SOLUTIONS TO MATH68181 EXTREME VALUES AND FINANCIAL RISK EXAM

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1 SOLUTIONS TO MATH68181 EXTREME VALUES AND FINANCIAL RISK EXAM

2 Solutions to Question A1 a) The joint cdf of X Y is F X,Y x, ) x 0 0 4uv + u + v + 1 dudv 4 u v + u + uv + u ] x dv x v + x + xv + x ] dv x v + x v + xv + xv ] x + x + x + x ] 4 3 marks) b) The marginal cdfs of X Y are F X x) F X,Y x, 1) 1 4 F Y ) F X,Y 1, ) 1 4 x + x + x + x ] ] xx + 1) + 1) marks) c) First note that w F X ) 1 F X belongs to the Weibull max domain of attraction since 1 F X 1 tx) lim t 0 1 F X 1 t) 1 1 tx) tx) t t) t) 1 tx) tx) t 0 1 t) t) 3tx t x t 0 3t t 3x tx t 0 3 t x 1

3 marks) d) First note that w F Y ) 1 F Y belongs to the Weibull max domain of attraction since 1 F Y 1 t) lim t 0 1 F Y 1 t) 1 1 t) t) t t) t) 1 t) t) t 0 1 t) t) 3t t t 0 3t t 3 t t 0 3 t marks) e) Use the formulas a n w F X ) F 1 X gives which has the valid root F X x) 1 1 n) bn 1 Inverting xx + 1) x + x p 0, p So, Hence, b n 1 x p F 1 X 1 1 ) ) 1 1 n n a n n n

4 marks) f) Use the formulas c n w F Y ) F 1 Y gives which has the valid root F Y ) 1 1 n) dn 1 Inverting + 1) + p 0, p So, Hence, d n p F 1 Y 1 1 ) ) 1 1 n n c n n n marks) g) The limiting cdf is lim F X,Y a n x + b n, c n + d n ) n 1 an x + 1) a n 4 n n + 1) + a n x + 1) a n + 1) + a n x + 1) a n + 1) + a n x + 1) a n + 1) ] n 1 n 4 a nx + 1) n a n n + 1) n 1 + a n + 1) 1 + a n x + 1) 1 + a n x + 1) 1 a n + 1) 1] n Now note that a n n n + ) 3n + 3

5 as n Hence, lim F X,Y a n x + b n, c n + d n ) n ) n ) n 1 n 4 n 3n x + 1 3n an + 1) 1 + a n x + 1) 1 + a n x + 1) 1 a n + 1) 1] n ) ) 1 4x 4 1 n 4 exp exp ) ) ) ) 1] n n 3 3 ) ) 1 4x 4 n 4 exp exp 4 n n 3 3 exp 4x ) h) Yes, the extremes are completel independent since 4x lim F X,Y a n x + b n, c n + d n ) exp n ) 3 ) ) 4x 4 exp exp 3 3 F X a n x + b n ) lim F Y c n + d n ) n n 5 marks) marks) 4

6 Solutions to Question A C u 1, u ) 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 ) 0 u C u 1, u ) 0 4 marks) SEEN a) for the copula defined b C u 1, u ) min C u, 0) min 0, 0) 0, ) u1 1 α u, u 1 u 1 β, we have C 0, u) min 0, 0) 0, C 1, u) min u, u 1 β) u, C u, 1) min u 1 α, u ) u, u 1 C u 1, u ) u C u 1, u ) u 1 u { u 1 α 1 u, if u α 1 u β u 1 u 1 β, if u α 1 > u β, { u 1 α 1 u, if u α 1 u β u 1 u 1 β, if u α 1 > u β, { 1 α)u α 1 u, if u α 1 u β u 1 β, if u α 1 > u β, { u 1 α 1, if u α 1 u β 1 β)u 1 u β, if u α 1 > u β, 0 0 5

7 4 marks) n, b) for the copula defined b C u 1, u ) u max 1 u 1 ) 1/n + u 1/n 1, 0)] we have n C u, 0) 0 max 1 u) 1/n + 0 1, 0)] 0, C 0, u) u C 1, u) u C u, 1) 1 C u 1, u ) u 1 u 1 0 { { C u 1, u ) u u 1 n max 1 0) 1/n + u 1/n 1, 0)] u u 0, n max 1 1) 1/n + u 1/n 1, 0)] u 0 u, n max 1 u) 1/n + 1 1, 0)] 1 1 u) u, u ] n 1 u 1 ) 1/n + u 1/n 1, if 1 u1 ) 1/n + u 1/n 1, u, if 1 u 1 ) 1/n + u 1/n < 1 ] n 1 1 u 1 ) 1/n + u 1/n 1 1 u1 ) 1/n 1, if 1 u 1 ) 1/n + u 1/n 1, 0, if 1 u 1 ) 1/n + u 1/n < 1 u ] n 1 u 1 ) 1/n + u 1/n 1, if 1 u1 ) 1/n + u 1/n 1, u, if 1 u 1 ) 1/n + u 1/n < 1 ] n 1 1 u 1 ) 1/n + u 1/n 1/n 1 1 u, if 1 u 1 ) 1/n + u 1/n 1, 1, if 1 u 1 ) 1/n + u 1/n < 1 )] n 1 1 u 1/n 1 u 1 ) 1/n + u 1/n 1, if 1 u1 ) 1/n + u 1/n 1, 1, if 1 u 1 ) 1/n + u 1/n < 1 0 ] c) for the copula defined b C u 1, u ) log α 1 + αu 1 1)α u 1), we have α 1 ] C u, 0) log α 1 + αu 1) 1 1) α 1 6 0, 4 marks)

8 C 0, u) log α ) ] αu 1) 0, α 1 C 1, u) log α 1 + α 1) ] αu 1) u log α 1 α α u, ] C u, 1) log α 1 + αu 1) α 1) u log α 1 α α u, C u 1, u ) 1 + αu 1 1) α u ] 1 1) α u 1 α u 1) u 1 α 1 α 1 0 C u 1, u ) 1 + αu 1 1) α u ] 1 1) α u α u 1 1) u α 1 α marks) d) for the copula defined b { u θ 1 1 ) δ + u θ 1 ) δ ] 1/δ + 1 } 1/θ, we have C u, 0) { u θ 1 ) δ + 0 θ 1 ) δ ] 1/δ + 1 } 1/θ 0, C 0, u) { 0 θ 1 ) δ + u θ 1 ) δ ] 1/δ + 1 } 1/θ 0, C 1, u) { 1 1) δ + u θ 1 ) δ ] 1/δ + 1 } 1/θ u, C u, 1) { u θ 1 ) δ + 1 1) δ ] 1/δ + 1 } 1/θ u, u 1 C u 1, u ) { u θ 1 1 ) δ + u θ 1 ) δ ] 1/δ + 1 } 1/θ 1 u θ 1 1 ) δ + u θ 1 ) ] δ 1/δ 1 u θ 1 1 ) δ 1 u θ

9 u C u 1, u ) { u θ 1 1 ) δ + u θ 1 ) δ ] 1/δ + 1 } 1/θ 1 u θ 1 1 ) δ + u θ 1 ) ] δ 1/δ 1 u θ 1 ) δ 1 u θ marks) 8

10 Solutions to Question A3 a) We can write for x > 0 > 0, where )] Gx, ) exp x + )A x + Aw) A 1 w), if 0 w w 1, A w), if w 1 < w w, A p w), if w p 1 w 1 We now check the conditions for A ) Clearl, A0) A 1 0) 1 A1) A p 1) 1 Also At) 0 since A i w) 0 for all w ever k Also since each A i w) 1, Aw) 1 Also since each A i w) maxw, 1 w), Aw) maxw, 1 w) Also since each A i w) is convex, A 1w), if 0 w w 1, A w), if w A 1 < w w, w) A pw), if w p 1 w marks) b) Note that Gx, 0) exp { x + 0)A 1 0 exp x) 9

11 G0, ) exp { 0 + )A p 1 exp ) So, the joint cdf is 1 exp x) exp ) + exp { x + )A 1 1 exp x) exp ) + exp { x + )A Gx, ) 1 exp x) exp ) + exp { x + )A p x+ x+ x+, if 0 x+ w 1,, if w 1 < x+ w,, if w p 1 x+ 1 c) the derivative of joint cdf with respect to x is ) exp x) + x+ A 1 A x+ 1 ) Gx, ) exp x) + x+ A A x+ x ) exp x) + A p x+ A p x+ x+ x+ x+ )] exp { x + )A 1 )] exp { x + )A )] exp { x + )A p so the conditional cdf of Y given X x is ) )] 1 + x+ A 1 A x+ 1 exp {x x + )A x+ 1 ) )] 1 + x+ G x) A A x+ exp {x x + )A x+ ) )] 1 + A p exp {x x + )A p x+ A p x+ x+ x+ x+ x+ x+ x+ x+ 1 marks), if 0 x+ w 1,, if w 1 < x+ w,, if w p 1 x+ 1,, if 0 x+ w 1,, if w 1 < x+ w,, if w p 1 x+ 1 d) the derivative of joint cdf with respect to is ) x exp ) + x+ A 1 + A x+ 1 ) Gx, ) x exp ) + x+ A + A x+ ) exp ) + + A p x x+ A p x+ 10 x+ x+ x+ )] exp { x + )A 1 )] exp { x + )A )] exp { x + )A p x+ x+ x+ 4 marks), if 0 x+ w 1,, if w 1 < x+ w,, if w p 1 x+ 1,

12 so the conditional cdf of X given Y is ) )] x 1 + x+ A 1 + A x+ 1 exp { x + )A x+ 1 ) )] x 1 + x+ Gx ) A + A x+ exp { x + )A x+ ) )] A p exp { x + )A p x x+ A p x+ x+ x+ x+ x+, if 0 x+ w 1,, if w 1 < x+ w,, if w p 1 x+ 1 e) the derivative of joint cdf with respect to x is { ) )] ) A 1 + A 1 x+ A 1 x+ A x+ x+ x+ x+ x x+ A 1 x x+ A x+ x+ x+ exp { x + )A 1, if 0 w x+ x+ 1, { ) )] ) )] A + A gx, ) exp { x + )A 1 { x+ A p x+ x+ ) A p exp { x + )A 1 x+ x+ x+, if w 1 < x+ w, )] x x+ A p x+ ) + A p x+ )], if w p 1 x+ 1 )] + x A x+) 1 + x A x+) + x A x+) p 4 marks) x+ x+ x+ 4 marks) 11

13 Solutions to Question B1 a) The cumulative distribution function of T conditional on N n is Pr T t N n) Pr max X 1,, X N ) t N n) Pr max X 1,, X n ) t N n) Pr X 1 t,, X n t) Pr X 1 t) Pr X n t) 1 + exp t)] exp t)] exp t)] n 4 marks) b) The unconditional cumulative distribution function of T is PrT t) Pr T t N n) θ1 θ) n 1 n1 1 + exp t)] n θ1 θ) n 1 n1 θ 1 + exp t)] exp t)] 1 n 1 θ) n 1 n1 θ 1 + exp t)] 1 { exp t)] 1 1 θ) } 1 θ θ + exp t) 4 marks) c) The unconditional probabilit densit function of T is f T t) θ exp t) θ + exp t)] 1 marks) 1

14 d) The moment generating function of T is M T s) where we have set θ θ expst t) θ + exp t)] dt expst t) θ + exp t)] dt 1 1 θ s 1+s 1 ) 1 s ) d 1 1 ] θ s 1+s 1 ) s d s 1 ) 1 s d 0 θ s B + s, 1 s) B1 + s, s)], θ θ+exp t) used the definition of the beta function 0 4 marks) e) Setting θ θ + exp t) p solving for t, we obtain value at risk of T as VaR p T ) log θ 1 p ] p 3 marks) f) The expected shortfall T is ES p T ) 1 p p 0 log θ log1 u) + log u] du p log θ 1 log1 u)du + 1 log udu p 0 p 0 log θ 1 p } {u log1 u)] p0 p + u 0 1 u du + 1 p {u log u] p0 p 0 log θ 1 { p } u p log1 p) + p 0 1 u du + 1 {p log p p} p p log θ 1 {p log1 p) p log1 p + log p 1 p log θ + log p 1 p log1 p) + p 13 } 1du

15 4 marks) 14

16 Solutions to Question B If there are norming constants a n > 0, b n a nondegenerate G such that the cdf of a normalized version of M n converges to G, ie ) Mn b n Pr x F n a n x + b n ) Gx) 1) a n as n then G must be of the same tpe as cdfs G G are of the same tpe if G x) Gax + b) for some a > 0, b all x) as one of the following three classes: I : Λx) 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 4 marks) The necessar sufficient conditions for the three extreme value distributions are: 1 F t + xγt)) I : γt) > 0 st lim t wf ) 1 F t) II : wf ) lim t 1 F tx) 1 F t) x α, x > 0, exp x), x R, III : wf ) < lim t 0 1 F wf ) tx) 1 F wf ) t) xα, x > 0 SEEN 4 marks) First, suppose that G belongs to the max domain of attraction of the Gumbel extreme value distribution Then, there must exist a strictl positive function sa ht) such that 1 G t + xht)) lim t wg) 1 Gt) e x 15

17 for ever x > 0 But 1 F t + xht)) lim t wf ) 1 F t) t wf ) t wg) t wg) 1 + xh t) ) f t + xht)) ft) 1 + xh t) ) { g t + xht)) 1 G t + xht))] λb G t + xht))] λ} a xh t) ) g t + xht)) gt) 1 G t + xht)) t wg) 1 Gt) g t) 1 G t)] λb 1 { 1 1 G t)] λ} a 1 ] λb 1 1 G t + xht)) {1 1 G t + xht))] λ 1 G t) 1 1 G t)] λ ] } λb 1 1 G t + xht)) {1 1 G t + xht))] λ a 1 1 G t) ] λb 1 G t + xht)) {1 1 G t + xht))] λ t wg) 1 G t) 1 1 G t)] λ ] λb { 1 G t + xht)) 1 1 λg t + xht))] t wg) 1 G t) 1 1 λg t)] ] λb { } a 1 1 G t + xht)) G t + xht)) t wg) 1 G t) G t) ] λb 1 G t + xht)) t wg) 1 G t) exp λbx) 1 1 G t)] λ } a 1 } a 1 } a 1 for ever x > 0, assuming wf ) wg) So, it follows that F also belongs to the max domain of attraction of the Gumbel extreme value distribution with ) lim P Mn b n x exp exp λbx)] n for some suitable norming constants a n > 0 b n a n 4 marks) 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 1 G tx) lim t 1 Gt) x β 16

18 for ever x > 0 But 1 F tx) lim t 1 F t) xf tx) t ft) { λb 1 xg tx) 1 G tx)] 1 1 G tx)] λ} a 1 { t g t) 1 G t)] λb G t)] λ} a 1 t xg tx) gt) t 1 G tx) 1 Gt) 1 G tx) 1 G t) 1 G tx) 1 G t) ] λb 1 {1 1 G tx)] λ ] λb 1 G tx) {1 1 G tx)] λ t 1 G t) 1 1 G t)] λ ] λb { 1 G tx) 1 1 λg tx)] t 1 G t) 1 1 λg t)] ] λb { } a 1 1 G tx) G tx) t 1 G t) G t) t x λbβ 1 G tx) 1 G t) ] λb } a G t)] λ ] λb 1 {1 1 G tx)] λ 1 1 G t)] λ } a 1 } a 1 } a 1 for ever x > 0 So, it follows that F also belongs to the max domain of attraction of the Fréchet extreme value distribution with ) lim P Mn b n x exp x λbβ) n for some suitable norming constants a n > 0 b n a n 4 marks) 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 wg) tx) lim t 0 1 G wg) t) xβ 17

19 for ever x > 0 1 F wf ) tx) lim t 0 1 F wf ) t) xf wf ) tx) t 0 f wf ) t) { λb 1 xg wf ) tx) 1 G wf ) tx)] 1 1 G wf ) tx)] λ} a 1 { t 0 g wf ) t) 1 G wf ) t)] λb G wf ) t)] λ} a 1 t 0 xg wf ) tx) g wf ) t) t 0 1 G wf ) tx) 1 G wf ) t) t 0 t 0 t 0 t 0 x λbβ 1 G wf ) tx) 1 G wf ) t) 1 G wf ) tx) 1 G wf ) t) ] λb 1 G wf ) tx) {1 1 G wf ) tx)] λ 1 G wf ) t) 1 1 G wf ) t)] λ ] λb { 1 G wf ) tx) 1 1 λg wf ) tx)] 1 G wf ) t) 1 1 λg wf ) t)] ] λb { } a 1 1 G wf ) tx) G wf ) tx) 1 G wf ) t) 1 G wf ) tx) 1 G wf ) t) ] λb G wf ) t) ] λb 1 {1 1 G wf ) tx)] λ } a G wf ) t)] λ ] λb 1 {1 1 G wf ) tx)] λ 1 1 G wf ) t)] λ } a 1 } a 1 } a 1 for ever x < 0, assuming wf ) wg) So, it follows that F also belongs to the max domain of attraction of the Weibull extreme value distribution with ) lim P Mn b n x exp x) λbβ) n for some suitable norming constants a n > 0 b n a n 4 marks) 18

20 Solutions to Question B3 a) Note that wf ) n Pr X wf )) 1 F wf ) 1) Pr X n) 1 F n 1) Pr X n) 1 Pr X 1) Pr X ) Pr X n 1) Pr X n) Pr X n) 1 Hence, there can be no non-degenerate limit 4 marks) b) Note that wf ) minn, K) Pr X wf )) 1 F wf ) 1) Pr X minn, K)) 1 F minn, K) 1) Pr X minn, K)) 1 Pr X max0, n + K N)) Pr X max0, n + K N) + 1) Pr X minn, K) 1 Pr X minn, K)) Pr X minn, K)) 1 Hence, there can be no non-degenerate limit 4 marks) 19

21 c) Note that wf ) Note that 1 F t + xht)) lim t 1 F t) t t t t exp x) 1 + xh t) ) f t + xht)) f t) 1 + xh t) ) exp b t + xht)) η exp bt + bxht))] ) 1 + xh t) ) 1 + xh t) exp bt η expbt)] exp {bxht) + η expbt) 1 exp bxht))]} exp {bxht) η expbt)bxht 1 if ht) So, F x) belongs to the Gumbel domain of attraction ηb expbt) 4 marks) d) Note that wf ) Then 1 F tx) lim t 1 F t) xf tx) t ft) ) xtx) α 1 exp β tx t t α 1 exp ) β t xtx) α 1 t t α 1 x α So, F x) belongs to the Fréchet domain of attraction 4 marks) e) Note that wf ) Then 1 F t + xht)) lim t 1 F t) t exp at axht))] b exp at)] b 1 1 b exp at axht))] t 1 1 b exp at)] exp at axht)) t exp at) exp axht)) t exp x) 0

22 if ht) 1 So, the cdf F x) belongs to the Gumbel domain of attraction a 4 marks) 1

23 Solutions to Question B4 a) If X is an absolutel continuous rom variable with cdf F ) then VaR p X) F 1 p) ES p X) 1 p p 0 F 1 v)dv marks) SEEN b) i) T is a N µ 1, σ1)+ +N µ k, σk ) N µ µ k, σ1 + + σk ) rom variable; marks) b) ii) Inverting Φ t µ 1 µ k σ σ k ) p, we obtain VaR p T ) µ µ k + σ σ k Φ 1 p) marks) b) iii) The expected shortfall is ES p T ) 1 p ] µ µ k + σ1 + + σk p Φ 1 v) dv 0 µ µ k + σ1 + + σk 1 p Φ 1 v)dv p 0 marks)

24 c) i) The joint likelihood function of µ 1, µ,, µ k σ1, σ,, σk is L ) µ 1, µ,, µ k, σ1, σ,, σk { ]} k n 1 exp πσi i1 j1 { k i1 1 π) n σ n i 1 π) nk σ n 1 σ n k X i,j µ i ) σi ]} exp 1 X i,j µ i ) exp σ i 1 j1 k i1 1 σ i ] X i,j µ i ) j1 marks) c) ii) The log likelihood function is log L nk logπ) n k log σ i 1 i1 k 1 σ i1 i X i,j µ i ) j1 The partial derivatives are log L µ i 1 σ i X i,j µ i ) 1 σi j1 ) ] X i,j nµ i j1 log L σ i n σ i + 1 σ 3 i X i,j µ i ) j1 Setting these to zero solving, we obtain µ i 1 n j1 X i,j σ i 1 n X i,j µ i ) j1 4 marks) 3

25 c) iii) µ i is unbiased consistent since E µ i ) 1 n E X i,j ) 1 n j1 µ i µ i j1 V ar µ i ) 1 n V ar X i,j ) 1 n j1 j1 σ i σ i n marks) c) iv) σ i is unbiased consistent since E σ ) ] i σ i n E 1 X i,j µ i ) σ i j1 σ i n E ] χ n 1)σ n 1 i n V ar σ i ) V ar σ i n 1 σ i ] X i,j µ i ) j1 σ4 i n V ar 1 σ i ] X i,j µ i ) j1 σ4 i n V ar ] χ σ 4 n 1 i n 1) n marks) c v) The maximum likelihood estimators of VaR p T ) ES p T ) are VaR p T ) 1 n k X i,j + 1 n i1 j1 k X i,j µ i ) Φ 1 p) i1 j1 ÊS p T ) 1 n k X i,j + 1 n i1 j1 k i1 j1 X i,j µ i ) 1 p p 0 Φ 1 v)dv marks) 4

26 Solutions to Question B5 a) Note that for t > min a 1, a,, a k ) F T t) PrT t) 1 PrT > t) 1 Pr min X 1, X,, X k ) > t) 1 Pr X 1 > t, X > t,, X k > t) 1 F t, t,, t) t 1 max, a 1 1 max t a,, t a k )] a 1, 1,, 1 a 1 a a k 1 min a 1, a,, a k )] a t a )] a t a 6 marks) b) The corresponding pdf is f T t) a min a 1, a,, a k )] a t a 1 for t > min a 1, a,, a k ) marks) c) Inverting 1 min a 1, a,, a k )] a t a p gives VaR p T ) min a 1, a,, a k ) 1 p) 1/a marks) 5

27 d) The expected shortfall is ES p T ) min a 1, a,, a k ) p min a 1, a,, a k ) p 1 1) a min a 1, a,, a k ) p 1 1) a p 0 1 u) 1 a du 1 u) 1 1 a ] p 1 p) 1 1 a 1 ] 0 marks) e) The likelihood function is n ) a 1 { n } L a, a 1,, a k ) a n min a 1, a,, a k )] na t i I t i > min a 1,, a k )] i1 n ) a 1 a n min a 1, a,, a k )] na t i {I min t 1,, t n ) > min a 1,, a k )]} As a function of min a 1,, a k ), it is increasing over, min t 1,, t n )) Hence, the maximum likelihood estimator of a 1, a,, a k are those values satisfing min a 1,, a k ) min t 1,, t n ) To find the maximum likelihood estimator of a, take the log of the likelihood log L a, a 1,, a k ) n log a + na log min a 1, a,, a k )] a + 1) The partial derivative with respect to a is log L a Setting this to zero gives { â n i1 n a + n log min a 1, a,, a k )] n log min a 1, a,, a k )] + i1 log t i i1 } 1 log t i This is a maximum likelihood estimator since log L a n a < 0 i1 log t i i1 6 8 marks)