) = exp { µ + Φ (u) }, VaR u (S ES]) + ES] = VaR u (S), Updated: November 0, 08 4 / 8

5 HS 08 Solution sheet 9 for all u (0, ). Since the distribution function of S is continuous and strictly increasing, according to Example 6.6 of the lecture notes, we have ES q (S ES]) = E S ES] S ES] VaR q (S ES])] = E S ES] S VaR q (S)] = E S S VaR q (S)] ES] = ES q (S) ES]. By definition of the mean excess function e S ( ) of S, we have ES q (S) = E S VaR q (S) S VaR q (S)]+VaR q (S) = e S VaR q (S)]+VaR q (S). Moreover, according to the formula given in Chapter 3..3 of the lecture notes, the mean excess function e S VaR q (S)] above level VaR q (S) for the log-normal distribution S is given by ] e S VaR q (S)] = ES] Φ log VaR q(s) µ ] VaR q (S). log VaR q(s) µ Φ Using the formula calculated above for VaR u (S) with u = q, we get ] ES q (S) = ES] Φ log VaR q(s) µ ] log VaR q(s) µ Φ ] = ES] Φ µ+ Φ ( q) µ ] Φ µ+ Φ ( q) µ ( Φ Φ ( q) ] ) = ES] Φ Φ ( q)] = ES] q ( Φ Φ ( q) ]). In particular, we have found For q = 99% we get ES q (S ES]) = q ES] ( Φ Φ ( q) ]) ES] Finally, π CoC is then given by = q ES] ( q Φ Φ ( q) ]) = { } q exp µ + ( q Φ Φ ( q) ]). ES 99% (S ES]) π CoC = ES] + r CoC ES 99% (S ES]) Note that we have π CoC ES] ES] = %. Thus, the loading π CoC ES] is given by approximately.6% of the pure risk premium. In particular, the cost-of-capital price in this example is higher using the expected shortfall risk measure at security level 99% than using the VaR risk measure at security level 99.5%. Updated: November 0, 08 5 / 8

6 HS 08 Solution sheet 9 (c) In parts (a) and (b) we found that VaR 99.5% (S ES]) < ES 99% (S ES]). Let q = 99%. Now the goal is to find u 0, ] such that which is equivalent to From part (b) we know that for all u (0, ), and that VaR u (S ES]) = ES q (S ES]), VaR u (S) = ES q (S). VaR u (S) = exp { µ + Φ (u) }, ES q (S) = q ES] ( Φ Φ ( q) ]). Hence, we can solve for u to get u = Φ log q ES] ( Φ Φ ( q) ])] µ 99.6%. We conclude that in this example the cost-of-capital price using the VaR risk measure at security level 99.6% is approximately equal to the cost-of-capital price using the expected shortfall risk measure at security level 99%. (d) Since S LN(µ, ) with µ = 0 and = 0.05 and U and V are assumed to be independent, we have U N (µ, ), V N (µ, ) and U + V N (µ, ). Let X N ( µ, ) for some µ R and > 0. Then, VaR q (X) can be calculated as X µ P X VaR q (X)] = q P VaR ] q(x) µ = q ] VaR q (X) µ Φ = q This implies that VaR q (X) = µ + Φ ( q). VaR q (U) + VaR q (V ) = µ + Φ ( q) + µ + Φ ( q) = µ + Φ ( q) and that Since VaR q (U + V ) = µ + Φ ( q). VaR q (U + V ) > VaR q (U) + VaR q (V ) Φ ( q) > Φ ( q) one can see that in this example for all q ( 0, ), and that Φ ( q) < 0, VaR q (U + V ) > VaR q (U) + VaR q (V ), VaR q (U + V ) < VaR q (U) + VaR q (V ), for all q (, ). In particular, the assertions on the exercise sheet follow readily. Updated: November 0, 08 6 / 8

7 HS 08 Solution sheet 9 Solution 9.3 Esscher Premium (a) Let α (0, r 0 ) and M S and M S denote the first and second derivative of M S, respectively. According to the proof of Corollary 6.6 of the lecture notes, the Esscher premium π α can be written as π α = M S (α) M S (α). Hence, the derivative of π α can be calculated as d dα π α = d M S (α) dα M S (α) = M S (α) ( ) M M S (α) S (α) = E S exp{αs} ] ( E S exp{αs}] M S (α) M S (α) M S (α) = x ] exp{αx} df (x) x exp{αx} df (x) M S (α) M S (α) = x df α (x) x df α (x)], where we define the distribution function F α by F α (s) = M S (α) s exp{αx} df (x), for all s R. Let X be a random variable with distribution function F α. Then, we get d dα π α = x df α (x) x df α (x)] = E X ] EX] = Var(X) 0. Hence, the Esscher premium π α is always non-decreasing. Moreover, if S is non-deterministic, then also X is non-deterministic. Thus, in this case we get d dα π α = Var(X) > 0. In particular, the Esscher premium π α then is strictly increasing in α. (b) Let α (0, r 0 ). According to Corollary 6.6 of the lecture notes, the Esscher premium π α is given by π α = d dr log M S(r). r=α For small values of α, we can use a first-order Taylor approximation around 0 to get π α d ( dr log M S(r) d + α r=0 dr log M S(r) = M S (0) r=0 M S (0) + α M S (0) ] ) M M S (0) S (0) M S (0) = ES] + α ( E S ] ES] ) = ES] + αvar(s). We conclude that for small values of α, the Esscher premium π α of S is approximately equal to a premium resulting from a variance loading principle. (c) Since S CompPoi(λv, G), we can use Proposition. of the lecture notes to get log M S (r) = λv M G (r) ], where M G denotes the moment generating function of a random variable with distribution function G. Since G is the distribution function of a gamma distribution with shape parameter γ > 0 and scale parameter c > 0, we have ( ) γ c M G (r) =, c r Updated: November 0, 08 7 / 8 )

8 HS 08 Solution sheet 9 for all r < c. In particular, also M S (r) is defined for all r < c, which implies that the Esscher premium π α exists for all α (0, c). Now let α (0, c). Then, the Esscher premium π α can be calculated as π α = d dr log M S(r) = d ( ) γ ] c r=α dr λv = d ( c r dr λv r ) γ ] c r=α = λv γ ( r ) γ = λv γ ( ) γ+ c. c c c c α r=α Note that since c > c α and γ > 0, we have ( ) γ+ c >, c α r=α and, thus, π α = λv γ c ( ) γ+ c > λv γ c α c = ES]. Solution 9.4 Utility Indifference Price and the Uncertainty in the Number of Claims In Exercise 9. we have shown that π = π (u, S, ) c 0 = { α log E exp α S }]. From this we can calculate { }] π = λv exp α log E α Y i = λv log E exp {αy i }] = α α λv log M Y (α). We then have We define i= i= π > π M Y (α) > log M Y (α). g(x) = x and h(x) = log x. For x = we have g() = 0 = h(). Since g is linear and h is strictly concave, we get g(x) > h(x) for all x in the domain of g and h. Since for the claim sizes we assume Y > 0, P-a.s., and since α > 0, we have M Y (α) >. If follows that M Y (α) > log M Y (α) and, thus, π > π. This does not come as a surprise: in the compound Poisson model we have randomness in the number of claims and under risk aversion we do not like this uncertainty. This explains why π > π. Updated: November 0, 08 8 / 8