SOLUTIONS TO MATH68181 EXTREME VALUES AND FINANCIAL RISK EXAM

Transcription

1 SOLUTIONS TO MATH68181 EXTREME VALUES AND FINANCIAL RISK EXAM

2 Solutions to Question A1 a) The marginal cdfs of F X,Y (x, y) = [1 + exp( x) + exp( y) + (1 α) exp( x y)] 1 are F X (x) = F X,Y (x, ) = [1 + exp( x)] 1 and F Y (y) = F X,Y (, y) = [1 + exp( y)] 1. b) F X belongs to the Gumbel max domain of attraction since 1 F X (t + x) t w(f X ) 1 F X (t) = t 1 [1 + e t x ] 1 1 [1 + e t ] 1 = t 1 [1 e t x ] 1 [1 e t ] = e x. c) F Y belongs to the Gumbel max domain of attraction since 1 F Y (t + x) t w(f Y ) 1 F Y (t) = t 1 [1 + e t y ] 1 1 [1 + e t ] 1 = t 1 [1 e t y ] 1 [1 e t ] = e y. SEEN d) Use the rule a n = γ ( 1 F 1 X Inverting (1 n 1 ) ) and b n = F 1 X (1 n 1 ). From part (b), γ(t) = 1. F X (x) = [1 + exp( x)] 1 = 1 n 1, 1

3 we obtain ( F ) 1 X 1 n 1 = log(n 1). e) Use the rule c n = γ ( 1 F 1 Y Inverting we obtain (1 n 1 ) ) and d n = F 1 Y (1 n 1 ). From part (c), γ(t) = 1. F Y (y) = [1 + exp( y)] 1 = 1 n 1, ( F ) 1 Y 1 n 1 = log(n 1). g) The iting cdf is F X,Y (a n x + b n, c n y + d n ) n = [1 + exp ( x log(n 1)) + exp ( y log(n 1)) + (1 α) exp ( x y 2 log(n 1))] n n [ ] n exp ( x) exp ( y) exp ( x y) = (1 α) n n 1 n 1 (n 1) 2 = exp { exp( x) exp( y)}. h) Yes, the extremes are completely independent since F X,Y (a n x + b n, c n y + d n ) = exp { exp( x) exp( y)} n = n F X (a n x + b n ) n F Y (c n y + d n ). (5 marks) 2

4 Solutions to Question A2 C (u 1, u 2 ) is a valid copula if C (u, 0) = 0, C (0, u) = 0, C (1, u) = u, C (u, 1) = u, u 1 C (u 1, u 2 ) 0 and u 2 C (u 1, u 2 ) 0. SEEN a) for the copula defined by C (u 1, u 2 ) = [α (min (u 1, u 2 )) m + (1 α)u m 1 u m 2 ] 1/m, we have C (u, 0) = [α 0 + (1 α) 0] 1/m = 0, C (0, u) = [α 0 + (1 α) 0] 1/m = 0, C (1, u) = [αu m + (1 α)u m ] 1/m = u, { C (u 1, u 2 ) = u 1 C (u, 1) = [αu m + (1 α)u m ] 1/m = u, [α + (1 α)u m 2 ] 1/m, if u 1 u 2, (1 α)u m 1 1 u 2 [α + (1 α)u m 2 ] 1/m 1 0, if u 2 u 1 and { C (u 1, u 2 ) = u 2 (1 α)u 1 u m 1 2 [α + (1 α)u m 2 ] 1/m 1, if u 1 u 2, [α + (1 α)u m 2 ] 1/m, if u 2 u

5 b) for the copula defined by C (u 1, u 2 ) = max (u 1 + u 2 1, 0), we have C (u, 0) = max (u + 0 1, 0) = 0, C (0, u) = max (0 + u 1, 0) = 0, C (1, u) = max (1 + u 1, 0) = u, C (u, 1) = max (u + 1 1, 0) = u, and { 1, if u1 + u C (u 1, u 2 ) = 2 1, u 1 0, if u 1 + u 2 < 1 { 1, if u1 + u C (u 1, u 2 ) = 2 1, u 2 0, if u 1 + u 2 < c) for the copula defined by C (u 1, u 2 ) = min ( ) ( ) u a 1, u b 2 min u 1 a 1, u2 1 b, we have C (u, 0) = min (u a, 0) min ( u 1 a, 0 ) = 0, C (0, u) = min ( 0, u b) min ( 0, u 1 b) = 0, C (1, u) = min ( 1, u b) min ( 1, u 1 b) = u, C (u, 1) = min (u a, 1) min ( u 1 a, 1 ) = u, and C (u 1, u 2 ) = u 1 C (u 1, u 2 ) = u 2 1, if u a 1 u b 2 and u 1 a 1 u2 1 b, au1 a 1 u 1 b 2, if u a 1 u b 2 and u 1 a 1 > u2 1 b, (1 a)u a 1 u b 2, if u a 1 > u b 2 and u 1 a 1 u2 1 b, 0, if u a 1 > u b 2 and u 1 a 1 > u2 1 b 0, if u a 1 u b 2 and u1 1 a u 1 b 2, (1 b)u a 1u b 2, if u a 1 u b 2 and u1 1 a > u 1 b 2, bu 1 a 1 u b 1 2, if u a 1 > u b 2 and u1 1 a u 1 b 2, 1, if u a 1 > u b 2 and u 1 a 1 > u 1 b

6 { [ d) for the copula defined by C (u 1, u 2 ) = exp ( log u 1 ) θ + ( log u 2 ) θ] } 1/θ, we have { [ C (u, 0) = exp ( log u) θ + ( ) θ] } 1/θ = 0, { [ C (0, u) = exp ( ) θ + ( log u) θ] } 1/θ = 0, { [ C (1, u) = exp 0 + ( log u) θ] } 1/θ = u, { [ ] } 1/θ C (u, 1) = exp ( log u) θ + 0 = u, [ C (u 1, u 2 ) = u 1 1 ( log u 1 ) θ 1 ( log u 1 ) θ + ( log u 2 ) θ] 1/θ 1 u 1 { [ exp ( log u 1 ) θ + ( log u 2 ) θ] } 1/θ 0 and [ C (u 1, u 2 ) = u 1 2 ( log u 2 ) θ 1 ( log u 1 ) θ + ( log u 2 ) θ] 1/θ 1 u 1 { [ exp ( log u 1 ) θ + ( log u 2 ) θ] } 1/θ 0. 5

7 Solutions to Question A3 a) We can write F (x, y) = exp { [ (x + y) 1 (θ + φ) y x + y + ]} θy2 (x + y) + φy3 2 (x + y) 3 for x > 0, y > 0, θ 0, φ 0, θ + 3φ 0, θ + φ 1 and θ + 2φ 1. This is in the form of [ ( )] y F (x, y) = exp (x + y)a x + y with A(w) = 1 (θ + φ)w + θw 2 + φw 3. We now check the conditions for A( ). Clearly, A(0) = 1 and A(1) = 1. Also A(t) 0 since θ + φ 1 implies 1 (θ + φ)w 1 for all w. Also A(w) 1 since A(w) 1 1 (θ + φ)w + θw 2 + φw 3 1 θ ( w 2 w ) + φ ( w 3 w ) 0 θw (w 1) + φw ( w 2 1 ) 0 (θ + φ + φw) w (w 1) 0. Note that max(w, 1 w) = w if w [1/2, 1]. So, for w [1/2, 1], A(w) max(w, 1 w) A(w) w 1 (θ + φ + 1)w + θw 2 + φw 3 0. Let g(w) = 1 (θ + φ + 1)w + θw 2 + φw 3. Note that g (w) = 2θw + 3φw 2 θ φ 1 0 for all w [1/2, 1]. But g(1) = 0 0, so A(w) max(w, 1 w) for all w [1/2, 1]. Note that max(w, 1 w) = 1 w if w [0, 1/2]. So, for w [0, 1/2], which holds since θ + φ 1. A( ) is convex since A(w) max(w, 1 w) A(w) 1 w (1 θ φ)w + θw 2 + φw 3 0, A (w) = 2θw + 3φw 2 θ φ 6

8 and A (t) = 2θ + 6φw 0 for all w since θ + 3φ 0. (6 marks) b) the joint cdf is F (x, y) = 1 exp( x) exp( y) + exp { [ (x + y) 1 (θ + φ) y x + y + ]} θy2 (x + y) + φy3. 2 (x + y) 3 c) the derivative of joint cdf with respect to x is [ ] F (x, y) θy 2 = exp( x) + F (x, y) x (x + y) + 2φy3 2 (x + y) 1, 3 so the conditional cdf if Y given X = x is [ ] θy 2 F (y x) = 1 + exp(x)f (x, y) (x + y) + 2φy3 2 (x + y) 1. 3 d) the derivative of joint cdf with respect to y is [ F (x, y) 2φy 3 (θ 3φ)y2 = exp( y) + F (x, y) + 2θy ] y (x + y) 3 (x + y) 2 x + y + θ + φ 1, so the conditional cdf if X given Y = y is [ 2φy 3 (θ 3φ)y2 F (x y) = 1 + exp(y)f (x, y) + 2θy ] (x + y) 3 (x + y) 2 x + y + θ + φ 1. 7

9 e) the derivative of joint cdf with respect to x and y is F (x, y) f(x, y) = x y [ ] [ θy 2 = F (x, y) (x + y) + 2φy3 2φy 3 2 (x + y) 1 (θ 3φ)y2 + 2θy ] 3 (x + y) 3 (x + y) 2 x + y + θ + φ 1 [ ] 2θy 2(3φ θ)y2 +F (x, y) + 6θy3. (x + y) 2 (x + y) 3 (x + y) 4 8

10 Solutions to Question B1 a) The mgf of X i is M Xi (t) = 0 λ exp [ (λ t)x] dx = [ ] λ exp [ (λ t)x] = 0 λ t λ 0 t λ = λ λ t. SEEN b) The mgf of T conditional on N = n is M T N=n (t) = E {exp [t (X X n )]} = E {exp [tx 1 ] exp [X n ]} = E {exp [tx 1 ]} E {exp [X n ]} = M X1 (t) M Xn (t) ( ) n λ = λ t c) ( λ λ t) n is the mgf of a gamma random variable with parameters λ and n. So, the conditional distribution of T is gamma with parameters λ and n. d) The conditional pdf of T is f T N=n (x) = λn x n 1 exp( λx). Γ(n) 9

11 So, the unconditional pdf of T is f T (x) = f T N=n (x)θ(1 θ) n 1 n=1 = λ n x n 1 exp( λx) θ(1 θ) n 1 Γ(n) n=1 = (λ(1 θ)x) n 1 θλ exp( λx) (n 1)! n=1 = θλ exp( λx) exp [λx(1 θ)] = θλ exp( θλx), an Exponential pdf with parameter θλ. e) The mean is 1/(θλ) and the variance 1/(θλ) 2. f) The unconditional cdf of T is F T (x) = 1 exp( θλx). Inverting 1 exp( θλx) = p, we obtain VaR p (T ) = 1 log(1 p). θλ 10

12 g) The expected shortfall T is ES p (T ) = 1 θλp = 1 θλp = 1 θλp = 1 θλp p log(1 t)dt 0 p {[t log(1 t)] p0 + 0 p { p log(1 p) + 0 { p log(1 p) p + } t 1 t dt } t dt 1 t p } 1 1 t dt = 1 θλp {p log(1 p) p + [ log(1 t)]p 0 } = 1 {p log(1 p) p log(1 p)}. θλp 0 11

13 Solutions to Question B2 If there are norming constants a n > 0, b n and a nondegenerate G such that the cdf of a normalized version of M n converges to G, i.e. ( ) Mn b n Pr x = F n (a n x + b n ) G(x) (1) a n as n then G must be of the same type as (cdfs G and G are of the same type if G (x) = G(ax + b) for some a > 0, b and all x) as one of the following three classes: I : Λ(x) = exp { exp( x)}, x R; { 0 if x < 0, II : Φ α (x) = exp { x α } if x 0 for some α > 0; { exp { ( x) III : Ψ α (x) = α } if x < 0, 1 if x 0 for some α > 0. SEEN (6 marks) The necessary and sufficient conditions for the three extreme value distributions are: 1 F (t + xγ(t)) I : γ(t) > 0 s.t. t w(f ) 1 F (t) II : w(f ) = and t 1 F (tx) 1 F (t) = x α, x > 0, = exp( x), x R, III : w(f ) < and t 0 1 F (w(f ) tx) 1 F (w(f ) t) = xα, x > 0. SEEN (6 marks) First, suppose that G belongs to the max domain of attraction of the Gumbel extreme value distribution. Then, there must exist a strictly positive function say h(t) such that 1 G (t + xh(t)) t w(g) 1 G(t) = e x 12

14 for every x (, ). But 1 F (t + xh(t)) t W (F ) 1 F (t) { 1 {1 [1 G (t + xh(t))] a } b = t w(f ) 1 {1 [1 G (t)] a } b { } aθ 1 G (t + xh(t)) = t w(f ) 1 G (t) = e aθx } θ for every x (, ), assuming w(f ) = w(g). So, it follows that F also belongs to the max domain of attraction of the Gumbel extreme value distribution with ( ) P Mn b n x = exp [ exp( aθx)] n for some suitable norming constants a n > 0 and b n. a n Second, suppose that G belongs to the max domain of attraction of the Fréchet extreme value distribution. Then, there must exist a β > 0 such that for every x > 0. But 1 F (tx) t 1 F (t) 1 G (tx) t 1 G(t) = x β { 1 {1 [1 G (tx)] a } b = t 1 {1 [1 G (t)] a } b { } aθ 1 G (tx) = t 1 G (t) = x βaθ for every x > 0. So, it follows that F also belongs to the max domain of attraction of the Fréchet extreme value distribution with ( ) P Mn b n x = exp ( x βaθ) n for some suitable norming constants a n > 0 and b n. a n } θ Third, suppose that G belongs to the max domain of attraction of the Weibull extreme value distribution. Then, there must exist a β > 0 such that 1 G (w(g) tx) t 0 1 G (w(g) t) = xβ 13

15 for every x > 0. But But t 0 1 F (w(f ) tx) 1 F (w(f ) t) = t 0 = t = x βaθ { 1 {1 [1 G (w(f ) tx)] a } b 1 {1 [1 G (w(f ) t)] a } b { } aθ 1 G (w(f ) tx) 1 G (w(f ) t) } θ for every x < 0, assuming w(f ) = w(g). So, it follows that F also belongs to the max domain of attraction of the Weibull extreme value distribution with ( ) P Mn b n x = exp ( ( x) βaθ) n for some suitable norming constants a n > 0 and b n. a n 14

16 Solutions to Question B3 a) Note that w(f ) = 1. Then t 0 1 F (1 tx) 1 F (1 t) = t 0 xf (1 tx) f(1 t) = t 0 x (1 tx) α 1 (tx) β 1 (1 t) α 1 t β 1 = x β. So, f(x) = Cx α 1 (1 x) β 1 belongs to the Weibull domain of attraction. b) Note that and So, 1/2, if k = 1, 0, F (k) = 1, if k 1, 0, otherwise 1, if k 0, 1 F (k 1) = 1/2, if k = 1, 2, 0, otherwise. p(k) 1 F (k 1) = Hence, there can be no non-degenerate it. 1/2, if k = 1, 1, if k = 1, 0, if k 2, k 1, 1, undefined, otherwise. c) Note that w(f ) =. Then 1 F (tx) t 1 F (t) = t xf (tx) f(t) = t x (1 + t 2 x 2 ) 1 (1 + t 2 ) 1 = x 1. 15

17 So, f(x) = π 1 (1 + x 2 ) 1 belongs to the Fréchet domain of attraction. d) Note that w(f ) =. It is easy to show that the corresponding cdf is { 0.5e x, if x < 0, F (x) = 1 0.5e x, if x 0. Take g(t) = 1. Then 1 F (t + xg(t)) t 1 F (t) e t x = t e t = e x. So, f = 0.5e x belongs to the Gumbel domain of attraction. e) Note that w(f ) =. Then 1 F (tx) t 1 F (t) = 1 exp ( (tx) 1 ) t 1 exp ( t 1 ) = t 1 (1 (tx) 1 ) 1 (1 t 1 ) (tx) 1 = = x 1. t t 1. So, the cdf F (x) = exp ( x 1 ) belongs to the Fréchet domain of attraction. 16

18 Solutions to Question B4 (a) If X is an absolutely continuous random variable with cdf F ( ) then VaR p (X) = F 1 (p) and ES p (X) = 1 p p 0 F 1 (v)dv. SEEN (b) (i) The corresponding cdf is for x > K; F (x) = x K ak a y a 1 dy = [ K a y a] x K = 1 Ka x a SEEN (b) (ii) Inverting we obtain VaR p (X) = K(1 p) 1/a. F (x) = 1 K a x a = p, SEEN (b) (iii) The expected shortfall is ES p (X) = K p p 0 (1 v) 1/a dv = K p [ (1 v) 1 1/a 1 1 a ] p 0 = ak ] [(1 p) 1 1 a 1. p(1 a) SEEN c) (i) The joint likelihood function of a and K is L (a, K) = n [ ak a x a 1 i I {x i K} ] ( n ) a 1 = a n K na x i I {min x i K}. 17

19 (1 marks) c) (ii) Note that L (a, K) is an increasing function over K [0, min x i ] and is zero elsewhere. So, the mle of K is K = min x i. c) (iii) The log-likelihood function corresponding to L (a, K) is So, log L (a, K) = n log a + na log K (a + 1) Setting this to zero gives d log L da n log x i + log I {min x i K}. = n n a + n log K log x i. [ n ] 1 â = n log x i n log min x i. c) (iv) The mle of VaR is VaR 0 (X) = K = min x i. By L Hopital s rule, the MLE of ES is also ÊS 0(X) = K = min x i. c (v) Let Z = min x i. The cdf of Z is F Z (z) = Pr (Z z) = 1 Pr (Z > z) = 1 Pr (min X i > z) = 1 Pr n (X > z) = 1 for z > K. The corresponding pdf is f Z (z) = nak na z na 1 18 ( ) na K z

20 for z > K. Z is a Pareto random variable. The expected value of Z is [ z E(Z) = nak na z na dz = nak na 1 na So, K and hence VaR0 (X), ÊS 0(X) are biased. K 1 na ] K. 19

21 Solutions to Question B5 a) The cdf of Y is for y > 0. F Y (y) = Pr(Y y) = Pr (max (X 1,..., X m ) y) = Pr (X 1 y,..., X m y) = Pr (X 1 y) Pr (X m y) ( ) ( ) y a y a = b a b a ( ) m y a = b a b) The corresponding pdf is for y > 0. f Y (y) = m b a ( ) m 1 y a. b a c) The nth moment of Y can be calculated as b E (Y n ) = y n m ( ) m 1 y a dy b a b a a b = m(b a) m (y a + a) n (y a) m 1 dy a b n ( ) n = m(b a) m a n k (y a) k+m 1 dy a k k=0 n ( n b = m(b a) )a m n k (y a) k+m 1 dy k k=0 a n ( [ ] n (y a) = m(b a) )a m n k k+m b k k + m k=0 a n ( n = m(b a) m n k (b a)k+m )a k k + m. k=0 20

22 So, E (Y ) = m(b a) m 1 k=0 1 k (b a)k+m a k + m and V ar (Y ) = m(b a) m 2 k=0 ( 2 2 k (b a)k+m )a k k + m E2 (Y ). d) Setting ( ) m y a = p b a gives VaR p (Y ) = a + (b a)p 1/m. e) The expected shortfall is p ES p (Y ) = 1 ( ) a + (b a)v 1/m dv p 0 = 1 ] p [av + (b a) v1+1/m p 1 + 1/m 0 = 1 ] [ap + (b a) p1+1/m p 1 + 1/m p 1/m = a + (b a) 1 + 1/m. 21

23 f) The likelihood and log likelihood functions are [ n ( ) m 1 m yi a L(a, b) = I {a < y i < b}] b a b a [ n ] m 1 n = m n (b a) mn (y i a) I {a < y i < b} [ n m 1 = m n (b a) mn (y i a)] I {min y i > a, max y i < b} and n log L(a, b) = n log m mn log(b a) + (m 1) log (y i a) + n log I {min y i > a, max y i < b}. The likelihood function is a decreasing function of b, so it MLE is max y i. derivative of the log-likelihood with respect to a is The partial log L(a, b) a = mn n b a (m 1) 1 y i a. So, the MLE of a is the root of mn n b a = (m 1) 1 y i a. (6 marks) 22