Credit Risk Modelling M.A. Fahrenwaldt Solutions to exercises
Credit Risk Modelling M.A. Fahrenwaldt
Solutions to exercises
- See handwritten solution
- See handwritten solution
- Default probability given by p = Φ(I) where
so that
which certainly satisfies dI/dσ_{V }> 0 if V_{0 }≥ B (initial assets greater than face value of debt) which would generally be the case.
- This is a very simple question. By row, the missing numbers are 0.05, 0.7, 0,0, 1. Squaring the matrix we get
and two-year default probabilities can be read from the final column.
- See R script R on qrmtutorial.org
- See R script R on qrmtutorial.org.
- See handwritten solution
- Under RT the pay-off of the bond is
I{_{τ>T}} + (1 − δ)I{_{τ}≤_{T}}
where the first term is the survival claim payoff and the second and third terms comprise the pay-off of the recovery. This simplifies to δI_{{τ>T} }+ (1 − δ) and hence the price of the bond is
where the second term is the value of a survival claim at time t, which is given in the notes. We thus obtain
.
1
The spread is
when τ > t.
When the hazard function under the risk-neutral measure Q is a constant γ^{Q}(t) = ¯γ^{Q }we get
.
If ¯γ^{Q }is small, note that we have
where we use the approximations exp(−x) ≈ 1 − x and ln(1 − x) ≈ −x for x small.
Under the RF recovery model the value of the survival claim is
.
To value the recovery we consider
Therefore when τ > t we have
.
When the hazard function under the risk-neutral measure Q is a constant γ^{Q}(t) = ¯γ^{Q }we get
.
If we assume a constant interest rate some further simplification is possible. In particular
.
See script HazardRateModelSpreads.R as well.
2
- See R script R on qrmtutorial.org.
- The result follows since
Z N∆t | = | ||
δγ¯^{Q}
0 |
_{e}−(r+¯γ^{Q})t dt | ||
= |
where we have made the change of variables u = t−(k−1)∆t. It follows from
that
Z N∆t | = | |||
δγ¯^{Q} | 0 | _{e}−(r+¯γ^{Q})t dt | ||
= | N
X −(r+¯γ^{Q})(k−1)∆t ∗ −(r+¯γ^{Q})∆t e x ∆te k=1 |
|||
= | N x∗∆t Xe−(r+¯γ^{Q})k∆t. |
k=1
Hence the same value of ¯γ^{Q }can make the value of the premium and default legs equal for a CDS with any maturity, that is any value of N.
- We need to compute moments of . We have m = 1000 and p_{i }= p = 0.01 for all i.
- δ_{i }= δ = 0.4 so . Obviously E(L) = δmp and var(L) = δ^{2}mp(1 − p). Numerical values are E(L) = 4 and sd(L) ≈ 1.
- We still have δ_{i }= δ = 0.4 and. So E(L) = δmp as before
but
cov(Y_{i},Y_{j})
= δ^{2}mp(1 − p) + δ^{2}m(m − 1)cov(Y_{1},Y_{2}) .
Now cov(Y_{1},Y_{2}) = ρ^{p}var(Y_{1})var(Y_{2}) = ρp(1 − p), so
var(L) = δ^{2}mp(1 − p) + δ^{2}m(m − 1)ρp(1 − p) = δ^{2}mp(1 − p)(1 + (m − 1)ρ) .
Numerical value is sd(L) ≈ 3.08.
- In this case , where E(∆_{i}) = δ = 0.4 for all i. Therefore
E(L) = δmp as before. For the variance:
m var(L) = ^{X}var(∆_{i}Y_{i}) + ^{X}cov(∆_{i}Y_{i},∆_{j}Y_{j})
.
Now the independence assumptions come into play:
)cov(Y_{1},Y_{2}) = mvar(∆_{1})E(Y_{1}) + mE(∆_{1})^{2 }var(Y_{1}) + m(m − 1)E(∆_{1})E(∆_{2})cov(Y_{1},Y_{2}) = mvar(∆_{1})p + mδ^{2}p(1 − p)(1 + (m − 1)ρ) .
The numerical value is sd(L) ≈ 3.10.
Note, how adding default dependence has profound effect. Adding independent stochastic LGDs has a minor effect. Relaxing the independence assumptions for LGDs would lead to a greater effect.
- The within-group asset correlation (say for first group) is given by
The between-group asset calculation is
E(X_{i}X_{j}) = E(b_{1}b_{2}F_{1}F_{2}) = ρb_{1}b_{2}
13.
Note that this appears to be a mixture model with a standard normal factor:
where
√
In an exchangeable default model we have d_{i }= d and b_{i }= ρ for all i. Moreover we can simply notation and introduce π_{k }for the joint default probability.
Thus we obtain
and, since for any i we know P(Y_{i }= 1) = Φ(d) = π, we infer
Asset correlation in the general one-factor model is given by ρ_{ij }= ρ(X_{i},X_{j}) = b_{i}b_{j }and this can be negative if one of b_{i }or b_{j }is negative, but not both. In the
√
equicorrelation model (where b_{i }= ρ for all i) this is not possible.
Negative default correlation is also obtained when ρ_{ij }< 0. Observe that
cov(Y_{i},Y_{j}) = E(Y_{i}Y_{j}) − E(Y_{i})E(Y_{j})
= P(Y_{i }= 1,Y_{j }= 1) − P(Y_{i }= 1)P(Y_{j }= 1)
= P(X_{i }≤ d_{i},X_{j }≤ d_{j}) − P(X_{i }≤ d_{i})P(Y_{j }≤ d_{j}) < 0
- Suppose that X_{1},…,X_{m }is distributed according to the Gumbel copula. and consider the threshold model (X_{i},d)_{1≤i≤m }where 0 < d < Clearly π = P(X_{i }≤ d) = d and
ktimes m−ktimes
where θ ≥ 1.
- In this question we have . Since L | Q = q ∼ B(1000,q) and
Q ∼ Beta(a,b) we calculate that
Since and
cov(Y_{i},Y_{j}) P(Y_{i }= 1,Y_{j }= 1) − π^{2 }E(Q^{2}) − π^{2 }1
ρY = varY_{i }= π − π2 = π − π2 = a + b + 1
we have two equations in two unknowns which are easily solved to give a = π(ρ^{−}_{Y }^{1 }− 1) and b = (1 − π)(ρ^{−}_{Y }^{1 }− 1). When π = 0.01 and ρ_{Y }= 0.005 we get a = 1.99 and b = 197.01.
- The distribution function is
The density is
where we have differentiated both sides of the equality Φ(Φ^{−1}(q)) = q to obtain
from which it follows that
.
Calculation of the mean follows in the solution of the next question.
- We consider the model where Y_{1},…,Y_{m }are conditionally independent given Q and all satisfy Y_{i }| Q = q ∼ Be(q). The mixing variable can be represented as Q = Φ(µ + σZ) where Z is standard normal.
In such a model
(1)
Now in Exercise 13 we constructed an exchangeable one-factor Gaussian threshold model with
which is obviously equivalent to
(2)
^{−1}(π)/√1 − ρ and σ = ^{p}ρ/(1 − ρ), The formulas for π_{k }coincide if µ = Φ which would imply that the models are equivalent. Solving for√ π and ρ gives
π = Φ(µ/ 1 + σ^{2}) and ρ = σ^{2}/(1+σ^{2})^{−1}. The former must be the mean of a probit-normal distribution.
6
- Given Ψ = ψ we assume the Y˜_{i }are conditionally independent Poisson: Y˜_{i }∼
Po(k_{i}ψ). Define the default indicator Y_{i }= 1_{{Y}˜_{i}>_{0}}. Then
p_{i}(ψ) = P(Y_{i }= 1 | Ψ = ψ) = 1 − P(Y^{˜}_{i }= 0 | Ψ = ψ) = 1 − exp(−k_{i}ψ) .
Now, if Ψ ∼ Ga(α,1), we can calculate that
,
from which it follows that
.
Inserting the gamma density f_{Ψ}(ψ) = ψ^{α}^{−1 }exp(−ψ)/Γ(α) we obtain
Now for q small we have the approximation −ln(1 − q) ≈ q from which it follows that the density of Q_{i }is closely approximated by a beta distribution Be(α,k_{i}^{−1}).
- Recall that the random variable N has a negative binomial distribution with parameters α > 0 and 0 < p < 1, written X ∼ NB(α,p), if its probability functions is
where for x ∈ R and k ∈ N0 denotes an extended binomial coefficient. It follows that the mgf is
where this is defined for t values such that (1 − p)e^{t }< 1.
The mgf of is therefore
To compute the moments of Z we will first compute the mgf and moments of X. We have
provided t < 1. Now the mean and variance of X are easily calculated from
.
Since
M_{Z}^{0 }(t) = p^{α}α(1 − p)(1 − (1 − p)M_{X}(t))^{−(α+1)}M_{X}^{0 }(t)
we obtain
Similarly, from
we get
from which it follows that
which is an example of var(Z) = var(N)E(X)^{2 }+ E(N)var(X).
Let X_{j }be a generic variable with the df of X_{ji}. The mgf of Z_{j }in terms of the mgf of X_{j }is
where
.
Putting everything together the mgf of Z is
.
- The loss distribution takes the form
L = M˜ =d M˜1 + M˜2
where M^{˜}_{1 }and M^{˜}_{2 }are two independent negative binomial variables. For j = 1,2 we know that
1000 !
M˜j | Ψ = ψ ∼ Po Xkiwijψj
i=1
1000 !
∼ Po 0.01ψ_{j }^{X}w_{ij}
i=1
Now = 750 and = 250 so we have that
M^{˜}_{j }| Ψ = ψ ∼ Po(λ_{j}ψ_{j})
where λ_{1 }= 7.5 and λ_{2 }= 2.5.
Now λ_{j}Ψ_{j }∼ Ga(α_{j},β_{j}/λ_{j}) from which it follows that M^{˜}_{j }∼ NB(α_{j},β_{j}/(β_{j }+ λ_{j})). Recalling the moments of negative binomial, which were computed in Exercise 19, we have that
from which we conclude that E (L) = λ_{1 }+ λ_{2 }= 10
sd(
Plugging in the values σ_{j}^{2 }= 2 gives UL = 135.
- This question shows that we can sometimes start with a mixture model and find a latent variable model that is equivalent. With ˜π_{k }as defined we calculate
π˜_{k }= E (P (Y_{1 }= 0,…,Y_{k }= 0 | Ψ = ψ))
k !
= E ^{Y}P(Y_{i }= 0 | Ψ = ψ)
i=1 k !
= E ^{X}exp(−κψ) = E (exp(−kκψ)) .
i=1
This integral can be evaluated to be
from which it follows that ˜π = (1 + κ)^{−α }so that κ = π˜^{−1/α }− 1 and
where
.
The interestimg question is how are the π_{k }(usual definition) related to π? In fact we have the formula
where denotes the so-called Clayton survival copula. For example, when k = 2 we can show that π_{2 }= P(Y_{1 }= 1,Y_{2 }= 1) = 1 − P(Y_{1 }= 0) − P(Y_{2 }= 0) + P(Y_{1 }= 0,Y_{2 }= 0)
= π1 + π1 − 1 + C1Cl/α(1 − π1,1 − π1)
= Cˆ1Cl/α(π1,π1)
- In the one-factor model we have
where Ψ, are iid standard normal. The variance of the systematic part is β_{i }= b^{2}_{i }. Standard manipulations allow us to write the conditional default probability as
where p_{i }is the unconditional default probability and we note that p_{i}(ψ) is an increasing function in ψ.
The loss coming from the ith obligor is L_{i }= 0.6e_{i}Y_{i }with conditional expectation
.
The total exposure when there are m obligors is a_{m }= (m/2)×5+(m/2)×1 =
3m. This gives an asymptotic loss operator
!
We can now compute that
which compares with a total exposure of 5000 × 5 + 5000 × 1 = 30000, all figures in £M.
- a)
Conditional independence of Y_{i }and Y_{j }given Ψ follows from independence of
Z_{i }and Z_{j}. The constants are µ_{i }= Φ^{−1}(p_{i})/^{p}1 − b^{2}_{i }and σ_{i }= b_{i}/^{p}1 − b^{2}_{i }.
- Imagine increasing the portfolio while retianing the structure. In a portfolioof size m we have where
!
For 50% of exposures we have e_{i }= 2, p_{i }= 0.01 and b_{i }= 0.6 giving l_{i}(ψ) =
A(ψ) and for 50% of exposures we have e_{i }= 4, p_{i }= 0.05 and b_{i }= 0.8 giving l_{i}(ψ) = B(ψ). A and B are fully determined functions of ψ which are aproximately A(ψ) = 1.2Φ(−2.91 + 0.75ψ) and B(ψ) = 2.4Φ(−2.74 + 1.33ψ). Since a_{m }= 3m, the asymptotic loss operator would be given by
and the VaR could be approximated by
VaR0.99(L(10000)) ≈ a10000¯l(Φ−1(0.99)) = 30000¯l(Φ−1(0.99)).
In our case this evaluates to approximately 8423.
- In this case we have
!
In this case the asymptotic loss operator is
^{¯}l_{2}(ψ) = 0.6^{−1}Φ(0.5 + ψ)^{¯}l(ψ)
where ^{¯}l(ψ) is the loss operator in c). So VaR is scaled by 0.6^{−1}Φ(0.5 + Φ^{−1}(0.99)). We get approximately 14005.