代写代考 SOLUTIONS TO CREDIT RISK EXERCISES

SOLUTIONS TO CREDIT RISK EXERCISES
to accompany the book QUANTITATIVE RISK MANAGEMENT: CONCEPTS, TECHNIQUES & TOOLS
. McNeil, Ru ̈diger Frey & Solutions to Additional Exercises
13. Denoting the gamma density by f, we easily compute the mgf of gamma to be 􏰏∞

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exp(tx)f(x)dx
xα−1 exp(−βx)βα
=β−t0 Γ(α)dx 􏰄 β 􏰅α
which is defined for t < β. The importance sampling density is gt(x) = exp(tx)f(x) = e−(β−t)xxα−1(β − t)α MX (t) Γ(α) which is the density of a gamma distribution with parameters α and β − t. The mean of this exponentially tilted gamma desnity is Egt (X) = α/(β − t) so clearly, as t → β, the mean can be made arbitrarily large, and, as t → −∞ the mean can be made arbitrarily small. 14. The moment generating function of M is 􏰍∞ etk e−μ μk k! = exp(−μ(1 − et)) and if we exponentially tilt M we introduce the new probability measure given so that, under Q, M is still Poisson, but with parameter etμ. The independent Poisson variables have joint mass function 􏰎k e−μiμli P(M1 = l1,...,Mk = lk) = i exp(tx) Γ(α) dx 􏰄 β 􏰅α 􏰏 ∞ xα−1e−(β−t)x(β − t)α Qt(M = k) = E = MM(t)P(M = k) P 􏰄 etM 􏰅 MM (t)I{M=k} = (etμ)k exp(−μet) andthemgfofL=􏰊ki=1eiMi is ML(t) = E 􏰀eLt􏰁 = 􏰎 E 􏰀eeiMit􏰁 = 􏰎 MMi (eit) The Qt probabilities are given by Qt(M1 =l1,...,Mk =lk)=E = 􏰎 exp(−μi(1 − eeit)) . i=1 P 􏰄 etL 􏰅 ML(t)I{M1=l1,...,Mk=lk} et􏰊ki=1 eili = 􏰋ki=1 e−μieμieeit P(M1 = l1,...,Mk = lk) 􏰎k 􏰆eteilie−μiμli 􏰇 =i i=1 e−μi eμi eei t li ! 􏰎k e−μi eei t (μi eei t )li = li! , so that,under Qt we have that Mi ∼ Poi(μieeit) and M1, . . . , Mk are indepen- We should find value of t such that EQt(L) = 􏰍eiEQt(Mi) = 􏰍eiμieeit = c i=1 i=1 for the optimum degree of tilting t. 15. In CreditRisk+ default counts are conditionally Poisson: Y ̃i |Ψ=ψ∼Poi(kiwiψ) . For simplicity we consider 1-factor case Y ̃i |Ψ=ψ∼Poi(kiψ) . We want to estimate tail probabilities for the portfolio loss L = 􏰊mi=1 eiY ̃i. We construct the inner IS estimator of θ(ψ) = P(L > c | Ψ = ψ) with the
following steps:
(a) We change the measure to Qt. Under Qt, conditional on Ψ, defaults counts Y ̃i are conditionally independent Poisson variables with mean μi = eeitkiψ. To get the optimal value for t we solve
mm EQt(L|Ψ=ψ)=􏰍eiEQt 􏰂Y ̃i |Ψ=ψ􏰃=􏰍eieeitkiψ=c.

(b) Generate defaults under Q . That is, generate Y ̃ (j), . . . , Y ̃ (j) and com- t1m
pute L(j) = 􏰊m eiY ̃(j) for j = 1,…,n1, where n1 is the number of i=1 i
simulations.
(c) Calculate estimator
θinner(ψ) =
I(L(j) > c)
n1 j=1 etL(j)
1 n1 M (t,ψ)
ML(t,ψ)=E exp 􏰍eiY ̃i |Ψ=ψ =􏰎exp􏰀−kiψ(1−eeit)􏰁 .
To construct the outer estimate consider the likelihood ratio
rs(x) = f(x) = MX(s) = βα gs(x) esx (β − s)αesx
where s is chosen so that the mean of Ψ under gs(x) represents a “bad” factor outcome. We would follow the following steps:
(a) Generate Ψ1, . . . , Ψn from a Ga(α, β − s) distribution.
(b) Calculate θˆinner(Ψ ) for j = 1, . . . , n where n is the number of simulations. n1 j
(c) Compute
θouter = θinner(Ψ )r (Ψ ) . n nn1 jsj

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