Actuarial models. Proof. We know that. which is. Furthermore, S X (x) = Edward Furman Risk theory / 72

Proof. We know that which is S X Λ (x λ) = exp S X Λ (x λ) = exp Furthermore, S X (x) = { x } h X Λ (t λ)dt, 0 { x } λ a(t)dt = exp { λa(x)}. 0 Edward Furman Risk theory 4280 21 / 72

Proof. We know that which is S X Λ (x λ) = exp S X Λ (x λ) = exp Furthermore, as required. { x } h X Λ (t λ)dt, 0 { x } λ a(t)dt = exp { λa(x)}. 0 S X (x) = E[S X Λ (x Λ)] = E[e ΛA(x) ] = M Λ ( A(x)), Edward Furman Risk theory 4280 21 / 72

Sometimes the conditional hazard function can be very simple. Edward Furman Risk theory 4280 22 / 72

Sometimes the conditional hazard function can be very simple. Take X Λ Exp(Λ). Then a(x) = 1. Example 1.9 Let Λ Ga(τ, α) and X Λ Wei(λ, γ) with ddf S(x) = e λx γ. Use the frailty mixture model to find the ddf of X. Solution We showed that S X (x) = M Λ ( A(x)). In this case Λ is a gamma rv, thus M Λ (t) = (1 αt) τ. Also, for Weibull with the the ddf above, a(x) = γx γ 1. Edward Furman Risk theory 4280 22 / 72

Solution Thus A(x) = x γ, and S X (x) = M Λ ( A(x)) = (1 + αx γ ) τ Definition 1.8 (Spliced distributions) An n component spliced distribution has the following pdf α 1 f 1 (x), if c 0 < x < c 1 α 2 f 2 (x), if c 1 < x < c 2 f (x) =. α n f n (x), if c n 1 < x < c n where for j = 1,..., n we have that α j > 0, α 1 + + α n = 1 and every f j is a legitimate pdf with the support on (c j 1, c j ). Edward Furman Risk theory 4280 23 / 72

Example 1.10 Show that the pdf below is a spliced pdf. { 0.01, if 0 x < 50 f (x) = 0.02, if 50 x < 75 Solution Indeed f (x) = { 0.5 0.02, if 0 x < 50 0.5 0.04, if 50 x < 75 where both f 1 and f 2 are uniform densities on different intervals. The splicing approach is in a way similar to the mixing one. Here however the processes generating losses differ with respect to loss amounts. This is not true for the mixing approach. Edward Furman Risk theory 4280 24 / 72

In addition to the methods described above, we can obtain new distributions as limiting cases of other ones. Edward Furman Risk theory 4280 25 / 72

In addition to the methods described above, we can obtain new distributions as limiting cases of other ones. Let X Ga(γ, 1). Then Y = θx 1/τ, τ > 0 is a generalized (actuaries call it) transformed gamma distribution. The density is { ( y ) τ } y τγ 1 τ f (y) = exp θ θ τγ Γ(γ), y > 0. For a > 0, b > 0, let the beta function be B(a, b) = 1 0 t a 1 (1 t) b 1 dt = Γ(a)Γ(b) Γ(a + b). Then let X Be(τ, α) be a beta random variable with the pdf f X (x) = 1 Be(τ, α) x τ 1 (1 x) α 1, 0 x 1. Edward Furman Risk theory 4280 25 / 72

Let Y = θ(x/(1 X)) 1/γ. Then, we have that the pdf of Y is the generalized beta of the second kind (actuaries call it the transformed beta) f Y (y) = 1 γ(x/θ) γτ Be(τ, α) x(1 + (x/θ) γ, y 0. ) α+τ Example 1.11 The generalized (transformed) gamma distribution is limiting case of the generalized (transformed) beta distribution when θ,α,θ/α 1/γ ξ. Solution The well-known Stirling s approximation says that for large α s, Γ(α) = e α α α 0.5 2π. Edward Furman Risk theory 4280 26 / 72

Solution Thus for large α s, θ = ξα 1/γ and using the pdf of the generalized beta rv, we have that f Y (y) = = = = Γ(α + τ) γ(x/θ) γτ Γ(α)Γ(τ) x(1 + (x/θ) γ ) α+τ e τ α (τ + α) τ+α 0.5 2πγy γτ 1 e α α α 0.5 2πΓ(τ)θ γτ (1 + y γ θ γ ) α+τ e τ (τ + α) τ+α 0.5 γy γτ 1 α α 0.5 Γ(τ)θ γτ (1 + y γ θ γ ) α+τ e τ (τ + α) τ+α 0.5 γy γτ 1 α τ+α 0.5 Γ(τ)ξ γτ (1 + (y/ξ) γ /α) α+τ = e τ (τ/α + 1) τ+α 0.5 γy γτ 1 Γ(τ)ξ γτ (1 + (y/ξ) γ /α) α+τ Edward Furman Risk theory 4280 27 / 72

Solution We also know that lim α ( 1 + τ ) α+τ 0.5 α Edward Furman Risk theory 4280 28 / 72

Solution We also know that and similarly lim α lim α ( 1 + τ ) α+τ 0.5 = e τ α (1 + (x/ξ)γ α ) α+τ Edward Furman Risk theory 4280 28 / 72

Solution We also know that and similarly lim α ( 1 + τ ) α+τ 0.5 = e τ α Hence lim α ) α+τ (1 + (x/ξ)γ = e (x/ξ)γ α lim f Y (y) = γy γτ 1 e (y/ξ)γ α ξ γτ, Γ(τ) which is the pdf of a generalized (transformed) gamma rv as needed. Edward Furman Risk theory 4280 28 / 72

Figure: Beta family of distributions Edward Furman Risk theory 4280 29 / 72

Definition 1.9 (Linear exponential family) A random variable X has a distribution belonging to the linear exponential family if its pdf can be reformulated in terms of a parameter θ as f (x; θ) = p(x)er(θ)x. q(θ) Here p(x) does not depend on θ, the function q(θ) is a normalizing constant, and r(θ) is the canonical parameter. The support of X must be independent of θ. Example 1.12 Gamma is linear exponential family, i.e., f (x; θ) = e x/θ x γ 1 θ γ Γ(γ) is in the linear exponential family with r(θ) = 1/θ, q(θ) = θ γ and p(x) = x γ 1 /Γ(γ). Edward Furman Risk theory 4280 30 / 72

Example 1.13 Normal random variable X N(θ, ν) is in the linear exponential family. Indeed { f (x; θ) = (2πν) 1/2 exp 1 } (x θ)2 2ν = (2πν) 1/2 exp { x 2 2ν + θ } ν x θ2 2ν { } (2πν) 1/2 exp x 2 2ν exp { θ ν x} = }, exp { θ 2 2ν that is linear exponential { family } with r(θ) = θ/ν, } p(x) = (2πν) 1/2 exp, q(θ) = exp x 2 2ν { θ 2 2ν Edward Furman Risk theory 4280 31 / 72

Proposition 1.7 Let X LEF (θ). Then its expected value is Proof. First note that Also, E[X] = q (θ) r (θ)q(θ). log f (x; θ) = log p(x) + r(θ)x log q(θ). θ f (x; θ) = 1 ( ) q(θ) 2 p(x)e r(θ) xr (θ)q(θ) p(x)e r(θ)x q (θ) = f (x; θ) ( r (θ)x q (θ)/q(θ) ) Edward Furman Risk theory 4280 32 / 72

Proof. We will now differentiate the above with respect to x. Then θ f (x; θ)dx = r (θ) xf (x; θ)dx q (θ) f (x; θ)dx q(θ) Note that the support of X is independent of θ by definition of the family and so is the range of x. Therefore θ f (x; θ)dx = r (θ) xf (x; θ)dx q (θ) q(θ) f (x; θ)dx that is equivalent to saying that θ (1) = r (θ)e[x] q (θ) q(θ). Or in other words Edward Furman Risk theory 4280 33 / 72

Proof. E[X] = which completes the proof. Proposition 1.8 q (θ) r (θ)q(θ) = µ(θ), Let X LEF(θ). Then its variance is Proof. Var[X] = µ (θ) r (θ). We will use the same technique as before. We have already shown that θ f (x; θ) = r (θ) (x µ(θ)) f (x; θ) Edward Furman Risk theory 4280 34 / 72

Proof. Differentiating again with respect to θ and using the already obtained first derivative of f (x; θ), we have that 2 f (x; θ) θ2 = r (θ) (x µ(θ)) f (x; θ) + r (θ) (x µ(θ)) f (x; θ) r (θ)µ (θ)f (x; θ) = r (θ) (x µ(θ)) f (x; θ) + (r (θ)) 2 (x µ(θ)) 2 f (x; θ) r (θ)µ (θ)f (x; θ) Also, because µ(θ) is the mean 2 θ 2 f (x; θ)dx = (r (θ)) 2 Var[X] r (θ)µ (θ). Recalling that the range of x does not depend on θ because of the support of X, we obtain that Edward Furman Risk theory 4280 35 / 72

Proof. from which as required. 2 θ 2 f (x; θ)dx = (r (θ)) 2 Var[X] r (θ)µ (θ). Var[X] = µ (θ) r (θ) Edward Furman Risk theory 4280 36 / 72

Proof. from which as required. 2 θ 2 f (x; θ)dx = (r (θ)) 2 Var[X] r (θ)µ (θ). Var[X] = µ (θ) r (θ) We shall further calculate the CTE of X at the confidence level x q = VaR q [X]. Edward Furman Risk theory 4280 36 / 72

Proof. from which as required. 2 θ 2 f (x; θ)dx = (r (θ)) 2 Var[X] r (θ)µ (θ). Var[X] = µ (θ) r (θ) We shall further calculate the CTE of X at the confidence level x q = VaR q [X]. Proposition 1.9 The CTE q [X] is given by CTE q [X] = µ(θ) + 1 r (θ) θ log S(x q; θ). Edward Furman Risk theory 4280 36 / 72

Proof. θ log S(x q; θ) Edward Furman Risk theory 4280 37 / 72

Proof. = θ log S(x q; θ) 1 S(x q ; θ) θ S(x q; θ) Edward Furman Risk theory 4280 37 / 72

Proof. = θ log S(x q; θ) 1 S(x q ; θ) θ S(x q; θ) = 1 S(x q ; θ) θ x q f (x; θ)dx Edward Furman Risk theory 4280 37 / 72

Proof. = = θ log S(x q; θ) 1 S(x q ; θ) θ S(x 1 q; θ) = S(x q ; θ) θ 1 S(x q ; θ) x q f (x; θ)dx. θ x q f (x; θ)dx Edward Furman Risk theory 4280 37 / 72

Proof. = = θ log S(x q; θ) 1 S(x q ; θ) θ S(x 1 q; θ) = S(x q ; θ) θ 1 S(x q ; θ) x q f (x; θ)dx. θ x q f (x; θ)dx We have already shown in Proposition 1.7 that Thus we have that θ log S(x q; θ) = θ f (x; θ) = f (x; θ) ( r (θ)x q (θ)/q(θ) ). Edward Furman Risk theory 4280 37 / 72

Proof. The latter equation reduces to θ log S(x q; θ) = r (θ)cte q [X] q (θ) q(θ), from which we easily deduce that Edward Furman Risk theory 4280 38 / 72

Proof. The latter equation reduces to θ log S(x q; θ) = r (θ)cte q [X] q (θ) q(θ), from which we easily deduce that CTE q [X] = 1 r (θ) θ log S(x q; θ) + q (θ) r (θ)q(θ) that completes the proof. Example 1.14 (CTE for gamma as a member of LEF) Let X Ga(γ, θ). Use the fact that it is in LEF to derive the CTE risk measure. Edward Furman Risk theory 4280 38 / 72

Solution Gamma is in LEF with r(θ) = 1/θ, p(x) = x γ 1 /Γ(γ) and q(θ) = θ γ. Thus we have that and also µ(θ) = r (θ) = θ 2, q (θ) r (θ)q(θ) = γθγ 1 θ γ 2 = γθ. Furthermore, we have that for the cdf of gamma G( ; γ, θ) because S(x q ; θ) = 1 G(x q ; γ, θ) = 1 G(x q /θ; γ) Edward Furman Risk theory 4280 39 / 72

Solution Gamma is in LEF with r(θ) = 1/θ, p(x) = x γ 1 /Γ(γ) and q(θ) = θ γ. Thus we have that and also µ(θ) = r (θ) = θ 2, q (θ) r (θ)q(θ) = γθγ 1 θ γ 2 = γθ. Furthermore, we have that for the cdf of gamma G( ; γ, θ) S(x q ; θ) = 1 G(x q ; γ, θ) = 1 G(x q /θ; γ) because gamma is in the scale family with scale parameter θ. Furthermore, θ x q/θ e x x γ 1 dx = x q e xq/θ (x/θ) γ 1 Γ(γ) θ 2. Γ(γ) Edward Furman Risk theory 4280 39 / 72

Solution (cont.) After all, we have the CTE risk measure given by CTE q [X] = 1 r (θ) θ log S(x q; θ) + q (θ) r (θ)q(θ) Edward Furman Risk theory 4280 40 / 72

Solution (cont.) After all, we have the CTE risk measure given by CTE q [X] = 1 r (θ) θ log S(x q; θ) + q (θ) r (θ)q(θ) 1 e xq/θ (x/θ) γ 1 = γθ + x q S(x q /θ; θ) Γ(γ) Edward Furman Risk theory 4280 40 / 72

Solution (cont.) After all, we have the CTE risk measure given by CTE q [X] = 1 r (θ) θ log S(x q; θ) + q (θ) r (θ)q(θ) 1 e xq/θ (x/θ) γ 1 = γθ + x q S(x q /θ; θ) Γ(γ) = γθ + x q h(x q /θ) Edward Furman Risk theory 4280 40 / 72

Solution (cont.) After all, we have the CTE risk measure given by CTE q [X] = 1 r (θ) θ log S(x q; θ) + q (θ) r (θ)q(θ) 1 e xq/θ (x/θ) γ 1 = γθ + x q S(x q /θ; θ) Γ(γ) = γθ + x q h(x q /θ) with h( ) a hazard function of a Ga(γ, 1) random variable. Which random variable we have already seen to possess a CTE of the form similar to the one above? Edward Furman Risk theory 4280 40 / 72

Definition 1.10 (Univariate elliptical family) We shall say that X E(a, b) is a univariate elliptical random variable if when X has a density, it is given by the form ( f (x) = c ( ) ) b g 1 x a 2, x R, 2 b where g : [0, ) [0, ) such that 0 x 1/2 g(x)dx <. Both the expectation and the variance can be infinite. For instance, to assure that the expectation is finite it must be 0 g(x)dx <, (change of variables in the integral for the mean). Edward Furman Risk theory 4280 41 / 72

Proposition 1.10 Let X E(a, b) that has a density as before. The normalizing constant is then ( 2 1 c = x g(x)dx) 1/2. Proof. From the fact that f (x) is a density it must be that f (x)dx = 0 Edward Furman Risk theory 4280 42 / 72

Proposition 1.10 Let X E(a, b) that has a density as before. The normalizing constant is then ( 2 1 c = x g(x)dx) 1/2. Proof. From the fact that f (x) is a density it must be that f (x)dx = c ( ( ) ) 1 x a 2 g dx b 2 b = 0 Edward Furman Risk theory 4280 42 / 72

Proposition 1.10 Let X E(a, b) that has a density as before. The normalizing constant is then ( 2 1 c = x g(x)dx) 1/2. Proof. 0 From the fact that f (x) is a density it must be that f (x)dx = c ( ( ) ) 1 x a 2 g dx b 2 b ( ) 1 ( ) 1 = c g 2 u2 du = 2c g 2 u2 du 0 noticing that we have an integral of an even function. Further by Edward Furman Risk theory 4280 42 / 72

Proof. simple change of variables z = u 2 /2, we have that f (x)dx = 2c = 2c From which as required. 0 0 (2z) 1/2 g (z) dz z 1/2 g (z) dz = 1. c = 1 ( 1 z 1/2 g (z) dz), 2 0 Edward Furman Risk theory 4280 43 / 72

Example 1.15 (Normal rv as a member of the elliptical family) Let g(x) = e x, then f (x) = c b exp { ( 1 2 ( ) )} x a 2, x R b which is a normal density, and the normalizing constant is c = 1 ( 1 x 1/2 e dx) x = 2 0 Edward Furman Risk theory 4280 44 / 72

Example 1.15 (Normal rv as a member of the elliptical family) Let g(x) = e x, then f (x) = c b exp { ( 1 2 ( ) )} x a 2, x R b which is a normal density, and the normalizing constant is c = 1 ( 2 0 ) 1 x 1/2 e x dx = 1 (Γ 2 ( )) 1 1 = 1. 2 2π Edward Furman Risk theory 4280 44 / 72

Example 1.16 ( ) p, Let g(x) = 1 + x k p where p > 1/2 and kp is a constant such that k p = The density is then f (x) = c b { (2p 3)/2, p > 3/2 1/2, 1/2 < p 3/2 (1 + ) p (x a)2 2k p b 2, x R. To find c, we have to evaluate the integral 0 z 1/2 g(z)dz = that after a change of variables becomes 0 z 1/2 (1 + z/k p ) p dz, Edward Furman Risk theory 4280 45 / 72

Solution z 1/2 g(z)dz = 0 or in other words 0 z 1/2 g(z)dz = k 1/2 p 0 k 1/2 p u 1/2 (1 + u) p du 0 u 1/2 1 du (1 + u) 1/2+p 1/2 Edward Furman Risk theory 4280 46 / 72

Solution z 1/2 g(z)dz = 0 or in other words 0 z 1/2 g(z)dz = k 1/2 p 0 = k 1/2 p B Thus the normalizing constant is c = k 1/2 p u 1/2 (1 + u) p du (1 + u) ( 1 2, p 1 2 0 u 1/2 1 1/2+p 1/2 du ). ( ( 1 2k p B 2, p 1 )) 1. 2 The density of a student-t rv is then Edward Furman Risk theory 4280 46 / 72

Solution f (x) = 1 b 2k p B ( 1 2, p 1 ) (1 + 2 ) p (x a)2 2k p b 2, x R. Elliptical distributions belong to the location-scale family (check at home). Note that for p, the student-t distribution above becomes a normal distribution. Also, for p = 1, we have the Cauchy distribution, which does not have a mean. Let G(x) = c x 0 g(y)dy and let G(x) = G( ) G(x), then we have the following proposition that establishes the form of the CTE risk measure with x q = VaR q [X]. Edward Furman Risk theory 4280 47 / 72

Proposition 1.11 Let X E(µ, β) (i.e., it has finite mean µ). Then CTE q [X] = µ + λβ 2, where λ = 1 β ( 1 G 2 ( ) ) xq µ 2 β S X (x q ). Proof. By definition, for z q = (x q µ)/β, we have that CTE q [X] = 1 S X (x q ) x q x c β g ( 1 2 ( ) ) x µ 2 dx β Edward Furman Risk theory 4280 48 / 72

Proposition 1.11 Let X E(µ, β) (i.e., it has finite mean µ). Then CTE q [X] = µ + λβ 2, where λ = 1 β ( 1 G 2 ( ) ) xq µ 2 β S X (x q ). Proof. By definition, for z q = (x q µ)/β, we have that ( 1 CTE q [X] = x c ( ) ) S X (x q ) x q β g 1 x µ 2 dx 2 β 1 ( ) 1 = c(µ + βz)g S X (x q ) 2 z2 dz. z q Edward Furman Risk theory 4280 48 / 72

Proof. Further, CTE q [X] = µ + 1 S X (x q ) z q ( ) 1 cβzg 2 z2 dz Edward Furman Risk theory 4280 49 / 72

Proof. Further, CTE q [X] = µ + = µ + 1 S X (x q ) 1 S X (x q ) z q 1 2 z2 q ( ) 1 cβzg 2 z2 dz cβg (u) du Edward Furman Risk theory 4280 49 / 72

Proof. Further, CTE q [X] = µ + = µ + 1 S X (x q ) 1 S X (x q ) = µ + β S X (x q ) z q 1 2 z2 q 1 2 z2 q ( ) 1 cβzg 2 z2 dz cβg (u) du cg (u) du Edward Furman Risk theory 4280 49 / 72

Proof. Further, CTE q [X] = µ + = µ + which completes the proof. 1 S X (x q ) 1 S X (x q ) = µ + β S X (x q ) z q 1 2 z2 q 1 2 z2 q = µ + β 2 1/β S X (x q ) G ( ) 1 cβzg 2 z2 dz cβg (u) du cg (u) du ( ) 1 2 z2 q, Edward Furman Risk theory 4280 49 / 72

We have so far seen distributions and families of distributions which are used to describe risks amounts. It is important to model the number of risks received by an insurer. For this purpose we shall turn to counting distributions. Recall that the pgf is P(z) = E[z X ] = x z x p(x) and d m [ d m dz m P(z) = E dz m zx ] = E[X(X 1) (X m+1)z X m ]. Edward Furman Risk theory 4280 50 / 72

Example 1.17 (Poisson distribution) We consider the Poisson X Po(λ), λ > 0 rv. The pmf is The pgf is P[X = x] = e λ λ x, x = 0, 1, 2,..., x! Edward Furman Risk theory 4280 51 / 72

Example 1.17 (Poisson distribution) We consider the Poisson X Po(λ), λ > 0 rv. The pmf is The pgf is P[X = x] = e λ λ x, x = 0, 1, 2,..., x! P(z) = E[z X e λ (zλ) x ] = x! x=0 = exp{ λ(1 z)}. Also: Edward Furman Risk theory 4280 51 / 72

Example 1.17 (Poisson distribution) We consider the Poisson X Po(λ), λ > 0 rv. The pmf is The pgf is P[X = x] = e λ λ x, x = 0, 1, 2,..., x! P(z) = E[z X e λ (zλ) x ] = x! x=0 = exp{ λ(1 z)}. Also: E[X] = d dz P(z) 1 = λ, Edward Furman Risk theory 4280 51 / 72

Proposition 1.12 Let X 1,..., X n be interdependent Poisson rv s. Then S = X 1 + + X n is Poisson with λ = λ 1 + + λ n. Proof. P S (z) = E[z S ] = E[z X 1+ +X n ] = Edward Furman Risk theory 4280 52 / 72

Proposition 1.12 Let X 1,..., X n be interdependent Poisson rv s. Then S = X 1 + + X n is Poisson with λ = λ 1 + + λ n. Proof. P S (z) = E[z S ] = E[z X 1+ +X n ] = n E[z X j ] j=1 Edward Furman Risk theory 4280 52 / 72

Proposition 1.12 Let X 1,..., X n be interdependent Poisson rv s. Then S = X 1 + + X n is Poisson with λ = λ 1 + + λ n. Proof. P S (z) = E[z S ] = E[z X 1+ +X n ] = n = exp{λ j (z 1)} = j=1 n E[z X j ] j=1 Edward Furman Risk theory 4280 52 / 72

Proposition 1.12 Let X 1,..., X n be interdependent Poisson rv s. Then S = X 1 + + X n is Poisson with λ = λ 1 + + λ n. Proof. P S (z) = E[z S ] = E[z X 1+ +X n ] = = j=1 which completes the proof. n E[z X j ] j=1 n n exp{λ j (z 1)} = exp λ j (z 1), j=1 Edward Furman Risk theory 4280 52 / 72

Proposition 1.13 Let the number of risks be X Po(λ). Also, let each risk be classified into m types with probabilities p j, j = 1,..., m independent of all others. Then we have that X j Po(λp j ), j = 1,..., m and they are independent. Proof. We have that P[X 1 = x 1,..., X m = x m ] = P[X 1 = x 1,..., X m = x m X = x]p[x = x]. Here the conditional probability is multinomial by its definition, thus for x = x 1 + + x m P[X 1 = x 1,..., X m = x m ]. = x! x 1! x m! px 1 1 pxm m e λ λ x x! Edward Furman Risk theory 4280 53 / 72

Proof. Hence P[X 1 = x 1,..., X m = x m ] Edward Furman Risk theory 4280 54 / 72

Proof. Hence = P[X 1 = x 1,..., X m = x m ] 1 x 1! x m! px 1 1 pxm m e λ λ x 1+ +x m Edward Furman Risk theory 4280 54 / 72

Proof. Hence = = P[X 1 = x 1,..., X m = x m ] 1 x 1! x m! px 1 1 pxm m e λ λ x 1+ +x m 1 x 1! x m! (λp 1) x1 (λp m ) xm e m j=1 λp j Edward Furman Risk theory 4280 54 / 72