Credit Risk Modelling M.A. Fahrenwaldt Solutions to exercises
1. See handwritten solution
2. See handwritten solution
3. Default probability given by p = Φ(I) where
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ln B − ln V0 − μV −1 1 √ I= √T σV+2σVT
dI μV +ln(V0/B) 1√ dσ=√2 +2T
which certainly satisfies dI/dσV > 0 if V0 ≥ B (initial assets greater than face
value of debt) which would generally be the case.
4. This is a very simple question. By row, the missing numbers are 0.05, 0.7, 0, 0, 1. Squaring the matrix we get
0.80 0.15 0.05 0.80 0.15 0.05 0.655 0.225 0.120 0.10 0.70 0.20 × 0.10 0.70 0.20 = 0.150 0.505 0.345 001 001 001
and two-year default probabilities can be read from the final column.
5. See R script 10_Merton_implied_credit_spreads.R on qrmtutorial.org
6. See R script 10_Rating_embedding.R on qrmtutorial.org.
7. See handwritten solution
8. 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
p1(t,T) = EQp0(t,T)δI{τ>T} +(1−δ)|Ht
= (1−δ)p0(t,T)+δEQp0(t,T)I{τ>T} |Ht
where the second term is the value of a survival claim at time t, which is given in the notes. We thus obtain
p (t,T)=(1−δ)p (t,T)+δp (t,T)I e−T γQ(s)ds. 1 00{τ>t}t
The spread is
1 p1(t,T) c(t,T) = −T−tln p0(t,T)
= − 1 ln1−δ+δe−T γQ(s)ds t
When the hazard function under the risk-neutral measure Q is a constant
when τ > t.
γQ(t) = γ ̄Q we get
c(t,T)=− 1 ln1−δ+δe−γ ̄Q(T−t). T−t
If γ ̄Q is small, note that we have
c(t,T)≈− 1 ln1−δγ ̄Q(T−t)≈δγ ̄Q
where we use the approximations exp(−x) ≈ 1 − x and ln(1 − x) ≈ −x for x
Under the RF recovery model the value of the survival claim is
e−T γQ(s)ds. t
E (1−δ)p0(t,τ)I{t<τ≤T} |Ht = (1−δ)I{τ>t} exp(−tγQ(s)ds)
To value the recovery we consider
EQ I{t<τ≤T}p0(t,τ)
Therefore when τ > t we have
γQ(s) −s γQ(u)du
T −t γQ(t) = γ ̄Q we get
−T γQ(s)ds ln e t
T −γ ̄Q(s−t) e ds .
c(t,T)=− 1 ln e−γ ̄Q(T−t) +(1−δ)γ ̄Q T −t
In particular
T e−γ ̄Q(s−t) eγ ̄QterT −(γ ̄Q+r)t −(γ ̄Q+r)T t p0(s,T)ds=γ ̄Q+r e −e .
See script HazardRateModelSpreads.R as well. 2
= (1−δ)I{τ>t}
T sQ p0(t,s)γQ(s)e− t γ (u)duds.
When the hazard function under the risk-neutral measure Q is a constant
If we assume a constant interest rate some further simplification is possible.
9. See R script 10_CDS_calibration.R on qrmtutorial.org.
10. The result follows since
= x∗∆t e−(r+γ ̄Q)k∆t. k=1
e−(r+γ ̄Q)t dt =
= e−(r+γ ̄Q)(k−1)∆tδγ ̄Q k=1
δγ ̄Q e−(r+γ ̄Q)t dt
x∗∆t e−(r+γ ̄ )∆t = δγ ̄Q N
e−(r+γ ̄Q)t dt
e−(r+γ ̄ )t dt.
k=1 t=(k−1)∆t
e−(r+γ ̄ )(k−1)∆te−(r+γ ̄
where we have made the change of variables u = t − (k − 1)∆t. It follows from
e−(r+γ ̄Q)u du = e−(r+γ ̄Q)(k−1)∆tx∗∆t e−(r+γ ̄Q)∆t
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.
11. We need to compute moments of L = mi=1 δiYi. We have m = 1000 and pi =p=0.01foralli.
a) δi = δ = 0.4 so L = δmi=1Yi. Obviously E(L) = δmp and var(L) = δ2mp(1 − p). Numerical values are E(L) = 4 and sd(L) ≈ 1.26.
b)Westillhaveδi =δ=0.4andL=δmi=1Yi. SoE(L)=δmpasbefore but
var(L)=δ2var Yi =δ2var(Yi)+δ2cov(Yi,Yj)
i=1 i=1 i̸=j
= δ2mp(1 − p) + δ2m(m − 1) cov(Y1, Y2) .
Now cov(Y1, Y2) = ρvar(Y1) var(Y2) = ρp(1 − p), so
var(L) = δ2mp(1 − p) + δ2m(m − 1)ρp(1 − p)
=δ2mp(1−p)(1+(m−1)ρ) . Numerical value is sd(L) ≈ 3.08.
c) In this case L = mi=1 ∆iYi, where E(∆i) = δ = 0.4 for all i. Therefore E(L) = δmp as before. For the variance:
var(L) = var(∆iYi) + cov(∆iYi, ∆j Yj )
= m E(∆21Y1) − E(∆1Y1)2 + m(m − 1) (E(∆1Y1∆2Y2) − E(∆1Y1)E(∆2Y2)) . 3
Now the independence assumptions come into play:
var(L) = m E(∆21)E(Y1) − E(∆1)2E(Y1)2 + m(m − 1)E(∆1)E(∆2) cov(Y1, Y2) = m var(∆1)E(Y1) + mE(∆1)2 var(Y1) + m(m − 1)E(∆1)E(∆2) cov(Y1, Y2) = m var(∆1)p + mδ2p(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.
12. The within-group asset correlation (say for first group) is given by E(XiXj) = E(b1F1B1F1) = b21
The between-group asset calculation is
E(XiXj) = E(b1b2F1F2) = ρb1b2
P(Xi1 ≤di1,…,Xik ≤dik)=P(bi1F +1−b2i1Zi1 ≤di1,…,bikF +1−b2ikZik)
=EPbi1F+1−b2i1Zi1 ≤di1,…,|F
di1 −bi1F =EPZi1 ≤ ,…,|F
1 − b 2i 1
k di−biF
=E PZij ≤ |F
j=1 1 − b2ij
∞k di −bix jj
= Φ1−b2 φ(x)dx −∞ j=1 ij
Note that this appears to be a mixture model with a standard normal factor:
P(Yi1 =1,…,Yik =1)=
dij −bijx pij(x)=Φ1−b2 .
∞k pij(x) φ(x)dx
In an exchangeable default model we have di = d and bi = √ρ for all i. More- over we can simply notation and introduce πk for the joint default probability.
Thus we obtain
∞ d−√ρxk
√1−ρ φ(x)dx
and,sinceforanyiweknowP(Yi =1)=Φ(d)=π,weinfer
∞ Φ−1(π) − √ρxk
Φ √1−ρ φ(x)dx
Asset correlation in the general one-factor model is given by ρij = ρ(Xi,Xj) = bibj and this can be negative if one of bi or bj is negative, but not both. In the equicorrelation model (where bi = √ρ for all i) this is not possible.
Negative default correlation is also obtained when ρij < 0. Observe that
cov(Yi, Yj ) = E(YiYj ) − E(Yi)E(Yj )
=P(Yi =1,Yj =1)−P(Yi =1)P(Yj =1)
=P(Xi ≤di,Xj ≤dj)−P(Xi ≤di)P(Yj ≤dj)<0
14. Suppose that X1, . . . , Xm is distributed according to the Gumbel copula. and
consider the threshold model (Xi, d)1≤i≤m where 0 < d < 1. Clearlyπ=P(Xi ≤d)=dand
πk = P(X1 ≤ d,...,Xk ≤ d)
= P(X1 ≤ π,...,Xk ≤ π)
= CGu(π,...,π, 1,...,1) θ
k times m−k times = exp−k(−lnπ)θ1/θ
= exp −k1/θ (− ln π) = exp k1/θ ln π
15. In this question we have L = 1000 Yi. Since L | Q = q ∼ B(1000,q) and
where θ ≥ 1.
Q ∼ Beta(a, b) we calculate that
1 1000 k 1000−k 1 P(L=k)= k q (1−q) β(a,b)q
1000β(k + a, 1000 − k + b) = k β(a,b)
Sincepi =E(Yi)=E(Q)= a =:πand a+b
we have two equations in two unknowns which are easily solved to give a =
a−1 b−1 (1−q) dq
π(ρ−1 −1) and b = (1−π)(ρ−1 −1). When π = 0.01 and ρY = 0.005 we get
, k=0,1,... .
cov(Yi,Yj) P(Yi =1,Yj =1)−π2
E(Q2)−π2 1 π−π2 =a+b+1
a = 1.99 and b = 197.01.
16. The distribution function is
FQ(q) = P(Q ≤ q) = P(Φ(μ + σZ) ≤ q)
The density is
Φ−1(q)−μ =PZ≤σ
Φ−1(q) − μ =Φσ.
Φ−1(q) − μ dΦ−1(q) 1 fQ(q)=φ σ dqσ
φΦ−1(q)−μ σ
= φ(Φ−1(q))σ ,
where we have differentiated both sides of the equality Φ(Φ−1(q)) = q to obtain
from which it follows that
φ Φ−1(q) dΦ−1(q) = 1 dq
dΦ−1(q) = 1 . dq φ (Φ−1(q))
Calculation of the mean follows in the solution of the next question.
17. We consider the model where Y1, . . . , Ym are conditionally independent given Q and all satisfy Yi | Q = q ∼ Be(q). The mixing variable can be represented as Q = Φ(μ + σZ) where Z is standard normal.
In such a model
πk = P(Y1 = 1,...,Yk = 1) = E(Qk)
= E (Φ(μ + σZ))k ∞
(Φ(μ + σx))k φ(x)dx . (1) Now in Exercise 13 we constructed an exchangeable one-factor Gaussian thresh-
old model with
∞ Φ−1(π) − √ρxk
√1−ρ φ(x)dx ∞ Φ−1(π) + √ρxk
Φ √1−ρ φ(x)dx. (2)
The formulas for πk coincide if μ = Φ−1(π)/√1 − ρ and σ = ρ/(1 − ρ), which would imply that the models are equivalent. Solving for π and ρ gives π=Φ(μ/√1+σ2)andρ=σ2/(1+σ2)−1. Theformermustbethemeanofa probit-normal distribution.
which is obviously equivalent to
18. Given Ψ = ψ we assume the Y ̃i are conditionally independent Poisson: Y ̃i ∼ Po(kiψ). Define the default indicator Yi = 1{Y ̃i>0}. Then
pi(ψ)=P(Yi =1|Ψ=ψ)=1−P(Y ̃i =0|Ψ=ψ) = 1 − exp(−kiψ) .
Now, if Ψ ∼ Ga(α, 1), we can calculate that
P(Qi ≤q)=P(pi(Ψ)≤q)
= P(1 − exp(−kiΨ) ≤ q)
−ln(1−q) =PΨ≤k,
from which it follows that
fQi (q) = fΨ ki ki(1 − q) .
−ln(1−q) 1
Inserting the gamma density fΨ(ψ) = ψα−1 exp(−ψ)/Γ(α) we obtain
1 (−ln(1−q))α−1 ln(1−q) 1
fQi (q) = Γ(α) kα−1 exp ki ki(1 − q)
= i (− ln(1 − q))α−1 (1 − q)1/ki−1 . Γ(α)
Now for q small we have the approximation − ln(1 − q) ≈ q from which it
follows that the density of Qi is closely approximated by a beta distribution
Be(α, k−1). i
19. 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
α+k−1 α k
P(N=k)= k p(1−p), k=0,1,2,...,
where x for x ∈ R and k ∈ N0 denotes an extended binomial coefficient. It k
follows that the mgf is MN (t) = E(etN )
∞ α + k − 1 α t k = kpe(1−p)
=∞ α+k−11−(1−p)etαet(1−p)k k=0
p α k 1−(1−p)et
p α = 1−(1−p)et
where this is defined for t values such that (1 − p)et < 1. 7
The mgf of Z = Ni=1 Xi is therefore MZ(t)=MN(lnMX(t))= 1−(1−p)MX(t)
To compute the moments of Z we will first compute the mgf and moments of
X. We have
MX (t) = E(exp(tX)) =
E(X2)=M′′(0)= θ(θ+1)(1−t)−(θ+2) =θ(θ+1). X t=0
etxe−xxθ−1/Γ(θ)dx
e−(1−t)xxθ−1(1 − t)θ/Γ(θ)dx provided t < 1. Now the mean and variance of X are easily calculated from
= (1 − t)−θ
= (1 − t)−θ
E(X) = MX′ (0) = θ(1 − t)−(θ+1)t=0 = θ
MZ′ (t) = pαα(1 − p)(1 − (1 − p)MX (t))−(α+1)MX′ (t) E(Z)=MZ′ (0)= α(1−p)MX′ (0)= α(1−p)E(X)= θα(1−p)
Similarly, from
M′′(t) = pαα(1−p)2(1−(1−p)M (t))−(α+2)(M′ (t))2+pαα(1−p)(1−(1−p)M (t))−(α+1)M′′ (t)
E(Z )=MZ(0)= p2 (MX(0)) + p MX(0)
2 ′′ α(α+1)(1−p)2 ′ 2 α(1−p) ′′
θ2α2(1 − p)2 θ2α(1 − p)2 θ2α(1 − p) θα(1 − p)
= p2 + p2 + p + p θ2α2(1 − p)2 θ2α(1 − p) θα(1 − p)
= p2 + p2 + p from which it follows that
var(Z) = θ2α(1 − p) + θα(1 − p) p2 p
whichisanexampleofvar(Z)=var(N)E(X)2 +E(N)var(X).
Let Xj be a generic variable with the df of Xji. The mgf of Zj in terms of the
mgf of Xj is
pj αj MZj(t)=MNj(lnMXj(t))= 1−(1−pj)MXj(t)
n MXj(t)=EetXj=etxbqjb .
b=1 Putting everything together the mgf of Z is
p p p αj j
MZ(t)= MZj(t)= 1−(1−p )n etxbq j=1 j=1 j b=1
20. 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 kiwijψj
∼Po 0.01ψw j ij
Now 1000 wi1 = 750 and 1000 wi2 = 250 so we have that
M ̃ j | Ψ = ψ ∼ P o ( λ j ψ j )
where λ1 = 7.5 and λ2 = 2.5.
Now λjΨj ∼ Ga(αj,βj/λj) from which it follows that M ̃j ∼ NB(αj,βj/(βj + λj)). Recalling the moments of negative binomial, which were computed in Exercise 19, we have that
E M ̃ = α j λ j = λ jβj
varM ̃=αj(βj+λj)λj =σ2(σ−2+λ)λ
j βj2 jjjj from which we conclude that
E(L)=λ1 +λ2 =10
sd(L) = σ2(σ−2 + 7.5) · 7.5 + σ2(σ−2 + 2.5) · 2.5
11 22 Plugging in the values σj2 = 2 gives UL = √135.
21. 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(Y1 =0,...,Yk =0|Ψ=ψ)) k
=E P(Yi=0|Ψ=ψ) i=1
= E exp(−κψ) = E (exp(−kκψ)) .
This integral can be evaluated to be
∞ exp(−kκψ)exp(−ψ)ψα−1
k times −1/α −α Cl
fact we have the formula
πk = C1/α(π,...,π)
where CˆCl denotes the so-called Clayton survival copula. For example, when
= (1 + kκ)
= (1 + kκ)−α
−α ∞ exp(−(1 + kκ)ψ)ψα−1(1 + kκ)α
Γ(α) dψ fromwhichitfollowsthatπ ̃=(1+κ)−α sothatκ=π ̃−1/α −1and
π ̃k = kπ ̃ −k+1 =C1/α(π ̃,...,π ̃) CCl(u ,...,u )=(u−θ +···+u−θ −d+1)−1/θ .
The interestimg question is how are the πk (usual definition) related to π? In
k = 2 we can show that
k times ˆCl
π2 =P(Y1 =1,Y2 =1)=1−P(Y1 =0)−P(Y2 =0)+P(Y1 =0,Y2 =0) =π +π −1+CCl (1−π,1−π)
1 1 1/α 1 1 =CˆCl (π,π)
22. In the one-factor model we have
Xi = biF+1−b2iεi
Xi = −biΨ+1−b2iεi
where Ψ, ε1, ε2, . . . are iid standard normal. The variance of the systematic part is βi = b2i . Standard manipulations allow us to write the conditional default probability as
Φ−1(pi) + √βiψ pi(ψ)=P(Yi =1|Ψ=ψ)=Φ √1−βi
where pi is the unconditional default probability and we note that pi(ψ) is an increasing function in ψ.
The loss coming from the ith obligor is Li = 0.6eiYi with conditional expecta-
li(ψ)=E(Li |Ψ=ψ)=0.6eiΦ √1−β 10
Φ−1(pi) + √βiψ
The total exposure when there are m obligors is am = (m/2)×5+(m/2)×1 = 3m. This gives an asymptotic loss operator
̄ 1 m l(ψ)= lim
Φ−1(0.01)+√0.8ψ
×0.6×5×Φ +4 ×0.6×1×Φ
√0.2 Φ−1(0.001) + √0.2ψ
+4 ×0.6×5×Φ m
= 3 0.9Φ √0.2
√ Φ−1(0.01) + √0.8ψ
√0.8 Φ−1(0.001) + √0.2ψ
+ 4 ×0.6×1×Φ
1 Φ−1(0.01) + √0.8ψ
Φ−1(0.001) + √0.2ψ
+0.9Φ √ 0.8
We can now compute that
10000 VaR L ≈
0.99 i i=1
30000 × ̄l Φ−1(0.999)
≈ 30000 × 31 × 0.77777 ≈ 7777
which compares with a total exposure of 5000 × 5 + 5000 × 1 = 30000, all figures in £M.
= Φ(μi + σiψ)
Conditional independence of Yi and Yj given Ψ follows from independence of
Zi and Zj. The constants are μi = Φ−1(pi)/1 − b2i and σi = bi/1 − b2i .
b) Imagine increasing the portfolio while retianing the structure. In a portfolio
P ( Y i = 1 | Ψ = ψ ) = P ( b i F + 1 − b 2i Z i ≤ d i | F = − ψ ) di +biψ
= P (Zi ≤ 1 − b2 ) i
of size m we have L(m) = mi=1 Li where
li(ψ)=E(Li |Ψ=ψ)=0.6eipi(ψ)=0.6eiΦ 1−b2 +1−b2ψ
For 50% of exposures we have ei = 2, pi = 0.01 and bi = 0.6 giving li(ψ) = A(ψ)andfor50%ofexposureswehaveei =4,pi =0.05andbi =0.8
Φ−1(p ) b ii
giving li(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 am = 3m, the asymptotic loss operator would be given by
̄ 1m 1m m
l(ψ) = lim li(ψ) = ( A(ψ) + B(ψ)) = (A(ψ) + B(ψ))/6
m→∞ am i=1 3m 2 2 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.
c) In this case we have
li(ψ)=E(Li |Ψ=ψ)=eiE(∆i |Ψ=ψ)pi(ψ)
Φ−1(p ) b ii
=eiΦ(0.5+ψ)Φ 1−b2 +1−b2ψ ii
In this case the asymptotic loss operator is
̄l2(ψ) = 0.6−1Φ(0.5 + ψ) ̄l(ψ)
where ̄l(ψ) is the loss operator in c). So VaR is scaled by 0.6−1Φ(0.5 + Φ−1(0.99)). We get approximately 14005.
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