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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
ACTL3162 General Insurance Techniques
5. Solvency and Ruin Theory∗
A/Prof Benjamin Avanzi
School of Risk and Actuarial Studies
UNSW Australia Business School
S2 2015
∗References: MW 10, 5.0–5.2 / FV / (A 13) / (D) / (K)1/67
mailto:[email protected]?subject=ACTL3162http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2319328http://dx.doi.org/10.1080/10920277.1998.10595667http://dx.doi.org/10.1080/10920277.1998.10595667http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2319328mailto:[email protected]?subject=ACTL3162
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Plan
1 Solvency considerationsBalance sheet and solvency
Risk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study2/67
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Balance sheet and solvency
Balance sheet
Balance sheet: we would like At ≥ Lt , withAt : assets at time t
Lt : liabilities at time t Let C t = At − Lt be the continuous time ‘surplus process’. At first sight it is
reasonable to require
Pr
inf t
C t ≥ 0 C 0 = c o = Prc 0
inf t
At − Lt ≥ 1− p .
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Balance sheet and solvency
Risk measures
The requirement above with time horizon 1,
Prc 0
[A1 − L1] ≥ 1− p ,
is that of a Value-at Risk VaR1−p (L1 − A1) on security level 1 − p More generally, one requires
ρ(L1 − A1) ≤ 0.
Solvency II uses VaR99.5%. There are other, arguably better ones,such as the TVaR (used, e.g., in the Swiss Solvency Test). For
given portfolio of insurance business, one can adjust c 0 and
investment strategy A
0 accordingly.4/67
ACTL3162 G l I T h i (2015) / 5 S l d R i Th
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Balance sheet and solvency
Market consistent values
The main difficulty is to model L1 − A1, both of which arestochastic and not independant.
A1 is typically well defined and corresponding to assets which
have a market value.
However, L1 does not have a market value. For comparability,we need to determine market-consistent values in amarked-to-model approach.
We splitL1 = X 1 + L
+1 ,
where X 1 are payments done over the next year, and L+1 what
remains to be paid at the end of the year.
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Balance sheet and solvency
Market consistent values: L+1
L+1 is hard to determine:
We need a market-consistent value for those outstandingclaims liabilities
This is different from the IBNR R(1) (we will discuss those inModule 7):
1 R(1) are calculated on a nominal basis (no discounting). Amarket-consistent value must include discounting.
2
R(1) are conditional expectations , given the information
available at time 1. A market-consistent value must include aloading to recognise the uncertaintly that is associated tothose future cash flows.
Think of it as a transfer value , or run-off value.
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Risk moduels
1 Solvency considerationsBalance sheet and solvency
Risk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study7/67
ACTL3162 General Insurance Techniques (2015) / 5 Solvency and Ruin Theory
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Risk moduels
Asset deficit ADt
LetADt = Lt −At = X t + L+t − At
be the asset deficit as of time t .
Obviously, we want ρ(ADt ) ≤ 0.In practice, its modelling is broken down into modules , whichare then re-aggregated using correlation matrices.
Most relevant to us:
Market risk: volatility of market prices of financial instruments
Insurance risk: typically split into branches. GI comprises: (i)reserve risk, (ii) premium risk.Credit risk: conterparty default riskOperational risk: “risk of loss arising from inadequate or failedinteral processes, or from personnel and systems,
or from external events”7/67
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Risk moduels
Asset deficit AD0 and its evolution over one time period
We decomposeAD0 = L0 − A0,
where
L0 = LPY0 + L
CY0 = L
+0
A0 = c 0 + APY0 + π
CY
At the end of the year, we have A1 and
L1 = X 1 + L+1 =
X PY1 + X
CY1 + X
Op1
+L+,PY1 + L
+,CY1
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Solvency considerations
Risk moduels
Asset deficit AD1
AD1 =X PY1 + X
CY1 + X
Op1
+L+,PY1 + L
+,CY1
− A1
(split into payments and outstanding loss liabilities)
=X PY1 + L
+,PY1
+X CY1 + L
+,CY1
+ X Op1 − A1
(split into PY risk and CY risk)
Market risk: all
Insurance risk: all but X Op1 and A1
Credit risk: mainly A0 (! reinsurance)
Operational risk: X Op19/67
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q ( ) / y y
Solvency considerations
Insurance liability variables
1 Solvency considerationsBalance sheet and solvency
Risk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study10/67
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q ( ) / y y
Solvency considerations
Insurance liability variables
Market-consistent values
We focus here on insurance liabilities:
LIns1 =X PY1 + L
+,PY1
+X CY1 + L
+,CY1
= L1 − X Op1
Introducing deflators ϕ’s (random) leads to
LIns1 = 1
ϕ1
s ≥1
E [ϕs X s |F 1 ] = X 1 + 1ϕ1
s ≥2
E [ϕs X s |F 1 ]
if ϕs and X s are uncorrelated :
= 1ϕ1
s ≥1
E [ϕs |F 1 ] E [X s |F 1 ]
=
s ≥1P (1, s )E [X s |F 1 ]
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Solvency considerations
Insurance liability variables
Simplification
Denote:
p (1, s ) = E [P (1, s ) |F 0 ]x s = E [X s
|F 0 ]
We then use the approximation
P (1, s )E [X s |F 1 ]= [p (1, s ) + (P (1, s )
−p (1, s ))] [x s + (E [X s
|F 1 ]−
x s )]
≈ p (1, s )x s (expected value as of time 0)+ (P (1, s )− p (1, s )) x s (discounting uncertainty)+p (1, s ) (E [X s |F 1 ]− x s ) (insurance cash flows uncertainty)
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Solvency considerations
Insurance liability variables
AD1 after simplification
We have then
AD1 =
s ≥1p (1, s )x s + (P (1, s )− p (1, s )) x s −A1
( = market and credit risks Z 1)
+s ≥1
p (1, s ) (E [X s |F 1 ]− x s )
( = insurance risk Z 2)
+X Op1( = operational risk Z 3)
Hereafter we focus on Z 2.
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Solvency considerations
Insurance liability variables
Insurance risk Z 2
We consider Z 2 = Z PY2 + Z CY2 where
Z PY2 =s ≥1
p (1, s )E X PYs |F 1
− x PYs
Z
CY
2 = s ≥1 p (1, s )E X CYs |F 1 − x CYs Assume that the proportion of cash flows paid in year s γ s isdeterministic. Then
Z PY2 = s ≥1
p (1, s )γ PYs X PY1 +R(1) −R(0)Z CY2 =
s ≥1 p (1, s )γ
CYs
[E [S 1 |F 1 ]− E [S 1 |F 0 ]]
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Solvency considerations
Insurance liability variables
Insurance risk on PY
LetCDR1 = −
X PY1 +R(1) −R(0)
,
which has expected value E [CDR1] = 0. Process:1 Calculate MSEP for each LoB
2 Specify a correlation matrix between the LoB
3 Aggregate (1) with the help of (2) and obtain an overall
variance4 Fit a translated gamma or lognormal to (3), assuming that the
mean is R(0), to approximate the distribution of CDR1
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Solvency considerations
Insurance liability variables
Insurance risk on CY
Aggregate claimsE [S 1 |F 1 ]
result from the premium exposure πCY . This is called the premiumliability . It is split into two independent random variables:
S lc: large claims (> threshold M ), modelled with compoundPoisson model (CRM) per LoB with Pareto claims severities.Then use Panjer or FFT.
S sc: small claims, whose moments are aggregated using an
appropriate correlation matrix, and then fit to with a gammaor lognormal distribution.
Note an assumption is required to aggregate Z CY2 and Z PY2 into Z 2.
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Ruin theory in discrete time
Surplus process and ruin
1 Solvency considerationsBalance sheet and solvencyRisk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study16/67
ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
R i h i di i
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Ruin theory in discrete time
Surplus process and ruin
Surplus process
We extend the definition of the surplus to more than one year, anddefine the surplus
C t = C (c 0)t = c 0 +
t
u =1(πu − S u ),with
c 0 is the initial capital
(πt , S t )t =1,2,3,... is an iid sequence with πt > 0 and S t ≥ 0.Furthermore, we assume that
X t = πt − S t
are independant and stationary increments.16/67
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R i th i di t ti
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Ruin theory in discrete time
Surplus process and ruin
Ruin
We hopeC t ≥ 0 for all t ≥ 0.
If not, then the ruin time τ is defined such that
τ = inf {s ∈ N0; C s
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Ruin theory in discrete time
Surplus process and ruin
Theorem 5.4
Depending on the sign of E [X 1], the process behaves differently:
1 if E [X 1] < 0, the process blows down to −∞;2 if E [X 1] > 0, the process blows up to
∞;
3 if E [X 1] = 0, the process either blows up or down to ±∞.So what?
1 if E [X 1] < 0 then ψ(c 0) = 1 for any c 0 ≥ 0;2 if E [X 1] > 0 (Net Profit Condition—NPC), then ψ(0) < 1.
Furthermore, it is obvious that under the NPC
ψ(c 0) ≤ ψ(0) ≤ 1.
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Ruin theory in discrete time
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Ruin theory in discrete time
Lundberg bound
1 Solvency considerationsBalance sheet and solvencyRisk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study19/67
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Ruin theory in discrete time
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Ruin theory in discrete time
Lundberg bound
Lundberg coefficient / Adjustment coefficient
Assume there exists an R > 0 such that
M −X 1 (R ) = M S 1−π1 (R ) = 1.Then, this R > 0 is called ‘Lundberg coefficient’. If it exists then itis unique.It can be shown that
ψ(c 0) ≤ e −Rc 0
.This is called Lundberg’s exponential bound .
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Ruin theory in continuous time
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Ruin theory in continuous time
Surplus process
1 Solvency considerationsBalance sheet and solvencyRisk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study20/67
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Ruin theory in continuous time
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y
Surplus process
Surplus process
We define now the continuous time surplus process
C (t ) = c 0 + πt − S (t ),
where
c 0 is the initial surplus;
π is the premium rate: c = (1 + θ)λE [Y 1]
θ is the relative security loading (θ > 0 under the NPC)
S (t ) = N (t )i =1 Y i are aggregate losses up to time t If furthermore, the losses {Y i } are iid and independent of {N (t )},and {N (t )} is a Poisson process, that is,
N (t )i =1 Y i is compound
Poisson, then
{C (t )} is called the Cramér-Lundberg process.20/67
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y
Surplus process
0 1 2 3 4
0
5
1 0
1 5
Poisson process N(t)
{N (t )}counting process
step function
Poisson process iff
incrementsN (t + h)−N (t ) ∼Poisson(λh)
⇐⇒
time between jumps W i ∼exponential(1/λ)
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Ruin theory in continuous time
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Surplus process
0 1 2 3 4
0
5
1 0
1 5
2 0
2 5
3 0
3 5
Compound Poisson process S(t)
We define
S (t ) =
N (t )
i =1 X i .{S (t )} is {N (t )} but
step i has heightX i instead of 1
Increments:S (t + h)− S (t ) ∼CPoisson(λh, P (x ))
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Surplus process
0 1 2 3 4
- 5
0
5
1 0
Cramer-Lundberg process U(t)
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The stability problem
1 Solvency considerationsBalance sheet and solvencyRisk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study24/67
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The stability problem
The stability problem
The “survival” of the insurance company will depend on certain(decision) variables:
initial surplus (c 0)
loading of premiums (θ)
reinsurance (e.g. α or d — see later)
What is the "best" way to choose/monitor these variables? Of course, this depends on what criterion the assessment is based:
probability of ruin—goes back to Lundberg (1909) and Cramér(1930, 1955)
utility—goes back to von Neumann and Morgenstern (1944)
present value of dividends—goes back to de Finetti (1957)
. . .
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The probability of ruin
1 Solvency considerationsBalance sheet and solvencyRisk moduelsInsurance liability variables
2 Ruin theory in discrete timeSurplus process and ruinLundberg bound
3 Ruin theory in continuous timeSurplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study25/67
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The probability of ruin
How to calculate the probability of ruin
There are different ways of calculating ψ:
analytically
ψ(u ) is very hard to calculate (in closed form), but possible forexponential and mixtures of exponential lossesψ(u , t ) is even more difficult to determine
using Panjer’s recursion via a special trick (assignment 2009,see reference D)
Monte-Carlo methods (simulations)In what follows, we assume we are using the Cramér-Lundbergmodel.
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Th b bili f i
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The probability of ruin
The adjustment coefficient
Consider the excess of losses over premiums over the interval [0, t ]:
S (t )− πt .
We define the adjustment coefficient R as the first positive solutionof
M S (t )−πt (R ) = E e R (S (t )−πt )
= e −R πt e λt [M Y 1(R )−1] = 1.
Furthermore
E e −RC (t )
= e −Rc 0 for all t ≥ 0
and thus e −RC (t )
is a martingale.26/67
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Th b bilit f i
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The probability of ruin
0.0 0.1 0.2 0.3 0.4 0.5 0.6
1 .
0
1 .
5
2 . 0
2 .
5
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The probability of ruin
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The probability of ruin
A Theorem 13.4.1
If {C (t )} is a Cramér-Lundberg process with θ > 0, then for c 0 ≥ 0
ψ(c 0) = e −Rc 0
E
e −RC (τ )|T < ∞
.
Since C (τ ) < 0, we have then (Lundberg’s exponential upperbound)
ψ(c 0) −y max =⇒ ψ(c 0) > e −R (c 0+y max)
Finally,e −R (c 0+y max) < ψ(c 0)
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The probability of ruin
Example
Assume Y 1 ∼ exp(β ). Find R and ψ(c 0).
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The probability of ruin
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The probability of ruin
Example
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Applications to reinsurance
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pp
1 Solvency considerationsBalance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study31/67
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Applications to reinsurance
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pp
Applications to reinsurance
We have now a model to study options about reinsurance
Based on the probability of ruin criterion, we will adjust the
adjustment coefficient (hence its name..) to meet a goal, suchas
maximise R ⇔ minimise ψ(c 0)find the cheapest reinsurance such that ψ(c 0) is inferior tosome level
Note that even if ψ(c 0) can’t be calculated, you can still playwith R and have qualitative results about ψ(c 0).
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Applications to reinsurance
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Numerical example (excess of loss)
Let X ∼ exp(1), θ = 1.25, θrins = 1.4. Consider nonproportionalreinsurance:
transferred loss = (X − d )+.We have then
πreins = 1.4λ ∞d
(x − d )e −x dx = 1.4λe −d
and
M Y ret (r ) = d 0 e rx e −x dx + ∞
d e rd
e −x
dx =
1
−re −d (1−r )
1− r .Hence, the (nonlinear) equation for R ret is
1 + (1.25
−1.4e −d )r
−
1− re −d (1−r )
1− r = 0.
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Applications to reinsurance
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0.7 0.8 0.9 1.0 1.1 1.2 1.3
0 .
3 1
0 .
3 2
0 .
3 3
0 .
3 4
0 .
3 5
d
R_
h
Using R, we have
d ∗ = 0.9632226
and
R ∗ret = 0.3493290,
which is much higher (better)
than the best we could achievewith proportional reinsurance.
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Applications to reinsurance
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A Theorem 14.5.1
Theorem 14.5.1 states that if
we are in a Cramér-Lundberg setting
we are considering two reinsurance treaties, one of which is
excess of loss
both treaties have same expected payments and samepremium loadings
then
the adjustment coefficient with the excess of loss treaty willalways be at least as good (high) as with any other type of reinsurance treaty
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de Finetti’s modification
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1 Solvency considerationsBalance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study35/67
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de Finetti’s modification
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Using the probability of ruin as a criterion presents some issues:
minimising ψ(c 0) supposes that companies should let theirsurplus grow without limit, which is not realistic
why should an older company hold more capital than a youngone, just because it is older?
furthermore, if some of the surplus is distributed from time totime, calculations of ψ(c 0) are wrong
Bruno de Finetti’s (1957) goal:
to propose an alternative formulation that would avoid
the misconceptions of the classical Cramér-Lundberg model and that would be sufficiently realistic and tractable to “study the practical problems regarding risk and reinsurance” (our translation)
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de Finetti’s modification
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Optimal dividend strategies
One has to find a rational way to model distribution of the surplusthe distribution of some of the surplus (to the shareholders) isconsidered as a dividend
the answer to how much and when dividends should bedistributed is called a dividend strategy
Consider the expected present value of dividends paid until ruin
This is the value of the company according to the Gordonmodel
shareholders (the decision makers) are likely to want to
maximise this value
This leads to the question of optimal dividend strategies.
optimal with respect to the expected present value of dividends, rather than the probability of ruin
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de Finetti’s modification
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Optimal dividends in the Cramér-Lundberg model
The optimal dividend strategy is a barrier strategy (in some cases):
0 1 2 3 4 5 6 7
0
5
1 0
1 5
Cramer-Lundberg process U(t)
0 1 2 3 4 5 6 7
0
5
1 0
1 5
Surplus X(t) and dividends D(t)
EPV of dividends: V (u ; b ) =
T
0 e −δt dD (t )
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin TheoryDependence modelling and copulas
Introduction to Dependence
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1 Solvency considerationsBalance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulas
Fitting copulas: case study38/67
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Introduction to Dependence
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Motivation
How does dependence arise?
Events affecting more than one variable
Underlying economic factors affecting more than one risk area
Reasons for modelling dependence:
Pricing:inflows and outflows of capital
Solvency assessment:bottom up: risks given
→ capital requirements
Capital allocation:top down: capital given → allocation per riskPortfolio structure: (or strategic asset allocation)how does the capital move compared to risks?
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Introduction to Dependence
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Examples
World Trade Centre causing losses to Property, Life, Workers’Compensation, Aviation insurers
Enron causing losses to the stock market and to Surety Bonds,
Errors & Omissions and Directors & Officers underwriters
Dot.com market collapse causing losses to the stock marketand to insurers of financial institutions and D&O writers
WTC / Enron / stock market losses causing impairment to
reinsurers solvency, so increasing credit risk on payments byreinsurers
Asbestos affecting many past liability years at once
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Introduction to Dependence
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Example of real actuarial data(Avanzi, Cassar and Wong, 2011)
Data were provided by the SUVA (Swiss workers compensationinsurer)
Random sample of 5% of accident claims in constructionsector with accident year 1999 (developped as of 2003)
Two types of claims: 2249 medical cost claims, et 1099 dailyallowance claims
1089 of those are common (!)
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Dependence modelling and copulas
Introduction to Dependence
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Scatterplot of the log of those 1089 common claims (LHS) etemprical copula (RHS):
There is obvious right tail dependence.
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Dependence modelling and copulas
Introduction to Dependence
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Correlation = dependance?
Correlation consumption of cheese (US) and deaths by becomingtangled in bedsheets (Tyler Vigen, 2015):
Correlation = 0.95!!
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Dependence modelling and copulas
Introduction to Dependence
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Common fallacies
Fallacy 1: a small correlation ρ(X 1, X 2) implies that X 1 and X 2 are close to being independent
wrong!
Independence implies zero correlation BUTA correlation of zero does not always mean independence.
See example 1 below.
Fallacy 2 : marginal distributions and their correlation matrix uniquely determine the joint distribution.
This is true only for elliptical families (including multivariatenormal), but wrong in general!
See example 2 below.
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Dependence modelling and copulas
Introduction to Dependence
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Example 1
Company’s two risks X 1 and X 2
Let Z ∼ N (0, 1) and Pr(U = −1) = 1/2 = Pr(U = 1)U stands for an economic stress generator, independent of Z
Consider:
X 1 = Z ∼ N (0, 1)and
X 2 = UZ ∼ N (0, 1).
Now Cov(X
1,X
2) = E
(X
1X
2) = E
(UZ 2
) = E
(U
)E
(Z 2
) = 0hence ρ(X 1, X 2) = 0. However, X 1 and X 2 are strongly dependent , with 50% probability co-monotone and 50%counter-monotone.
This example can be made more realistic44/67
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Dependence modelling and copulas
Introduction to Dependence
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Example 2
Marginals and correlations—not enough to completely determine joint distribution
Marginals: Gamma(5, 1)
Correlation: ρ = 0.75
Different dependence structures: Normal copula vsCook-Johnson copula
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Dependence modelling and copulas
What is a copula?
1 Solvency considerations
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1 Solvency considerationsBalance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulasFitting copulas: case study
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Dependence modelling and copulas
What is a copula?
Skl ’ i h
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Sklar’s representation theorem
The copula “couples”, “links”, or “connects” the joint distribution toits marginals.Sklar (1959): There exists a copula function C such that
F (x 1, x 2, ..., x n) = C (F 1 (x 1) , F 2 (x 2) , ..., F n (x n))
where F k is the marginal for X k , k = 1, 2, ...,n. Equivalently,
Pr (X 1 ≤ x 1, ..., X n ≤ x n) = C (Pr (X 1 ≤ x 1) , ..., Pr (X n ≤ x n)) .
Under certain conditions, the copula
C (u 1, ...,u n) = F F −11 (u 1) , ..., F
−1n (u n)
is unique, where F −1k denote the respective quantile
functions. This is one way of constructing copulas.47/67 ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Dependence modelling and copulas
What is a copula?
E l
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Example
Let
F (x , y ) =
(x +1)(e y −1)x +2e y −1 (x , y ) ∈ [−1, 1]× [0,∞]1− e −y (x , y ) ∈ (1,∞]× [0,∞]
0 elsewhereHence
F (x ) = x + 1
2 , x ∈ [−1, 1]
F −1(u ) = 2u − 1 = x G (u ) = 1 = e −y , y ≥ 0
G −1(u ) = − ln(1− u ) = y
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Dependence modelling and copulas
What is a copula?
E l
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Example
Finally,
C (u , v ) = (2u − 1 + 1)[(1− v )−1 − 1]
2u − 1 + 2(1− v )−1
− 1=
2u (1− 1 + v )(2u − 2)(1− v ) + 2
= 2uv
2u
−2uv
−2 + 2v + 2
= uv u + v − uv
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Dependence modelling and copulas
What is a copula?
D it i t d ith l
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Density associated with a copula
For continuous marginals with respective pdf f 1,...f n, the jointpdf of X can be written as
f (x 1, ...,x n) = f 1 (x 1)
· · ·f n (x n)
×c (F 1 (x 1) , ...,F n (x n))
where the copula density c is given by
c (u 1, ..., u n) = ∂ nC (u 1, ..., u n)
∂ u 1∂ u 2
· · ·∂ u n
.
Observe that the copula c distorts the independence to inducethe actual dependence structure.
If independent, c = 1.
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Dependence modelling and copulas
Archimedean copulas
1 Solvency considerations
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Balance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulasFitting copulas: case study
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Dependence modelling and copulas
Archimedean copulas
The family of Archimedean copulas
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The family of Archimedean copulas
C is Archimedean if it has the form
C (u 1
, ..., u n
) = ψ−
1 (ψ (u 1
) +· · ·
+ ψ (u n
))
for all 0 ≤ u 1, ...,u n ≤ 1 and for some function ψ (called thegenerator) satisfying:
ψ (1) = 0;
ψ is decreasing; andψ is convex.
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Dependence modelling and copulas
Archimedean copulas
The Clayton copula
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The Clayton copula
The Clayton copula is defined by
C (u 1, ..., u n) = n
k =1u −θk − n + 1
−1/θ.
Archimedean type with:
ψ (t ) = t −θ − 1, θ > 1ψ−1 (s ) = (1 + s )−1
/θ
The case of n = 2:
C (u 1, u 2) =u −θ1 + u
−θ2 − 1
−1/θ.
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Dependence modelling and copulas
Archimedean copulas
The Frank copula
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The Frank copula
The Frank copula is defined by
C (u 1
, ...,u n
) = 1
log θ log1 +
nk =1 (θ
u k
−1)
(θ − 1)n−1 .Archimedean type with:
ψ (t ) = − log
θt −1θ−1
, θ ≥ 0
ψ−1
(s ) = − 1
log θ log 1− e −s 1− e −θTry the case of n = 2.
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Dependence modelling and copulas
Simulation of copulas
1 Solvency considerationsB l h d l
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Balance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulasFitting copulas: case study
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ACTL3162 General Insurance Techniques (2015) / 5. Solvency and Ruin Theory
Dependence modelling and copulas
Simulation of copulas
Simulating random variables with a dependent structure
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Simulating random variables with a dependent structure
We will introduce the general conditional distribution method
The overarching idea is (for the bivariate case)
simulate two independent uniform random variable u and t
‘tweak’ t into a v ∈ [0, 1] so that it has the right dependencestructure (w.r.t. u ) with the help of the copulamap u and v into marginal x and y using their distributionfunction
However, there are some specific, more efficient algorithms
that are available for certain types of copulas (see referencebooks)
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Dependence modelling and copulas
Simulation of copulas
Preliminary: the conditional distribution function
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Preliminary: the conditional distribution function
We will need the conditional distribution function for V givenU = u , which is denoted by c u (v ) :
c u (v ) = Pr[V ≤ v |U = u ]= lim
∆u →0C (u + ∆u , v )− C (u , v )
∆u
= ∂ C (u , v )
∂ u
.
In particular, we will need its inverse.
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Dependence modelling and copulas
Simulation of copulas
Example
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Example
For the copula
C (u , v ) = uv
u + v − uv we have
c u (v ) = v (u + v − uv )− uv (1− v )
(u + v − uv )2 =
v
u + v − uv 2≡ t
c −1
u (t ) =
√ tu
1−√ t (1− u ) ≡ v
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Dependence modelling and copulas
Simulation of copulas
The conditional distribution method
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The conditional distribution method
Goal: generate a pair of pseudo-random variables (X , Y ) with d.f.’sF and G , respectively, with dependence structure described by thecopula C .Algorithm
1 Generate two independent uniform (0, 1) pseudo-randomvariable u and t
2 Set v = c −1u (t )
3 Map (u , v ) into (x , v ):
x = F −1(u )
v = G −1(v )
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Dependence modelling and copulas
Simulation of copulas
Example
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Example
Let X and Y be exponential with mean 1 and standard Normal,respectively. Furthermore, the copula describing their dependence issuch as in the previous example:
C (u , v ) = uv
u + v − uv Furthermore, you are given the following pseudo-random(independent) uniforms:
0.3726791, 0.6189313, 0.75949099, 0.01801882
Simulate two pairs of outcomes for (X , Y ).
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Dependence modelling and copulas
Simulation of copulas
Exercise
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Use of the conditional distribution method yields1 We can use the uniforms given in the question such that
(u 1, t 1) = (0.3726791, 0.6189313)
(u 2, t 2) = (0.75949099, 0.01801882)
2 Set v i = u i √ t i 1−(1−u i )√ t i for i = 1, 2:v 1 = 0.5788953
v 2 = 0.1053509
3 Mapping (u i , v i ) into (x i , y i ) using
x i = F −1(u i ) = − ln(1− u i ) and y i = Φ−1(v i ) we have
(x 1, y 1) = (0.4662971, 0.8648739)
(x 2, y 2) = (
−0.3247659,
−1.251638)
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Dependence modelling and copulas
Fitting copulas: case study
1 Solvency considerationsBalance sheet and solvency
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Balance sheet and solvencyRisk moduels
Insurance liability variables2 Ruin theory in discrete time
Surplus process and ruinLundberg bound
3 Ruin theory in continuous time
Surplus processThe stability problemThe probability of ruinApplications to reinsurancede Finetti’s modification
4 Dependence modelling and copulasIntroduction to DependenceWhat is a copula?Archimedean copulasSimulation of copulasFitting copulas: case study
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Dependence modelling and copulas
Fitting copulas: case study
Insurance company losses and expenses
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p y p
Data consists of 1 500 general liability claims
Provided by the Insurance Services Office, Inc.
X 1 is the loss, or amount of claims paid.
X 2 is the ALAE, or allocated loss adjustment expense.
Policy contains policy limits, and hence, censoring.
δ is the indicator for censoring so that the observed data
consists of
(x 1i , x 2i , δ i ) for i = 1, 2, ..., 1500.
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Dependence modelling and copulas
Fitting copulas: case study
Summary statistics of data
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y
Policy Loss LossLoss ALAE Limit (Uncensored) (Censored)
Number 1 500 1 500 1 352 1 466 34Mean 41 208 12 588 559 098 37 110 217 491Median 12 000 5 471 500 000 11 048 100 000Std Deviation 102 748 28 146 418 649 92 513 258 205Minimum 10 15 5 000 10 5 000Maximum 2 173 595 501 863 7 500 000 2 173 595 1 000 000
25th quantile 4 000 2 333 300 000 3 750 50 00075th quantile 35 000 12 577 1 000 000 32 000 300 000
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Fitting copulas: case study
Figure 2: loss vs ALAE
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2 4 6 8 10 12 14
4
6
8
1 0
1 2
LOSS vs ALAE on a log scale
log(LOSS)
l o g ( A L A E )
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Dependence modelling and copulas
Fitting copulas: case study
Maximum likelihood estimation
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Case 1: loss variable is not censored, i.e. δ = 0.
f (x 1, x 2) = f 1 (x 1) f 2 (x 2) ∂ 2
∂ x 1∂ x 2C (F 1 (x 1) , F 2 (x 2))
Case 2: loss variable is censored, i.e. δ = 1.
∂
∂ x 2P (X 1 > x 1, X 2 ≤ x 2) = ∂
∂ x 2[F 2 (x 2)− F (x 1, x 2)]
= f 2 (x 2)− ∂
∂ x 2 F (x 1, x 2)
= f 2 (x 2)
1− ∂
∂ x 2C (F 1 (x 1) , F 2 (x 2))
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Dependence modelling and copulas
Fitting copulas: case study
Choice of marginals and copulas
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Pareto marginals: F k (x k ) = 1−
λk λk +x k
θk for k = 1, 2.
For the copulas, several candidates were used:
Copula C (u 1, u 2) C 2 (u 1, u 2) = ∂ C (u 1, u 2)∂ u 2C 12 (u 1, u 2) = ∂
2
C (u 1, u 2)∂ u 1∂ u 2
Independent u 1 × u 2 u 1 1
Claytonu −α1 + u
−α2 − 1
−1/α(C /u 2)
α+1 (α + 1) C α · (C /u 1u 2)α+1
Gumbel-Hougaard exp− ((− log u 1)α + (− log u 2)α)1/α
log u 2log C
α−1C
u 2
1
C C 1C 2 [1 + (α− 1) / (− log C )]
Frank 1
α log
1 +
(e αu 1 − 1) (e αu 2 − 1)e α − 1
e αu 1 (e αu 2 − 1)(e α − 1) + (e αu 1 − 1) (e αu 2 − 1)
α (e α − 1) e α(u 1+u 2)[(e α − 1) + (e αu 1 − 1) (e αu 2 − 1)]2
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Dependence modelling and copulas
Fitting copulas: case study
Parameter estimates
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Independence Clayton Gumbel-Hougaard Frank
Parameter Estimate s.e. Estimate s.e Estimate s.e. Estimate s.e.
Loss (X 1) λ1 14 552 1 404 14 000 2 033 14 001 1 292 14 323 1 359θ1 1.139 0.067 1.143 0.093 1.120 0.062 1.106 0.064
ALAE (X 2) λ2 15 210 1 661 16 059 2 603 14 122 1 409 16 306 1 762θ2 2.231 0.178 2.315 0.261 2.108 0.151 2.274 0.181
Dependence α na na 1.563 0.047 1.454 0.034 -3.162 0.175
Loglik -31 950.81 -32 777.89 -31 748.81 -31 778.45
AIC 42.61 43.71 42.34 42.38
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Dependence modelling and copulas
Fitting copulas: case study
AIC criterion
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Akaike Information Criterion (AIC)
In the absence of a better way to choosing/selecting a copula
model, one may use the AIC criterion defined by
AIC = (−2 + 2m) /n
where is the value of maximised log-likelihood, m is the
number of parameters estimated, and n is the sample size.
Lower AIC generally is preferred.
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Dependence modelling and copulas
Fitting copulas: case study
Summary
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To find the distribution of the sum of dependent random variableswith copulas (one approach):
Fit marginals independently
Describle/fit dependence with a copula (roughly)
Get a sense of data (scatter plots, dependence measures)Choose candidate copulasFor each candidate, estimate parameters via MLEChoose a copula based on nll(highest) or AIC(lowest)
Perform simulations to look at the distributions of aggregates
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