B1. Bahnemann

Question 1

Distribution review (Possion, Negative Binomial, Exponential, pareto)

Answer 1

Poisson:
- f(n) = lamda^n * exp(-lamda)/(fact(n))
- mean = variance = lamda
Negative Binomial:
-f(n) = combin(r+n-1, n) p^n (1-p)^r
- mean = pr/(1-p)
- variance = pr/(1-p)^2
Exponential
-f(x) = exp(-x/beta) / beta
-F(x) = 1- exp(-x/betal)
-mean = beta
-variance= beta^2
Pareto
-f(x) = alpha* beta^alpha / (x+beta)&^(alpha+1)
- F(x) = 1- (beta/x+Beta)^alpha
- mean = beta/ (alpha - 1)
- variance = alpha beta ^2 / [ (alpha -1)^2 (alpha -2)]

Question 2

Panjer’s Recursive Algorithm

Answer 2

• approximate a real aggregate loss distribution given an equal-spaced discrete severity distribution
• count distribution N must satisfies f(n) = (na+b)/b * f(n-1)
• both Poisson (a = 0, b = lamda) and Negative Binomial (a = p, b = (r-1)p)satisfies this
• f(0) = 0
• fs(mh) = Prb(S=mh) = summation from K=1 to m (a+ bk/m) fx(kh)fs(mh-kh)

Question 3

Expected excess claim severity

Answer 3

E(Xa) = e(a) = (E[X] - E[X;a])/ (1-F(a))

Question 4

excess claim severity variance

Answer 4

VAR(X^2 a;L] = E[X^2 a;L] - E[Xa;L]^2
E[X^2 a;L] = [E(X^2;a+L) - E(x^2;a) - 2a*[E(x;a+L) - E(x;a)) ]/(1-F(a))

Question 5

expected excess claim severity in a layer

Answer 5

E[Xa;L] = (E[X;a+L] - E[X;a]) / (1-F(a))

Question 6

Expected excess claim count

Answer 6

E[Na] = pE[N]

Question 7

excess claim count variance

Answer 7

Var(Na) = P^2 var(N) + p*(1-p) E(N)

Question 8

Expected aggregate losses in a layer

Answer 8

E[S] = E[N]*[E(X;a+L) - E(X;a)]

Question 9

aggregate losses in a layer variance

Answer 9

Var(S) = E[X^2;a+L] - E[X^2;a] -2aE[S] + r*E(S)^2
r - claim contagion parameter accounts for claims not being independent

Question 10

Variance approach for risk load

Answer 10

risk load = k[E(X^2;L) + (var(n)/E(n)-1)*E(X;L)^2]
premium for basic limit = E(N)[(E(X;b) + K E(X^2;b)]

Question 11

Standard deviation approach for risk load

Answer 11

risk load = k/sqrt(E[N])) * sqrt( E[X^2;L] + (var(n)/E(n)-1)*E(X;L)^2]

Question 12

risk load for a layer of coverage

Answer 12

risk load = k* (E(X^2;a+L) - E(X^2;a) - 2a*(E(X;a+L) - E(X;a))

Question 13

Straight deductible premium formula with a basic limit b and deductible b

Answer 13

P = Pb (1-C(d))
C(d) = (E(x;d) + F(d)ALAE)/ (E(x;b) + ALAE)
general formula P = frequency * (E[X;b] - E[x;d] + (1-F(d))ALAE) *(1+ULAE)

Question 14

Franchise deductible Premium formula

Answer 14

P = frequency* [E(x;b) - E(x;b) + (1-F(d))(d+ALAE)) (1+ULAE)
C(d) = [E(x;d) - d(1-F(d)) + F(d) ALAE]/ (E(X;b)+ALAE)

Question 15

Diminishing deductible net loss

Answer 15

If d<x<D, X = (D/(D-d)) * (X-d)
if x>D, X = x

Question 16

Impact of inflation in the fixed excess layer count

Answer 16

=E(TXa;L) / E(Xa;L)
E(TXa;L) = T* [E(X;(a+L)/T) - E(x;a/T)] /(1-F(a/T))

Question 17

Impact of inflation on excess loss severity

Answer 17

=[1-F(a/T) ]/[1-F(a)]

Question 18

Impact of inflation on aggregate losses in the layer

Answer 18

T*[E(X;(a+L)/T) - E(x;a/T)] / [E(X;a+L) - E(X;a)]

Question 19

Pure premium after inflation

Answer 19

frequency* inflation rate * [E(X;L/T) - E(X;d/T) +(1-F(d/T))*ALAE)] *(1+ULAE)

Question 20

Trend impact net of deductible and capped by the limit

Answer 20

T[E(X;L/T) - E(X;d/T) +(1-F(d/T))ALAE)] / [E(X;L) - E(X;d) +(1-F(d))*ALAE)]