Credit Risk Modelling M.A. Fahrenwaldt Solutions to exercises

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

so that

which certainly satisfies dI/dσV > 0 if V0 B (initial assets greater than face value of debt) which would generally be the case.

  1. 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.

  1. See R script R on qrmtutorial.org
  2. See R script R on qrmtutorial.org.
  3. See handwritten solution
  4. Under RT the pay-off of the bond is

I{τ>T} + (1 − δ)I{τT}

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 γQ(t) = ¯γQ we get

 .

If ¯γQ 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 γQ(t) = ¯γQ we get

 .

If we assume a constant interest rate some further simplification is possible. In particular

 .

See script HazardRateModelSpreads.R as well.

2

  1. See R script R on qrmtutorial.org.
  2. The result follows since
Z Nt =
δγ¯Q

0

e−(rγQ)t dt
    =

where we have made the change of variables u = t−(k−1)∆t. It follows from

that

Z Nt =
δγ¯Q 0 e−(rγQ)t dt
      = N

X −(rγQ)(k−1)∆t     ∗                   −(rγQ)∆t e      x te

k=1

      = N x∗∆t Xe−(rγQ)kt.

k=1

Hence the same value of ¯γQ can make the value of the premium and default legs equal for a CDS with any maturity, that is any value of N.

  1. We need to compute moments of . We have m = 1000 and pi = p = 0.01 for all i.
    1. δi = δ = 0.4 so . Obviously E(L) = δmp and var(L) = δ2mp(1 − p). Numerical values are E(L) = 4 and sd(L) ≈ 1.
    2. We still have δi = δ = 0.4 and. So E(L) = δmp as before

but

cov(Yi,Yj)

= δ2mp(1 − p) + δ2m(m − 1)cov(Y1,Y2) .

Now cov(Y1,Y2) = ρpvar(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.

  1. In this case , where E(∆i) = δ = 0.4 for all i. Therefore

E(L) = δmp as before. For the variance:

m var(L) = Xvar(∆iYi) + Xcov(∆iYi,jYj)

 .

Now the independence assumptions come into play:

)cov(Y1,Y2) = mvar(∆1)E(Y1) + mE(∆1)2 var(Y1) + m(m − 1)E(∆1)E(∆2)cov(Y1,Y2) = mvar(∆1)p + 2p(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.

  1. The within-group asset correlation (say for first group) is given by

The between-group asset calculation is

E(XiXj) = E(b1b2F1F2) = ρb1b2

13.

Note that this appears to be a mixture model with a standard normal factor:

where

In an exchangeable default model we have di = d and bi = ρ for all i. Moreover we can simply notation and introduce πk for the joint default probability.

Thus we obtain

 

and, since for any i we know P(Yi = 1) = Φ(d) = π, we infer

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

  1. Suppose that X1,…,Xm is distributed according to the Gumbel copula. and consider the threshold model (Xi,d)1≤im where 0 < d < Clearly π = P(Xi d) = d and

ktimes                                                                                                          mktimes

where θ ≥ 1.

  1. In this question we have . Since L | Q = q ∼ B(1000,q) and

Q ∼ Beta(a,b) we calculate that

Since and

cov(Yi,Yj)    P(Yi = 1,Yj = 1) − π2                                    E(Q2) − π2                                           1

ρY =        varYi            =                 π π2                              =          π π2 = a + b + 1

we have two equations in two unknowns which are easily solved to give a = π(ρY 1 − 1) and b = (1 − π)(ρY 1 − 1). When π = 0.01 and ρY = 0.005 we get a = 1.99 and b = 197.01.

  1. 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.

  1. 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

(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 = pρ/(1 − ρ), The formulas for πk coincide if µ = Φ which would imply that the models are equivalent. Solving for√         π and ρ gives

π = Φ(µ/ 1 + σ2) and ρ = σ2/(1+σ2)−1. The former must be the mean of a probit-normal distribution.

6

  1. 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

 ,

from which it follows that

 .

Inserting the gamma density fΨ(ψ) = ψα−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 Qi is closely approximated by a beta distribution Be(α,ki−1).

  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)et < 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

MZ0 (t) = pαα(1 − p)(1 − (1 − p)MX(t))−(α+1)MX0 (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 Xj be a generic variable with the df of Xji. The mgf of Zj in terms of the mgf of Xj is

where

.

Putting everything together the mgf of Z is

 .

  1. 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                                                           0.01ψj Xwij

i=1

Now  = 750 and    = 250 so we have that

M˜j | Ψ = ψ ∼ Po(λjψj)

where λ1 = 7.5 and λ2 = 2.5.

Now λjΨj ∼ Ga(αjjj) from which it follows that M˜j ∼ NB(αjj/(βj + λj)). Recalling the moments of negative binomial, which were computed in Exercise 19, we have that

from which we conclude that E (L) = λ1 + λ2 = 10

sd(

Plugging in the values σj2 = 2 gives UL =                                            135.

  1. This question shows that we can sometimes start with a mixture model and find a latent variable model that is equivalent. With ˜πk as defined we calculate

π˜k = E (P (Y1 = 0,…,Yk = 0 | Ψ = ψ))

k                               !

= E YP(Yi = 0 | Ψ = ψ)

i=1  k !

= E Xexp(−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− 1 and

where

 .

The interestimg question is how are the πk (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 π2 = P(Y1 = 1,Y2 = 1) = 1 − P(Y1 = 0) − P(Y2 = 0) + P(Y1 = 0,Y2 = 0)

= π1 + π1 − 1 + C1Cl(1 − π1,1 − π1)

= Cˆ1Cl(π11)

  1. In the one-factor model we have

where Ψ, are iid standard normal. The variance of the systematic part is βi = b2i . Standard manipulations allow us to write the conditional default probability as

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 expectation

 .

The total exposure when there are m obligors is am = (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.

  1. a)

Conditional independence of Yi and Yj given Ψ follows from independence of

Zi and Zj. The constants are µi = Φ−1(pi)/p1 − b2i and σi = bi/p1 − b2i .

  1. Imagine increasing the portfolio while retianing the structure. In a portfolioof size m we have where

!

For 50% of exposures we have ei = 2, pi = 0.01 and bi = 0.6 giving li(ψ) =

A(ψ) and for 50% of exposures we have ei = 4, pi = 0.05 and bi = 0.8 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

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.

  1. In this case we have

!

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.