Solutions to the exercises of Modern Actuarial Risk Theory, 1st and 2nd edition R. Kaas, M.J. Goovaerts, J. Dhaene, M. Denuit, 06/01/2003. 1.2.1 Prove Jensen’s inequality: if v(x) is convex, then E[v(X)] ≥ v(E[X]). Use the following definition of convexity: a function v(x) is convex if, and only if, for every x0 a line l0(x) = a0x+b0 exists, such that l0(x0) = v(x0) and moreover l0(x) ≤ v(x) for all x [usually, l0( ) is a tangent line of v( )]. Pay special attention to the case v(x) = x2. Take x0 = E[X], then from v(x) ≥ l0(x) ∀x, we have E[v(X)] ≥ E[l0(X)] = E[a0X+b0] = a0E[X]+b0 = l0(E[X]) = v(E[X]). • The tangent line in x0 satisfies l0(x0) = v(x0) and l0′(x0) = v′(x0). If v(x) = x2, then l0(x) = x02+2x0(x−x0) must hold. We have v(x)−l0(x) = (x−x0)2 ≥ 0, so E[X2] = E[v(X)] ≥ E[l0(X)] = µ2. • But also: E[X2] = Var[X] + (E[X])2 ≥ (E[X])2. (1)
2.4.15 Show that the skewness of Z = X+2Y is zero if X ~ binomial(8,p) and Y ~ Bernoulli(1−p). For which p is Z symmetric? In Exer. 2.4.11 we proved that the 3rd cumulant of Z is the sum of those of X and of 2Y, hence 8p(1−p)(1−2p) + 23(1−p)p(1−2(1−p)). This equals 0 for all p. • Necessary for symmetry is Pr[Z = 0] = Pr[Z = 10]. We have Pr[Z = 0] = Pr[X = 0,Y = 0] = (1−p)8p and Pr[Z = 10] = p8(1−p). They are equal only if p ∈ {0,½,1}; for all these values, Z is symmetrical. 2.4.16 For which values of δ does X−δY have skewness 0 if X ~ gamma(2,1) and Y ~ exponential(1), X⊥Y? The 3rd central moment of X−δY equals 2α/β3 α=2,β=1 + (−δ)3×2α/β3 α=1,β=1 = 4 − 2δ3, 3 which is zero only if δ = 2 .
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3.5.4 In case of a compound Poisson distribution for which the claims have mass points 1,2,...,m, determine how many multiplications are needed to calculate the probability F(t) using Panjer’s recursion. Distinguish the cases m < t and m ≥ t. Write f(s) = −λs Σ1s h p(h) f(s−h). Compute λ h p(h) beforehand for all h, which takes 2min{m,t} multiplications. For m ≥ t, we need ½t(t+1) more, for m < t we need ½m(m+1)+m(t−m) more. So the number of operations increases linearly with t if the maximal possible claim is finite, quadratically otherwise. 3.5.5 Prove that E[N] = (a+b)/(1−a) if qn = Pr[N=n] satisfies (3.26). ∞ (3.26) ∞ b nq n = na qn 1 = E[N] = n n 1 n 1 ∞ a b a (n 1)qn 1 a b = aE[N]+a+b ⇒ E[N] = . 1 a n 1 (62)
3.9.10 In Sections 3.9 and 3.10, the retention is written as µ+kσ, so it is expressed in terms of a number of standard deviations above the expected loss. In practice, the retention is always expressed as a percentage of the expected loss. Consider two companies for which the risk of absence due to illness is to be covered by stop-loss insurance. This risk is compound Poisson distributed with parameter λi and exponentially distributed individual losses X with E[X]=1000. Company 1 is small: λ1=3; company 2 is large: λ2=300. What are the net stop-loss premiums for both companies in case the retention d equals 80%, 100% and 120% of the expected loss respectively? Express these amounts as a percentage of the expected loss and use the CLT. Write Si for the risks covered, then Si ~ cP(λi,X), so µi = 1000λi, σi2 = (1000 2)2λi, since E[X2] = Var[X] + E[X]2. 0.8µ = µ+zσ holds for z = −0.2µ/σ = −0.1 (2λi); 1.2µ = µ+zσ for z = 0.2µ/σ = +0.1 (2λi). → (92)
4.8.11 The ruin processes of company 1 and 2 are both compound Poisson with intensities λ1 = 1 and λ2 = 8, claims
~ exponential(3) and exponential(6), and
loading factors θ1 = 1 and θ2 = ½. The claims processes of both companies are independent. These companies decide to merge, without changing their premiums. Determine the intensity, claims db and loading factor of the ruin process for the merged company. Assume that both companies have an initial capital equal to 0, then so does the merged company. Compare the probabilities of the following events (continuous infinite ruin probabilities): "both companies never go bankrupt" with "the merged company never goes bankrupt". Argue that, regardless of the values of the initial capitals u1 and u2 for the separate companies, and consequently of u1+u2 for the merged company, the following holds: the event "both companies never go bankrupt" has a smaller probability than "the merged company never goes bankrupt".
c = c1+c2, λ = λ1+λ2 = 9, p(x) = 1⁄9p1(x) + 8⁄9p2(x). Finish this exercise by yourself.
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7.3.1 Derive the formula Cov[X,Y] = E[Cov[X,Y Z]] + Cov[E[X Z],E[Y Z]] for the decomposition of covariances into conditional covariances. Use E[W] = E[E[W Z]]: E[Cov[X, Y Z]] = E[E[XY Z]] − E[E[X Z] E[Y Z]] = E[XY] − E[X] E[Y] + E[X] E[Y] − E[E[X Z] E[Y Z]] = Cov[X, Y] − {E[E[X Z] E[Y Z]] − E[E[X Z]] E[E[Y Z]]} = Cov[X, Y] − Cov[E[X Z], E[Y Z]]. Note: X ≡ Y reduces to Var[X] = E[Var[X Z]] + Var[E[X Z]].
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8.6.2 Check the validity of the entries in Table E for all distributions listed. Verify the reparametrizations, the canonical link, the cumulant function, the mean as a function of θ and the variance function. Also determine the function c(y;φ). yθ b(θ) c(y,φ)} . In general: f Y(y; θ,φ) = exp{ φ •
Normal: fY(y;µ,σ²) =
1 σ 2π
2
2
e−½ (y−µ) /σ .
Equating corresponding terms in the log-densities leads to φ = σ2, θ = µ; b(θ) = ½θ2; this leaves c(y,φ) =
+ log(2πφ) 2 φ 1 y²
⇒ θ(µ) = µ, so µ(θ) = θ; [also, µ(θ) = b′(θ) checks out] ⇒ V(µ) = b″(θ(µ)) = 1
→ (183)
10.3.6 As the previous exercise, but now for the case that the probability mass of F on (a,b) has been concentrated on an appropriate d, i.e., such that G(x) is constant both on [a,d) and [d,b). Also consider that case that all mass on the closed interval [a,b] is concentrated. Now we have F ≥ G on (−∞,d), F ≤ G on (d,∞), so F is ‘thicker-tailed’ than G, and G ≤SL F. Concentration gives smaller stop-loss premiums. 10.3.7 Show that the cdf F with the following differential is indirectly more dangerous than the uniform(0,3) cdf: dx/3 for x ∈ (0,1) ∪ (2,3) 1 dF(x) = ⁄12 for x ∈ {1,2} 1 ⁄6 for x = 1½. If H is the uniform(0,3) cdf, look at G with G = F on (−∞,1½), G = H on [1½,∞). F is thicker-tailed than G, which in turn has only one sign change with H. But F and H have 3 changes of sign in their difference. (213)
10.6.4 Consider n married couples with one-year life insurances, all having probability of death 1% for her and 2% for him. The amounts insured are unity for both sexes. Assume that the mortality between different couples is independent. Determine the distribution of the individual model for the total claims, as well as for the collective model approximating this, a) assuming that the mortality risks are also independent within each couple, and b) that they are comonotonic. In case a) we have S ~ B1+B2 with Bj ~ bin(n,j×0.01) independent, so S has a cdf which is the convolution of two binomial db’s. • Just as in Example 10.6.7 for the comonotonic db one may derive that Pr[X1=X2=0] = 0.98, Pr[X1=0,X2=1] = 0.01 = Pr[X1=1,X2=1]. So in case b) we have S ~ ∑Ci with the C’s independent and Pr[C=0] = 0.98, Pr[C=1] = Pr[C=2] = 0.01, hence S ~ compound binomial(n, 0.02) with p(1) = p(2) = ½. • The collective model for case a) is N3 ~ Poisson(0.03n). → (243)
