Solutions of the Financial Risk Management Examination

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1 Solutions of the Financial Risk Management Examination Thierry Roncalli January 9 th 03 Remark The first five questions are corrected in TR-GDR and in the document of exercise solutions, which is available in my web page. The Basle II regulation Market risk 3 Credit risk 4 Counterparty credit risk 5 Operational risk 6 Value at risk of a long/short portfolio The principal reference for this exercise is TR-GDR pages We note P A t resp. P B t the value of the stock A resp. B at the date t. The portfolio value is: P t x A P A t + x B P B t with x A and x B the number of stocks A and B. We deduce that the PnL between t and t + is: PnL t; t + P t + P t x A P A t + P A t + x B P B t + P B t x A P A t R A t; t + + x B P B t R B t; t + with R A t; t + and R B t; t + the asset returns of A and B between the dates t and t +.. We have x A +, x B and P A t P B t 00. It comes that: PnL t; t + 00 R A R B We have R A R B N 0, A B with: A B % Thierry Roncalli, La Gestion des Risques Financiers, Economica, deuxième édition, 009. The direct link is

2 The annual volatility of the long/short portfolio is then equal to 0%. To compute the value at risk for a time horizon of one day, we consider the square root rule TR-GDR, page 74. We obtain 3 : VaR D Φ A B The probability to lose.89 euros per day is equal to %. 60. We have PnL t; t + 00 R A R B. We use the historical data to calculate the scenarios of asset returns R A, R B. We then deduce the empirical distribution of the PnL. Finally, we compute the corresponding empirical quantile. With 50 scenarios, the % decile is between the second and third worst cases: VaR D [ The probability to lose.905 euros per day is equal to %. This result is very similar to the one calculated with the gaussian VaR. 3. The PnL formula becomes TR-GDR, pages 9-95 : with C A t the call option price. We have: PnL t; t + P A t + P A t P B t + P B t C A t + C A t C A t + C A t P A t + P A t where is the delta of the option. For the analytical VaR, we obtain: PnL t; t + 50 R A 00 R B VaR D Φ 0.99 A/B because: A/B The daily 99% VaR decreases from.89 euros to.50 euros. 3 because Φ 0.99 Φ 0.0.

3 7 Parameter estimation for operational risk. a The density of the gaussian distribution Y N µ, is: g y π exp y µ Let X LN µ,. We have: It comes that: with y ln x. f x X e Y f x g y dy dx π exp y µ x x π exp ln x µ b The log-likelihood function of the sample {L,..., L n } is: c We have: L µ, ln n f L i ln f L i n ln π n ln ln L i {ˆµ, ˆ} arg max L µ, ln Li µ We notice that: max L µ, max n ln π n ln ln L i ln Li µ max n ln π n ln xi µ with x i ln L i. We recognize the gaussian log-likelihood function. ˆµ x i ln L i n n ˆ x i ˆµ ln L i ln L i n n n d Using Bayes formula, we have: Pr {X x X H} Pr {H X x} Pr {X H} F x F H F H 3

4 with F the cdf of X. It comes that the conditional density is: f H x x Pr {X x X H} f x F H Φ ln H µ x π exp It comes that the log-likelihood function of the sample {L,..., L n } is: L µ, ln n f H L i n ln π n ln ln L i ln H µ n ln Φ ln x µ ln Li µ. a By definition, we have: λ λn Pr {N t n} e E [N t n Pr {N t n} n0 n0 λ λn ne λe λ λ n0 λ n b We have: [ m E N t i i0 m n0 i0 n0 λ λn n i e λ λn n n n m e The term of the sum is equal to zero if n 0,,..., m. It comes that: [ m λ λn E N t i n n n m e i0 nm+ e λ nm+ λ m+ e λ λ n n m! nm+ λ m+ e λ n 0 λ n n! λ n m n m! 4

5 with n n m +. It comes that: [ m E N t i i0 λ m+ e λ e λ λ m+ var N t E [ Nt E [N t E [ Nt N t + E [Nt E [N t E [N t N t + E [N t E [N t Using the formula with m, we finally obtain: var N t λ + + λ λ λ c The estimator based on the first moment is: ˆλ T T t N t whereas the estimator based on the second moment is: ˆλ T N t T N t T T t 3. Let L be the random sum: N t L where L i LN µ,, L i L j and N t P λ. i0 L i t a We have: E [L E [ Nt L i i0 E [N t E [L i λ exp µ + b Because n x i n x i + i j x ix j, it comes that: E [ L N t N t N t E L i + L i L j i0 i j E [N t E [ L i + E [Nt N t E [L i L j E [N t E [ L i + E [ N t E [Nt E [L i E [L j 5

6 We have: E [N t λ E [ Nt var N t + E [N t λ + λ E [ L i var L i + E [L i e µ+ e E [L i E [L j e µ+ e µ+ e µ+ and: E [ L λe [ L i + λ + λ λ E [L i E [L j λe [ L i + λ E [L i E [L j λe µ+ + λ e µ+ var L E [ L E [L λe µ+ + λ e µ+ λ e µ+ λe µ+ + e µ+ e µ+ c We have: { It comes that: and: var L E [L E [L λe µ+ var L λe µ+ λeµ+ λ e µ+ e λ ln λ + ln var L ln E [L µ ln E [L ln λ ln E [L + ln E [L 3 ln λ ln var L Let ˆλ be an estimated value of λ. We finally obtain: and ˆµ ln m + ln m 3 ln ˆλ ln V ˆ ln ˆλ + ln V ln m where m and V are the empirical mean and variance of aggregated losses. 8 Copula functions. a If P is a lower triangular matrix, then the matrix Σ may be decomposed as: Σ P P 6

7 This is the Cholesky decomposition. In our case, we have: 0 P ρ ρ We verify that: 0 P P ρ ρ ρ ρ Σ ρ 0 ρ The copula of U is the copula of the gaussian standardized vector X X, X with correlation ρ. Let n and n be two random drawing of N 0,. Using the Cholesky decomposition 4, we can simulate X by the following way: x n x ρn + ρ n We deduce that we can simulate U with the relations: u Φ x Φ n u Φ x Φ ρn + ρ n b We have: C u, u Φ Φ u, Φ u ; ρ Φ u Φ u ρx Φ ρ u 0 Φ ϕ x dx Φ u ρφ u du ρ because x Φ u, du ϕ x dx, Φ 0 and Φ Φ u u. The conditional copula function C is then equal to: C u, u C u, u u Φ u ρφ u Φ ρ c We know that Pr {U u } u and C u, u Pr {U u U u }. Because C U, and C u, U are two independent uniform random variables, we obtain the following algorithm: i. we simulate two independent uniform variates v and v ; ii. set u equal to v ; iii. set u equal to K u v where K u u C u, u. 4 We have N µ, Σ µ + P N 0, I. 7

8 If we apply this algorithm to the Normal copula, we obtain: Φ u ρφ u Φ v ρ It comes that: Φ u ρφ u + ρ Φ v u v u Φ ρφ v + ρ Φ v We see that this algorithm is a special case of the Cholesky algorithm if we take n Φ v and n Φ v. Whereas n and n are directly simulated in the Cholesky algorithm with a gaussian random generator, they are simulated using the inverse transform in the conditional distribution method. d We know that: e λ τ U [0, τ λ ln u τ λ ln u. a If ρ, the Normal copula becomes the upper Frechet copula. It means that τ and τ are comonotonic. Because U U, we have: or: We have E [τ τ /λ and: e λ τ e λ τ E [τ τ E λ τ λ τ [ λ τ τ λ λ var τ + E [τ λ λ λ λ + λ λ λ ρ τ, τ E [τ τ E [τ E [τ τ τ + λ λ λ λ λ λ 8

9 b If ρ, the Normal copula becomes the lower Frechet copula. It means that τ and τ are counter-monotonic. Because U U, we have: e λ τ e λ τ or: τ ln e λ τ The correlation is equal to if τ is a decreasing linear function of τ. The function f t λ ln e λ t is a decreasing function, but not a linear function. 9 Credit spreads. We have TR-GDR, page 47 : Let U S τ. We have U [0, and: λ ρ τ, τ > F t e λt S t e λt f t λe λt Pr {U u} Pr {S τ u} Pr { τ S u } S S u u We deduce that S τ U [0, TR-GDR, page 48. It comes that τ S U with U U [0,. Let u be a uniform random variate. Simulating τ is equivalent to transform u into t: t λ ln u. We have TR-GDR, pages : P 4 s N P + R N with s the spread, N the notional, P the premium leg and P + the protection leg. The quarterly payment of the premium leg explains the factor /4 in the formula of P. We deduce the flow chart given in Figure. 3. The ATM margin or spread is the value of s such that the CDS price is zero TR-GDR, page 40: P t E [P P + 0 We have the following triangle relation TR-GDR, page 40: s λ R 9

10 Figure : Flow chart from the viewpoint of the protection buyer 4sN euros every three months if the default does not occur { }} { 3M 6M 9M Y τ R N euros at the default time τ t 4. Let PD be the annual default probability. We have PD S e λ λ because λ is generally small λ 0%. 5. We have: PD λ s R PD % 67 bps 5% 0 Extreme value theory and stress-testing. See TR-GDR, page -9.. We have TR-GDR, pages 3-33: 3. See TR-GDR, page 39. G n x Pr {max X,..., X n x} Pr {X x,..., X n x} n Pr {X i x} x µ Φ 4. a An extreme value EV copula C satisfies the following relation: for all t > 0. n C u t, u t C t u, u b The product copula is an EV copula because: C u t, u t uu t t u u t [ C u, u t 0

11 c We have: C u t, u t u t θ u t θ min t t u θ u θ u θ u θ min C t u, u u tθ, utθ min u θ, uθ u θ, uθ t t d The upper tail dependence λ is defined as follows: λ lim u + u + C u, u u It indicates the probability to have an extreme in one direction knowing that we have already an extreme in the other direction. If λ 0, extremes are independent and the copula of extreme values is the product copula. If λ, extremes are comonotonic and the copula of extreme values is the upper Fréchet copula. Moreover, the upper tail dependence of the copula between the random variables is equal to the upper tail dependence of the copula between the extremes. e If θ > θ, we obtain using L Hospital s rule: u + u θ u θ min u θ, u θ λ lim u + u lim u + u + u θ u θ u θ u lim u + u + u θ u lim u θ u θ lim u + u θ + θ u θ θ If θ > θ, λ θ. We deduce that the upper tail dependence of the the Marshall-Olkin copula is min θ, θ. f If θ 0 or θ 0, λ 0. It comes that the copula of the extremes is the product copula. Extremes are then not correlated.