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) [1 + ex( t)] 1 [1 + ex( t)] 1 [1 + ex( t)] n. b) The unconditional cumulative distribution function of T is Pr(T t) Pr (T t N n) θ(1 θ) n 1 n1 [1 + ex( t)] n θ(1 θ) n 1 n1 θ [1 + ex( t)] 1 [1 + ex( t)] 1 n (1 θ) n 1 n1 θ [1 + ex( t)] 1 { 1 [1 + ex( t)] 1 (1 θ) } 1 θ θ + ex( t). c) The unconditional robability density function of T is f T (t) θ ex( t) [θ + ex( t)] 2. (1 marks) 1

d) The moment generating function of T is M T (s) where we have set y θ θ ex(st t) [θ + ex( t)] 2 dt ex(st t) [θ + ex( t)] 2 dt 1 ( 1 θ s y 1+s (1 y) 1 s 1 y 1 ) dy y [ 1 1 ] θ s y 1+s (1 y) s dy y s (1 y) 1 s dy θ s [B(2 + s, 1 s) B(1 + s, 2 s)], θ θ+ex( t) and used the definition of the beta function. e) Setting θ θ + ex( t) and solving for t, we obtain value at risk of T as [ VaR (T ) log θ 1 ]. (3 marks) f) The exected shortfall T is ES (T ) 1 [ log θ log(1 u) + log u] du log θ 1 log(1 u)du + 1 log udu log θ 1 } {[u log(1 u)] + u 1 u du + 1 {[u log u] log θ 1 { } u 1 + 1 log(1 ) + 1 u du + 1 { log } log θ 1 { log(1 ) log(1 )} + log 1 log θ + log 1 log(1 ) +. 2 } 1du

3

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 ) G(x) (1) a n as n then G must be of the same tye as (cdfs G and G are of the same tye if G (x) G(ax + b) for some a >, b and all x) as one of the following three classes: I : Λ(x) ex { ex( x)}, x R; { if x <, II : Φ α (x) ex { x α } if x for some α > ; { ex { ( x) III : Ψ α (x) α } if x <, 1 if x for some α >. SEEN The necessary and sufficient conditions for the three extreme value distributions are: 1 F (t + xγ(t)) I : γ(t) > s.t. t w(f ) 1 F (t) II : w(f ) and t 1 F (tx) 1 F (t) x α, x >, ex( x), x R, III : w(f ) < and t 1 F (w(f ) tx) 1 F (w(f ) t) xα, x >. SEEN First, suose that G belongs to the max domain of attraction of the Gumbel extreme value distribution. Then, there must exist a strictly ositive function say h(t) such that 1 G (t + xh(t)) t w(g) 1 G(t) e x 4

for every x >. But 1 F (t + xh(t)) t w(f ) 1 F (t) t w(f ) t w(g) t w(g) ( 1 + xh (t) ) f (t + xh(t)) f(t) ( 1 + xh (t) ) { g (t + xh(t)) [1 G (t + xh(t))] λb 1 1 [1 G (t + xh(t))] λ ( 1 + xh (t) ) g (t + xh(t)) g(t) 1 G (t + xh(t)) t w(g) 1 G(t) g (t) [1 G (t)] λb 1 { 1 [1 G (t)] λ [ ] λb 1 1 G (t + xh(t)) {1 [1 G (t + xh(t))] λ 1 G (t) 1 [1 G (t)] λ [ ] } λb 1 1 G (t + xh(t)) {1 [1 G (t + xh(t))] λ a 1 1 G (t) [ ] λb 1 G (t + xh(t)) {1 [1 G (t + xh(t))] λ t w(g) 1 G (t) 1 [1 G (t)] λ [ ] λb { 1 G (t + xh(t)) 1 [1 λg (t + xh(t))] t w(g) 1 G (t) 1 [1 λg (t)] [ ] λb { 1 G (t + xh(t)) G (t + xh(t)) t w(g) 1 G (t) G (t) [ ] λb 1 G (t + xh(t)) t w(g) 1 G (t) ex( λbx) 1 [1 G (t)] λ 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 ex [ ex( λbx)] n for some suitable norming constants a n > and b n. a n Second, suose 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 G(t) x β 5

for every x >. But 1 F (tx) t 1 F (t) xf (tx) t f(t) { λb 1 xg (tx) [1 G (tx)] 1 [1 G (tx)] λ { t g (t) [1 G (t)] λb 1 1 [1 G (t)] λ t xg (tx) g(t) t 1 G (tx) 1 G(t) [ 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 { 1 G (tx) G (tx) t 1 G (t) G (t) t x λbβ [ 1 G (tx) 1 G (t) ] λb 1 [1 G (t)] λ ] λb 1 {1 [1 G (tx)] λ 1 [1 G (t)] λ 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 ex ( x λbβ) n for some suitable norming constants a n > and b n. a n Third, suose 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 (w(g) tx) t 1 G (w(g) t) xβ 6

for every x >. 1 F (w(f ) tx) t 1 F (w(f ) t) xf (w(f ) tx) t f (w(f ) t) { λb 1 xg (w(f ) tx) [1 G (w(f ) tx)] 1 [1 G (w(f ) tx)] λ { t g (w(f ) t) [1 G (w(f ) t)] λb 1 1 [1 G (w(f ) t)] λ t xg (w(f ) tx) g (w(f ) t) t 1 G (w(f ) tx) 1 G (w(f ) t) t t t t x λbβ [ 1 G (w(f ) tx) 1 G (w(f ) t) [ 1 G (w(f ) tx) 1 G (w(f ) t) [ ] λb 1 G (w(f ) tx) {1 [1 G (w(f ) tx)] λ 1 G (w(f ) t) 1 [1 G (w(f ) t)] λ [ ] λb { 1 G (w(f ) tx) 1 [1 λg (w(f ) tx)] 1 G (w(f ) t) 1 [1 λg (w(f ) t)] [ ] λb { 1 G (w(f ) tx) G (w(f ) tx) 1 G (w(f ) t) [ 1 G (w(f ) tx) 1 G (w(f ) t) ] λb G (w(f ) t) ] λb 1 {1 [1 G (w(f ) tx)] λ 1 [1 G (w(f ) t)] λ ] λb 1 {1 [1 G (w(f ) tx)] λ 1 [1 G (w(f ) t)] λ for every x <, 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 ex ( ( x) λbβ) n for some suitable norming constants a n > and b n. a n 7

Solutions to Question 3 a) Note that w(f ) n and Pr (X w(f )) 1 F (w(f ) 1) Pr (X n) 1 F (n 1) Pr (X n) 1 Pr (X 1) Pr (X 2) Pr (X n 1) Pr (X n) Pr (X n) 1. Hence, there can be no non-degenerate it. b) Note that w(f ) min(n, K) and Pr (X w(f )) 1 F (w(f ) 1) Pr (X min(n, K)) 1 F (min(n, K) 1) Pr (X min(n, K)) 1 Pr (X max(, n + K N)) Pr (X max(, n + K N) + 1) Pr (X min(n, K) 1 Pr (X min(n, K)) Pr (X min(n, K)) 1. Hence, there can be no non-degenerate it. 8

c) Note that w(f ). Note that 1 F (t + xh(t)) t 1 F (t) t t ( t t ( ex( x) ( 1 + xh (t) ) f (t + xh(t)) f (t) ( 1 + xh (t) ) ex [b (t + xh(t)) η ex (bt + bxh(t))] ) 1 + xh (t) ) 1 + xh (t) ex [bt η ex(bt)] ex {bxh(t) + η ex(bt) [1 ex (bxh(t))]} ex {bxh(t) η ex(bt)bxh(t)} 1 if h(t). So, F (x) belongs to the Gumbel domain of attraction. ηb ex(bt) d) Note that w(f ). Then 1 F (tx) t 1 F (t) xf (tx) t f(t) ( ) x(tx) α 1 ex β tx t t α 1 ex ( ) β t x(tx) α 1 t t α 1 x α. So, F (x) belongs to the Fréchet domain of attraction. e) Note that w(f ). Then 1 F (t + xh(t)) t 1 F (t) t 1 [1 + ex ( at axh(t))] b 1 [1 + ex ( at)] b 1 [1 b ex ( at axh(t))] t 1 [1 b ex ( at)] ex ( at axh(t)) t ex ( at) ex ( axh(t)) t ex( x) 9

if h(t) 1. So, the cdf F (x) belongs to the Gumbel domain of attraction. a 1

Solutions to Question 4 (a) If X is an absolutely continuous random variable with cdf F ( ) then VaR (X) F 1 () and ES (X) 1 F 1 (v)dv. SEEN (b) (i) T is a N (µ 1, σ1)+ 2 +N (µ k, σk 2) N (µ 1 + + µ k, σ1 2 + + σk 2 ) random variable; (b) (ii) Inverting Φ ( t µ 1 µ k σ 2 1 + + σ 2 k ), we obtain VaR (T ) µ 1 + + µ k + σ 2 1 + + σ 2 k Φ 1 (). (b) (iii) The exected shortfall is ES (T ) 1 [ ] µ 1 + + µ k + σ1 2 + + σk 2 Φ 1 (v) dv µ 1 + + µ k + σ1 2 + + σk 2 1 Φ 1 (v)dv. 11

c) (i) The joint likelihood function of µ 1, µ 2,..., µ k and σ1, 2 σ2, 2..., σk 2 is L ( ) µ 1, µ 2,..., µ k, σ1, 2 σ2, 2..., σk 2 { [ ]} k n 1 ex 2πσi i1 { k i1 1 (2π) n σ n i 1 (2π) nk σ n 1 σ n k (X i,j µ i ) 2 2σi 2 [ ]} ex 1 (X i,j µ i ) 2 ex [ 2σ 2 i 1 2 k i1 1 σ 2 i ] (X i,j µ i ) 2. c) (ii) The log likelihood function is log L nk log(2π) n k log σ i 1 2 i1 k 1 σ 2 i1 i (X i,j µ i ) 2. The artial derivatives are log L µ i 1 σ 2 i (X i,j µ i ) 1 σi 2 [( ) ] X i,j nµ i and log L σ i n σ i + 1 σ 3 i (X i,j µ i ) 2. Setting these to zero and solving, we obtain µ i 1 n X i,j and σ 2 i 1 n (X i,j µ i ) 2. 12

c) (iii) µ i is unbiased and consistent since E ( µ i ) 1 n E (X i,j ) 1 n µ i µ i and V ar ( µ i ) 1 n 2 V ar (X i,j ) 1 n 2 σ 2 i σ2 i n. c) (iv) σ 2 i is unbiased and consistent since E ( σ ) [ ] 2 i σ2 i n E 1 (X i,j µ i ) 2 σ 2 i σ2 i n E [ ] χ 2 (n 1)σ 2 n 1 i n and V ar ( σ i 2 ) V ar [ σ 2 i n 1 σ 2 i ] (X i,j µ i ) 2 σ4 i n 2 V ar [ 1 σ 2 i ] (X i,j µ i ) 2 σ4 i n V ar [ ] χ 2 2σ 4 2 n 1 i (n 1). n 2 c (v) The maximum likelihood estimators of VaR (T ) and ES (T ) are VaR (T ) 1 n k X i,j + 1 n i1 k (X i,j µ i ) 2 Φ 1 () i1 and ÊS (T ) 1 n k X i,j + 1 n i1 k i1 (X i,j µ i ) 2 1 Φ 1 (v)dv. 13

Solutions to Question 5 a) Note that for t > min (a 1, a 2,..., a k ). F T (t) Pr(T t) 1 Pr(T > t) 1 Pr (min (X 1, X 2,..., X k ) > t) 1 Pr (X 1 > t, X 2 > t,..., X k > t) 1 F (t, t,..., t) [ ( t 1 max, a 1 [ 1 max t a 2,..., t a k )] a ( 1, 1,..., 1 a 1 a 2 a k 1 [min (a 1, a 2,..., a k )] a t a )] a t a (6 marks) b) The corresonding df is f T (t) a [min (a 1, a 2,..., a k )] a t a 1 for t > min (a 1, a 2,..., a k ). c) Inverting 1 [min (a 1, a 2,..., a k )] a t a gives VaR (T ) min (a 1, a 2,..., a k ) (1 ) 1/a. 14

d) The exected shortfall is ES (T ) min (a 1, a 2,..., a k ) min (a 1, a 2,..., a k ) ( 1 1) a min (a 1, a 2,..., a k ) ( 1 1) a (1 u) 1 a du [ (1 u) 1 1 a ] [(1 ) 1 1 a 1 ]. e) The likelihood function is ( n ) a 1 { n } L (a, a 1,..., a k ) a n [min (a 1, a 2,..., a k )] na t i I [t i > min (a 1,..., a k )] i1 ( n ) a 1 a n [min (a 1, a 2,..., 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 2,..., a k are those values satisfying 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 2,..., a k )] (a + 1) The artial derivative with resect to a is log L a Setting this to zero gives { â n i1 n a + n log [min (a 1, a 2,..., a k )] n log [min (a 1, a 2,..., a k )] + i1 log t i. i1 } 1 log t i. This is a maximum likelihood estimator since 2 log L a 2 n a 2 <. i1 log t i. i1 15 (8 marks)