Banhemman

Question 1

Name 2 count distributions

Answer 1

1. Poisson
2. Negative Binomial

Question 2

For Poisson distribution, identify f(n), E(N) and V(N)

Answer 2

E(N) = V(N) = u
f(n) = u^n * exp(-u) / n!

Question 3

For Negative Binomial distribution, identify f(n), E(N) and V(N)

Answer 3

f(n) = (r+n-1 chose n)p^n(1-p)^r
E(N) = pr/(1-p)
V(N) = pr/(1-p)^2

Question 4

For Uniform distribution, identify f(x), F(x), E(X) and V(X)

Answer 4

f(x) = 1/(b-a)
F(x) = (x-a)/(b-a)
E(X) = (b+a)/2
V(X) = (b-a)^2 / 12

Question 5

For Exponential distribution, identify f(x), F(X), E(X) and V(X)

Answer 5

f(x) = exp(-x/b) / b
F(x) = 1 - exp(-x/b)
E(X) = b
V(X) = b^2

Question 6

For Gamma distribution, identify E(X) and V(X)

Answer 6

E(X) = ab
V(X) = a*(b^2)

Question 7

For Lognormal distribution, identify E(X) and V(X)

Answer 7

E(X) = exp(u+s^2/2)
V(X) = (exp(s^2) - 1)*exp(2u+s^2)

u = ln(mean) - s^2/2
s^2 = ln(CV^2+1)

Question 8

For Shifted Pareto distribution, identify f(x), F(X), E(X) and V(X)

Answer 8

f(x) = a(b^a)/(x+b)^(a+1)
F(x) = 1 - (b/(x+b))^a
E(X) = b/(a-1)
V(X) = a(b^2)/((a-1)^2*(a-2))

Question 9

Discuss 4 methods for estimating distribution parameters

Answer 9

1. Methods of moments
   Compute moments for different values of m, set them equal to theoretical values and use sample data to solve for parameters
2. Maximum likelihood
   For sample data with n observations, create log-likelihood function, create log-likelihood function sum of lnf(xi), take derivatives with respect to each parameter, set equal to 0 and solve for estimated parameters.
3. Minimum chi-squared
   Create m ranges of claim size intervals. Tabulate actual claim counts in each range and expected counts using target distribution with initial seed parameters. Calculate Chi-squared values:
   X^2 = sum of (A - E)^2 / E
   And use computer to iterate different parameter choices until X^2 is minimized.
4. Minimum distance
   Create m ranges of claim size intervals. Tabulate cum % of actual claim counts from sample data for each range and expected cum % using target distribution with initial seed parameters. Calculate:
   D =(Sum of (Fn(Ck) - Fo(Ck))^2)^0.5
   Use computer to iterate different parameter choices until D is minimized.

Question 10

Define truncation (aka discarding)

Answer 10

Usually in case of claims below deductible.

They do not even appear in insurer’s data

Question 11

Define censoring (aka capping)

Answer 11

Usually in case of a limit

Question 12

Define Shifting

Answer 12

Usually with a straight deductible.

For claims larger than deductible they get reduced (shifted) by deductible amount.

Question 13

Explain Panjer’s recursive algorithm

Answer 13

Recursive algorithms can be used to approx real aggregate loss distributions.

Panjer’s recursive algorithm generates an aggregate loss distribution given equally-spaced discrete severity distribution with spacing h and count distribution N that satisfies fN(n) = (na+b)fN(n-1)/n

Both Poisson (a=0, b=u) and NB (a=p, b=(r-1)p) satisfies the relationship.

Assuming fx(0) = 0, we have
fs(0) = fN(0)
fs(mh) = P(S = mh) = sum over k of (a+bk/m)fx(kh)fs(mh-kh)

Question 14

Given Y = max(0, X-a), calculate
1. Mean insurance pmt
2. Prob insurer pays 0
3. Limited E(XS) value
4. pdf of Xa

Answer 14

1. E(Y) = E(X) - E(X;a)
2. Fx(a) = P(X<=a)
3. E(Xa;l) = E(X;a+l) - E(X;a) / (1-Fx(a))
4. fXa(x) = fx(x+a)/(1-Fx(a))

Question 15

Explain how severity curve can be used to fit theoretical curve

Answer 15

We can calculate XS severity at various limits using empirical data and then compare that to theoretical curve

Straight line going up -> Pareto -> slope is 1/(a-1) and intercept is b/(a-1)

Increasing XS Sev -> Weibull

Hockey stick -> Lognormal

Decreasing XS Sev -> Gamma

Flat -> exponential -> b = XS Sev

Question 16

Describe the difference between interval of losses and layer of losses

Answer 16

A range of losses can be defined on an interval basis or layer basis.

Interval mean = sev = Tot loss/n

Layer mean = E(X;a+l) - E(X;a) / (1-Fx(a))

We need to price for losses in layer

Question 17

Calculate the variance for a layer of losses

Answer 17

V(Xa;l) = E(Xa^2;l) - E^2(Xa;l)

E(Xa^2;l) = E(X^2;a+l) - E(X^2;a) - 2a(E(X;a+l) - E(X;a)) / (1-Fx(a))

Question 18

Calculate coefficient of variation for layer of losses

Answer 18

CV(X) = V(Xa;l)^0.5/E(Xa;l)

Question 19

Given distribution, calculate expected number of claims above retention

Answer 19

N = ground-up claim counts
Na = counts for which x>a

p = 1 - Fx(a)
E(Na) = pE(N)
V(Na) = p^2V(N) + p(1-p)E(N)

Binomial:
Na follows Bin(k,p)
p = 1-Fx(a)
fNa(n) = sum over k of P(n excess counts given N=k)*P(N=k)

Poisson:
Na follows Poisson(pu)
E(Na) = pu = V(Na)

NB:
Na follows Bin(a,pv)
E(Na) = pE(N)
V(Na) = p^2V(N) + p(1-p)E(N)

Question 20

Calculate impact of severity inflation in XS layer

Answer 20

tau =1 + i%

Impact of inflation:
tau = E(tXa;l)/E(Xa;l) = t(E(X;(a+l)/t) - E(X;a/t))Sx(a)/(E(X;a+l) - E(X;a))*Sx(a/t)

Impact of severity inflation on aggregate losses:
tau(s) = t(E(X;(a+l)/t) - E(X;a/t))/(E(X;a+l) - E(X;a)) = tautau(n)

Impact of severity inflation on XS counts:
tau(n) = Sx(a/t)/Sx(a)

Question 21

Calculate Mean and Variance of Aggregate Losses in layer

Answer 21

E(S) = E(N)(E(X;a+l) - E(X;a))
V(S) = E(N)(E(X^2;a+l) - E(X^2;a)) - 2aE(S) + vE^2(s)
v is the claim contagion parameter: accounts for claim counts not being independent of each other

if n follows poisson, v=0

Question 22

Describe Risk Charge

Answer 22

Final rate por a policy needs to incorporate all expenses and profit as well as charge for risk.

The risk charge is a premium amount used to cover contingencies such as:
1. Random deviations of losses from expected values (process risk)
2. Uncertainty in selection of parameters describing loss process (param risk)

Question 23

Calculate the final rate and policy premium

Answer 23

N = cc
m = exposures
phi = fréquenté

E(N) = mphi
E(S) = E(N)E(Y)

Pure premium = p = phi*E(Y)

Final rate = R = (p+f)/(1-v)
f are fixed expenses
v are variable expenses

Policy premium = P = mR

Basic Limit P = mphi(E(X;b)*(1+u) + f) / (1-v)

Question 24

Define Loss Cost Multiplier (LCM)

Answer 24

1 / (1-v)

Loads all other costs on top of p to get final rate

Question 25

Describe 2 approaches to determine ILFs

Answer 25

1. Empirical data directly
2. Theoretical curve fit to empirical data (more common for highest limits with little empirical data)

Question 26

In determining ILFs, 3 assumptions are commonly made, explain.

Answer 26

1. All UW expenses and profit are variable and do not vary by limit
   In practice, profit loads might be higher for higher limits since they are more volatile
2. Freq and Sev are independent
3. Freq is the same for all limits
   This might not be true in practice due to adverse or favourable selection. GLMs or LR methods will reflect those differences but ILF approach will not.

Question 27

Calcule ILF

Answer 27

ILF(L) = E(X;L) / E(X;b)

Question 28

Calculate premium for a layer using ILFs

Answer 28

Pa,l = Pb * (I(a+l) - I(a))

Only accurate if:
1. Limit applied to loss & ALAE
2. Insurer covering XS layer does not cover ALAE
3. ALAE is proportional to loss

If limit applies to loss only and ALAE is not proportional, use this instead:
Pa,l = LCME(N)(E(X;a+l) - E(X;a) + (1-Fx(a))*e)

Error caused by not using this formula is usually small compared to risk load in practice

Question 29

Describe the consistency test for a set of ILFs

Answer 29

Premium should decrease as attachment point increases at a decreasing rate

I(L) = E(X;L)+e / E(X;b)+e
I’(L) = 1-Fx(L) / E(X;b)+e >= 0 (increasing)
I’‘(L) = -fx(L)/E(X;b)+e <= 0 (decreasing rate)

An exception to this could occur if one of the ILF assumptions was violated (ex: if liability lawsuits were influenced by size of limit so freq is not the same for all limits)

Question 30

Describe a situation where it could acceptable to not meet consistency test

Answer 30

If adverse selection is considered, the second condition could be overcome.

This will occur when risks that have large loss potential purchase higher limits or when court cases and settlements tend to loss at policy limits when settling.

Question 31

Explain 2 approaches to calculate risk loads for ILFs

Answer 31

To consider higher risk for higher limits, we include a risk charge rho(L) to obtain risk loaded ILFs.

1. Miccolis approach:
   rho(L) = kV(s)/E(N) = k(E(X^2;l)+dE^2(X;l))
   k is an arbitrary constant
   d = V(N)/E(N) - 1
   If N follows Poisson, d=0 so risk load independent of N
2. ISO approach:
   rho(L) = kV(s)^0.5 / E(N) = k(E(X^2;l) + d*E^2(X;l))^0.5 / E^0.5(N)

If k not given but Pb is, Pb = E(N)*(E(X;b) + rho(b)) -> solve for k

The risk load is higher with var method for higher limits (so higher prep) since variance increases faster than standard deviation.

Question 32

Calculate risk load for layer of coverage

Answer 32

rho(a,l) = k*(E(X^2;a+l) - E(X^2;a) - 2a(E(X;a+l) - E(X;a))

P = E(X;a+l) - E(X;a) + rho(a,l)

Question 33

State 2 advantages of using ISO approach over Miccolis approach

Answer 33

1. More fitted for thick-tailed distributions like Pareto
2. Express risk load in currency unit rather than currency squared

Question 34

Calculate ILF in case of both per-claim limit and aggregate limit

Answer 34

E(S_l) = E(N)*E(X;l) if only per-claim

With both limits:
I(l,L) = E(S_l;L) / E(N)*E(X;b)

Question 35

If straight deductible, calculate premium

Answer 35

Xd = max(X-d, 0)

Pd,b = Pb(1-C(d))
C(d) is the LER aka deductible credit factor
C(d) = E(X;d) + Fx(d)e / E(X;b)+e

Pure premium = p_d,b = phi(E(X;b) - E(X;d) + (1-Fx(d))e)(1+u)

Premium = P_d,l = Pb(I(l) - C(d))

Question 36

If franchise deductible, calculate pure premium and LER

Answer 36

Xd = if(X>d, X, 0)

Loss is truncated but not shifted by d so net loss Xd = X

Pure Premium = p_d,b = phi(E(X;b) - E(X;d) + (1-Fx(d))(d+e))*(1+u)

C(d) = (E(X;d) - d(1-Fx(d)) + Fx(d)*e) / (E(X;b) + e)

Question 37

If diminishing deductible, calculate LER

Answer 37

Xd = max(min(D(X-d)/(D-d), x), 0)

Loss below d is fully eliminated, deductible declines linearly from d to D and loss above D is paid in full (d disappears at D)

C(d,D) = (D(E;d)/(D-d) - dE(X;D)/(D-d))/E(X;b)

Question 38

Describe the LER relation between the 3 deductible types

Answer 38

Straight > diminishing > franchise

Because with franchise, losses above d are paid in full.
In contrast, a straight deductible reduces loss by d.
Diminishing a kinda of a mix of the 2.

Question 39

Calculate Pure Premium after inflation

Answer 39

p = phitau(E(X;l/t) - E(X;d/t) + (1-Fx(d/t))e)(1+u)

Question 40

Estimate trend impact net of deductible and capped by limit

Answer 40

tau = t(E(X;l/t) - E(X;d/t) + (1-Fx(d/t))e)/(E(X;l) - E(X;d) + (1-Fx(d))*e)

Question 41

State 2 reasons why loss cost per straight deductible can increase more than ground-up severity trend

Answer 41

1. For losses above deductible, trend is entirely in excess layer
2. Losses just under deductible are pushed into excess layer by trend, creating new losses for excess layer.

Question 42

Insurer believes average increase in XS losses due to inflation is too high. Describe 2 differences in assumptions the insurer may have with the consulting firm.

Answer 42

1. If insurer assumes a heavier tail distribution, more loss would be in XS which reduces the impact of XS inflation.
2. Insurer may assume a higher average severity, this would essentially act like an inflation factor shifting more of XS to higher limits so impact of trend on this particular layer is lessened. Shape of the loss distribution could also affect trend in this layer.