Chapter17 - Ruin Theory Flashcards
Describe N(t) and state its distribution.
N(t) is the number of claims arising within a time period of [0,t] for all t>0.
If N(1) ~ Poi(lambda) then N(t)~Poi(lambda*t)
(This is called the Poisson Process)
State the conditions for the claim number process.
- N(0) = 0
- N(t) must be integer value for all t>0
- N(s)<N(t) for s<t - claims are non-decreasing over time
4.When s<t, N(t)-N(s) represent the number of claims over the period (s,t]
What type of Process is the Poisson Process?
A counting process
State the conditions for a claim number process to be a Poisson process
- N(0) = 0 and N(s)<=N(t) when s<t
- Maximum of one claim in a very short time interval h
- The number of claims in a time interval of length h does not depend on when that time interval starts
4.For s<t , the number of claims in the time interval (s,t] is independent of the number of claims up to time s
What is the distribution of T1, the time until the first claim?
T1~exp(lambda)
P(T1<= t) = 1-exp(-lamba*t)
What is meant by the compound Poisson process?
The Poisson process for the number of claims will be combined with a claim amount distribution to give a compound Poisson process for the aggregate claims.
State the 3 assumptions for the aggregate claims process S(t)
to be a compound Poisson process.
- The random variables Xi are independently and identically distributed
- The random variables Xi are independent of N(t) for all t>0
- The stochastic process N(t) is a Poisson process whose parameter is lambda
What is the kth moment about zero of Xi’s
m_k = E[x^k]
Define the moment generating function Mx(r)
Mx(r)=E[exp(-r*x))
Where Mx(r) is dependent on the distribution of x
What is the mean and variance of S(t)
If S(t)=x1+x2+…+xN then (N, number of claims and N~Poi(lambda)
E[s(t)] = lamdaE[x] = lambdam1
Var[s(t)]=lamdaE[x^2]=lambdam2
Define and state Lundbergs inequality
Lundberg’s inequality tells us that we can find an upper bound for the probability of Ultimate ruin (Y(U))
Y(U) <=exp(-R*U)
R - the adjustment coefficient
U initial surplus
What does large values of R, the Adjustment Coefficient, imply.
R measures risk - it in a inverse measure of risk. So a large R implies the probability of ultimate ruin, Y(U), reduces.
What is the equation to calculate R?
lambda*Mx(r) = lambda + CR
Where C is the the premium income per unit of time
What is the upper bound for R?
lambda + CR = lambdaMx(R)
lambda + CR = lambdaE[exp(Rx)]
lambda + CR = lambdaE[1+ Rx + (Rx)^2/2
lambda + CR = lambda(E(1) + RE(x) + 1/2R^2E(x^2)
lambda + CR = lambda(1+ Rm1 + 1/2R^2m2)
R< 2(c-lambam1)/lambdam2
If c = (1+o)lamda*m1 where o is the insurers loading factor, then
R<(2om1)/m2
When can you define a low bound for R?
When the distribution of X, the individual claim amounts, has an upper limit, say M.
What is the lower bound for R?
M is the upper bound to the distribution of X. e.g. X~uni(0,100), M=100
R>1/Mlog(c/lamda*m1)
If c=(1+o)lamda*m1 then
R>1/Mlog(1+o)
Define proportional reinsurance.
alpha is the retained proportion that the insurer will pay of every claims.
Y=alphaX
Z=(1-alpha)X
What is the mean and variance of the claim amount paid by the direct insurer and reinsurer under proportional reinsurance?
Y = amount paid by direct insurer
Z = amount paid by reinsurer
E[Y] = E[alphaX] = alphaE[X}
E[Z] = E[(1-alpha)X] = (1-alpha)^2E[X]
Var[Y] = Var[alphax] = alpha^2Var[X]
Var[Z] = Var[(1-alpha)X] = (1-alpha)^2Var[X]