SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

Solutions to Question 1 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) t + a 2a t + a ) 2a n t + a. 2a 4 marks) b) The unconditional cumulative distribution function of T is PrT t) Pr T t N n) n exp ) n! n ) n t + a n exp ) 2a n! n ) n 1 t + a) exp ) n! 2a n ) t + a) exp + 2a ) t a) exp. 2a 4 marks) c) The unconditional probability density function of T is f T t) ) t a) 2a exp. 2a 1 marks) 1

d) The moment generating function of T is M T s) 2a exp ) a exp 2 a 2a exp 2a exp 2 ) s + 2a ) s + 2 2a st + t 2a ) 1 [ exp ) dt st + t 2a ) 1 [ exp sa + 2 )] a a ) exp sa )]. 2 4 marks) e) Setting ) t a) exp. p 2a and solving for t, we obtain value at risk of T as VaR p T ) a + 2a log p. 3 marks) f) The expected shortfall T is ES p T ) a + 2a p p log xdx a + 2a p {[t log t]p p} a + 2a {p log p p} p a + 2a {log p 1}. 4 marks) 2

Solutions to Question 2 If there are norming constants a n >, 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 ) Gx) 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) Gax + b) for some a >, b and all x) as one of the following three classes: I : Λx) exp { exp x)}, x R; { if x <, II : Φ α x) exp { x α } if x for some α > ; { exp { x) III : Ψ α x) α } if x <, 1 if x for some α >. SEEN 4 marks) The necessary and sufficient conditions for the three extreme value distributions are: 1 F t + xγt)) I : γt) > s.t. t wf ) 1 F t) II : wf ) and t 1 F tx) 1 F t) x α, x >, exp x), x R, III : wf ) < and t 1 F wf ) tx) 1 F wf ) t) xα, x >. 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 strictly positive function say ht) such that 1 G t + xht)) t wg) 1 Gt) e x 3

for every x, ). But 1 F t + xht)) t wf ) 1 F t) t wf ) t wf ) t wf ) t wf ) t wf ) t wf ) exp bx) 1 + xh t) ) f t + xht)) ft) 1 + xh t) ) g t + xht)) G a 1 t + xht)) [1 G t + xht))] b 1 g t) G a 1 t) [1 G t)] b 1 1 + xh t) ) g t + xht)) [1 G t + xht))] b 1 1 + xh g t) [1 G t)] b 1 t) ) g t + xht)) [1 G t + xht))] b 1 g t) [1 G t)] b 1 1 G t + xht)) [1 G t + xht))] b 1 1 G t) [1 G t)] b 1 [ ] b 1 G t + xht)) 1 G t) for every x, ), assuming wf ) wg). 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 bx)] n for some suitable norming constants a n > and 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 > such that 1 G tx) t 1 Gt) x β 4

for every x >. But 1 F tx) t wf ) 1 F t) xf tx) t wf ) ft) xg tx) G a 1 tx) [1 G tx)] b 1 t wf ) g t) G a 1 t) [1 G t)] b 1 xg tx) [1 G tx)] b 1 t wf ) g t) [1 G t)] b 1 t wf ) t wf ) t wf ) x bβ xg tx) [1 G tx)] b 1 g t) [1 G t)] b 1 1 G tx) [1 G tx)] b 1 1 G t) [1 G t)] b 1 [ ] b 1 G t + xht)) 1 G t) for every x >. 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 bβ) n for some suitable norming constants a n > and 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 > such that 1 G wg) tx) t 1 G wg) t) xβ 5

for every x >. But t 1 F wf ) tx) 1 F wf ) t) t xf wf ) tx) fwf ) t) t xg wf ) tx) G a 1 wf ) tx) [1 G wf ) tx)] b 1 g t) G a 1 wf ) t) [1 G wf ) t)] b 1 t xg wf ) tx) [1 G wf ) tx)] b 1 g wf ) t) [1 G wf ) t)] b 1 xg wf ) tx) [1 G wf ) tx)] b 1 t g wf ) t) [1 G wf ) t)] b 1 1 G wf ) tx) [1 G wf ) tx)] b 1 t 1 GwF ) t) [1 G wf ) t)] b 1 [ ] b 1 G wf ) tx) t 1 GwF ) t) x bβ for every x <, assuming wf ) wg). 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) bβ) n for some suitable norming constants a n > and b n. a n 4 marks) 6

Solutions to Question 3 a) Note that wf ) N and Pr X wf )) 1 F wf ) 1) Pr X N) 1 F N 1) Pr X N) 1 Pr X 1) Pr X 2) Pr X N 1) 1 N 1 1 N 1 N 1 N 1 N 1 N 1 N 1. Hence, there can be no non-degenerate it. 4 marks) b) Note that wf ) n and Pr X wf )) 1 F wf ) 1) Pr X n) 1 F n 1) Pr X n) 1 Pr X n 1) Pr X n) Pr X n) 1. Hence, there can be no non-degenerate it. c) Note that wf ) b. The corresponding cdf is F x) log x log a log b log a. 7

Note that t 1 F wf ) tx) 1 F wf ) t) t 1 F b tx) 1 F b t) 1 logb tx) log a log b log a t 1 logb t) log a log b log a log b logb tx) t log b logb t) t x. x b tx 1 b t So, F x) belongs to the Weibull domain of attraction. 4 marks) d) Note that wf ). Then 1 F t + xgt)) t 1 F t) { [ 1 1 exp t + xgt)) 2 ]} a t 1 {1 exp [ t 2 ]} a { [ 1 1 a exp t + xgt)) 2 ]} t 1 {1 a exp [ t 2 ]} [ t + xgt)) 2 ] exp t exp [ t 2 ] exp [ 2txgt) x 2 g 2 t) ] t exp x) if gt) 1/2t). So, F x) [1 exp x 2 )] a belongs to the Gumbel domain of attraction. 4 marks) 8

e) Note that wf ). Then 1 F tx) t 1 F t) t 1 1 + [1 exp tx) 1 )] a 1 1 + [1 exp t 1 )] a [1 exp tx) 1 )] a t [1 exp t 1 )] a [ 1 exp tx) 1 ) t 1 exp t 1 ) [ ] 1 1 + tx) 1 a t 1 1 + t 1 x a. ] a. So, the cdf F x) 1 [1 exp x 1 )] a belongs to the Fréchet domain of attraction. 4 marks) 9

Solutions to Question 4 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 F 1 v)dv. 2 marks) SEEN b) i) The corresponding cdf is for a < x < a; F x) x a 3y 2 2a 3 dy 3 2a 3 [ y 3 3 ] x a 3 2a 3 [ ] x 3 + a 3 x3 + a 3 3 2a 3 2 marks) b) ii) Inverting we obtain VaR p X) 2p 1) 1/3 a. F x) x3 + a 3 2a 3 p, 1 marks) b) iii) The expected shortfall is ES p X) 1 p p 2v 1) 1/3 dv 1 p [ 32v 1) 4/3 8 ] p 3a [ 2p 1) 4/3 1 ]. 8p 2 marks) 1

c) i) The likelihood function of a is L a) 3 n 2 n a 3n 3 n 2 n a 3n 3 n 2 n a 3n 3 n 2 n a 3n 3 n 2 n a 3n n [ X 2 i I { a < X i < a} ] i1 n ) 2 X i i1 n i1 I { a < X i < a} n ) 2 X i I {max X 1,..., X n ) < a, min X 1,..., X n ) > a} i1 n ) 2 X i I {a > max X 1,..., X n ), a > min X 1,..., X n )} i1 n ) 2 X i I {a > max [max X 1,..., X n ), min X 1,..., X n )]}. i1 3 marks) c) ii) Note that L a) is a decreasing function over a and a > max [max X 1,..., X n ), min X 1,..., X n )]. Hence, the mle of a is max [max X 1,..., X n ), min X 1,..., X n )]. c) iii) The mles of VaR and ES are VaR p X) 2p 1) 1/3 â and ÊS p X) 3â [ 2p 1) 4/3 1 ], 8p where â max [max X 1,..., X n ), min X 1,..., X n )]. 1 marks) 2 marks) 11

c iv) Let Z max [max X 1,..., X n ), min X 1,..., X n )]. The cdf of Z is F Z z) Pr max [max X 1,..., X n ), min X 1,..., X n )] z) Pr max X 1,..., X n ) z, min X 1,..., X n ) z) Pr max X 1,..., X n ) z, min X 1,..., X n ) z) Pr X 1 z,..., X n z, X 1 z,..., X n z) Pr z X 1 z,..., z X n z) Pr n z X z) [Pr X z) Pr X z)] n [ ] z 3 n a 3 for z >. The corresponding pdf is f Z z) 3na 3n z 3n 1 for z >. 3 marks) c v) The expected value of Z is a [ z EZ) 3na 3n z 3n dz 3na 3n 3n+1 Hence, â is biased for a. 3n + 1 ] a 3n a3n+1 dz 3na 3n + 1 3na 3n + 1. 2 marks) c vi) Since [ ] Bias VaRp X) 2p 1) 1/3 Bias [â], we see that VaR p X) is biased for VaR p X). 1 marks) c vii) Since ] Bias [ÊSp X) 3 [ 2p 1) 4/3 1 ] Bias [â], 8p 12

we see that ÊS px) is biased for ES p X). 1 marks) 13

Solutions to Question 5 a) Note that F α 1 + 1 x 1 ) α 1 k x i, i1 2 F x 1 x 2 αα + 1) 2 1 + 1 ) α 2 k x i, i1 3 F αα + 1)α + 3) 1 + 1 x 1 x 2 x 3 3 ) α 3 k x i, i1 and so on. In general, as required. f x 1, x 2,..., x k ) αα + 1) α + k 1) k 1 + 1 ) α k k x i, i1 4 marks) b) The pdf of S is f S s) as required. s s x1 s x1 x k 1 f x 1, x 2,..., t x 1 x 2 x k 1 ) dx k dx 2 dx 1 s s x1 k αα + 1) α + k 1) k αα + 1) α + k 1) k s s x1 s x1 x k 1 αα + 1) α + k 1) 1 + s ) α k s k αα + 1) α + k 1) s 1 k + s ) α k, k! k s x1 x k 1 s x1 ) α k 1 + 1 x i dx k dx 2 dx 1 i1 1 + s ) α k dxk dx 2 dx 1 s x1 x k 1 dx k dx 2 dx 1 14 4 marks)

c) The cdf of S is F S s) as required. αα + 1) α + k 1) k! k αα + 1) α + k 1) k! s x k 1 + x ) α k dx /s+) 1 s/s+) y α 2 1 y) k dy αα + 1) α + k 1) 1 y) α 2 y k dy k! αα + 1) α + k 1) B s/s+) k + 1, α 1), k! 4 marks) d) The nth moment of S is E S n ) αα + 1) α + k 1) k! k n+1 αα + 1) α + k 1) k! x n+k 1 + x ) α k dx 1 1 y) n+k y α n 2 dy n+1 αα + 1) α + k 1) B α n 1, n + k + 1). k! 4 marks) e) The joint likelihood function is L α, ) αn α + 1) n α + k 1) n k!) n kn Its log is The joint likelihood function is [ ] α n α + 1) n α + k 1) n log L α, ) log k!) n + k kn n ) k [ n s i The partial derivatives of this with respect to α and are log L α, ) α n α + n α + 1 + + 15 i1 i1 1 + s i n log s i α + k) i1 n n α + k 1 i1 log ) ] α k n i1 1 + s i. log ) 1 + s i ).

and log L α, ) kn + α + k 2 n s i 1 + s ) 1 i. i1 The maximum likelihood estimates of α and are the simultaneous solutions of n α + n α + 1 + + n α + k 1 n log 1 + s ) i i1 and kn α + k 2 n s i 1 + s ) 1 i. i1 4 marks) 16