SOLUTIONS TO MATH38181 EXTREME VALUES EXAM

Transcription

1 SOLUTIONS TO MATH388 EXTREME VALUES EXAM

2 Solutions to Question If there re norming onstnts n >, b n nd nondegenerte G suh tht the df of normlized version of M n onverges to G, i.e. ( ) Mn b n Pr x F n ( n x + b n ) G(x) () n s n then G must be of the sme type s (dfs G nd G re of the sme type if G (x) G(x + b) for some >, b nd ll x) s one of the following three lsses: I : Λ(x) 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 <, if x for some α >. The neessry nd suffiient onditions for the three extreme vlue distributions re: F (t + xγ(t)) I : γ(t) > s.t. t w(f ) F (t) II : w(f ) nd t F (tx) F (t) x α, x >, exp( x), x R, III : w(f ) < nd t F (w(f ) tx) F (w(f ) t) xα, x >. First, suppose tht G belongs to the mx domin of ttrtion of the Gumbel extreme vlue distribution. Then, there must exist stritly positive funtion, sy h(t), suh tht G (t + xh(t)) t w(g) G(t) exp( x)

3 for every x (, ). But, using L Hopitl s rule, we note tht F (t + xh(t)) t w(f ) F (t) ) t w(f ) t w(g) t w(g) t w(g) t w(g) t w(g) t w(g) exp( x) ( + xh (t) ( + xh (t) f (t + xh(t)) f(t) ) { log G (t + xh(t))]} g (t + xh(t)) ( ) + xh (t) g (t + xh(t)) ( + xh (t) { log G(t)]} g(t) { log G (t + xh(t))] g(t) ) g (t + xh(t)) g(t) ( ) + xh (t) g (t + xh(t)) ( + xh (t) g(t) G (t + xh(t)) G(t) g(t) ) g (t + xh(t)) log G(t)] G (t) G (t + xh(t)) ( + xh (t) } ( ) + xh (t) g (t) ) g (t + xh(t)) g (t + xh(t)) g(t) ( ) + xh (t) g (t + xh(t)) for every x (, ). So, it follows tht F lso belongs to the mx domin of ttrtion of the Gumbel extreme vlue distribution with Pr { n (M n b n ) x} exp { exp( x)} n for some suitble norming onstnts n > nd b n. Seond, suppose tht G belongs to the mx domin of ttrtion of the Fréhet extreme vlue distribution. Then, there must exist β >, suh tht G(tx) t G(t) xβ g(t) 2

4 for every x >. But, using L Hopitl s rule, we note tht F (tx) t F (t) t xf(tx) f(t) x { log G(tx)]} g(tx) t { log G(t)]} g(t) { } xg(tx) log G(tx)] t g(t) log G(t)] { xg(tx) G(t) xg(tx) t g(t) G(tx) g(t) { } xg(tx) g(t) xg(tx) t g(t) xg(tx) g(t) xg(tx) t g(t) G(tx) t G(t) x β } for every x >. So, it follows tht F lso belongs to the mx domin of ttrtion of the Fréhet extreme vlue distribution with Pr { n (M n b n ) x} exp ( x β) n for some suitble norming onstnts n > nd b n. Third, suppose tht G belongs to the mx domin of ttrtion of the Weibull extreme vlue distribution. Then, there must exist α >, suh tht G(w(G) tx) t G(w(G) t) xα 3

5 for every x >. But, using L Hopitl s rule, we note tht F (w(f ) tx) t F (w(f ) t) t xf(w(f ) tx) f(w(f ) t) x { log G(w(F ) tx)]} g(w(f ) tx) t { log G(w(F ) tx)]} g(w(f ) t) { } xg(w(f ) tx) log G(w(F ) tx)] t g(w(f ) t) log G(w(F ) t)] { xg(w(f ) tx) G(w(F ) t) xg(w(f ) tx) t g(w(f ) t) G(w(F ) tx) g(w(f ) t) { xg(w(f ) tx) g(w(f ) t) xg(w(f ) tx) t g(w(f ) t) xg(w(f ) tx) g(w(f ) t) xg(w(f ) tx) t g(w(f ) t) G(w(F ) tx) t G(w(F ) t) x α. } } So, it follows tht F lso belongs to the mx domin of ttrtion of the Weibull extreme vlue distribution with Pr { n (M n b n ) x} exp { ( x) α } n for some suitble norming onstnts n > nd b n. 4

6 Solutions to Question 2 i) Note tht w(f ) nd tke γ(t). Then F (t + x) t F (t) exp ( t x)] α α exp ( t x) t exp ( t)] α t α exp ( t) exp( x).. So, the exponentited exponentil df F (x) exp( x)] α belongs to the Gumbel domin of ttrtion. ii) Note tht w(f ) nd tke γ(t) /β. Then F (t + x/β) t F (t) { exp( βt x)} α p + p { exp( βt)} α t p + p { exp( βt x)} α { exp( βt)} α α exp( βt x) t α exp( βt) exp( x). So, the exponentited exponentil geometri df F (x) Gumbel domin of ttrtion. { exp( βx)} α p+p{ exp( βx)} α belongs to the iii) the orrespond pdf is f(x) kβ( p)k exp( kβx) p exp( βx)] k+. Note tht w(f ) nd tke γ(t) /(kβ). Sine f(x) kβ( p) k exp( kβx) s x, F (t + xγ(t)) t F (t) t f(t + x/(kβ)) f(t) exp( x). So, the exponentil-negtive binomil distribution df F (x) ( p)k exp( kβx) p exp( βx)] k the Gumbel domin of ttrtion. belongs to iv) For the degenerte distribution, p(k) {, if k k,, if k k. So, F (k) {, if k k,, if k < k, nd Pr(X k) F (k ) /, if k k, /, if k < k < k +, /, if k k +, /, if k < k. 5

7 Hene, there n be no sequenes n > nd b n suh tht (M n b n )/ n hs non-degenerte iting distribution. v) For the Poisson distribution, Pr(X k) F (k ) λ k /k! jk λj /j! + jk+ k!λj k /j!. The term in the denomintor of the lst term n be rewritten s λ j (k + )(k + 2) (k + j) ( ) j λ λ/k k λ/k j (when k > λ) nd the bound tends to s k nd so it follows tht p k /( F (k )). Hene, there n be no sequenes n > nd b n suh tht (M n b n )/ n hs non-degenerte iting distribution. j 6

8 Solutions to Question 3 If X is n bsolutely ontinuous rndom vrible with df F ( ) then VR p (X) F (p) nd gives Setting ES p (X) p p x b p F (v)dv. VR p (X) + p(b ). So, p ES p (X) + v(b )] dv p ] p + p2 (b ) p 2 + p (b ). 2 i) The joint likelihood funtion of nd b is L (, b) b I{ < X < b} b I{ < X 2 < b} b I{ < X n < b} n I{ < X (b ) n i < b} i (b ) n I{mx (X, X 2,..., X n ) < b}i{ < min (X, X 2,..., X n )}, where I{ } denotes the inditor funtion. ii) Note tht (b ) n is n inresing funtion of over (, min(x, X 2,..., X n )). So, the mximum of L(, b) will be ttined t min(x, X 2,..., X n ). iii) Note tht (b ) n is deresing funtion of b over (mx(x, X 2,..., X n ), ). So, the mximum of L(, b) will be ttined t b mx(x, X 2,..., X n ). iv) The mle of Vlue t Risk is VR p (X) min (X, X 2,..., X n ) + p mx (X, X 2,..., X n ) min (X, X 2,..., X n )]. 7

9 The mle of Expeted Shortfll is ÊS p (X) min (X, X 2,..., X n ) + p 2 mx (X, X 2,..., X n ) min (X, X 2,..., X n )]. nd v) Let Z min(x, X 2,..., X n ). Then for < z < b, so nd E(Z) n n nb b + F Z (z) ( ) n b z b (b z)n f Z (z) n (b ) n b b b (b z)n z (b ) dz n (b z)n (b (b z)) (b ) dz n (b z) n (b ) dz n n n(b ) n +. vi) Now let Z mx(x, X 2,..., X n ). Then ( z F Z (z) b for < z < b, so E(Z) n n ) n (z )n f Z (z) n (b ) n b b b (z )n z (b ) dz n b (z )n (z + ) (b ) dz n n (z ) n dz + n (b ) n n(b ) n + + n + + nb n +, Hene, VR p (X) nd ÊS p(x) re bised. b (b z) n (b ) n dz (z ) n (b ) n dz 8

10 Solutions to Question 4 ) The df of M mx(x, Y, Z) is F M (m) Pr(X < m, Y < m, Z < m) Pr(X > m) Pr(Y > m) Pr(Z > m) + Pr(X > m, Y > m) + Pr(X > m, Z > m) + Pr(Y > m, Z > m) Pr(X > m, Y > m, Z > m) + m ] + m ] + m ] + + m b + m ] b + + m + m ] + + m b + m ] + m + m b + m ]. b) Differentiting F M (m) with respet to m gives the pdf of M s f M (m) d + m ] d + + m ] d + + m ] ( d b b + ) + m b + m b + ) + m + m ] ( d b + ) + m b + m ] +d + b + ) + m + m b + m ]. ] ) The nth moment of M n be lulted s E (M n ) d + d b m n + m ] dm m n + m ] dm b ] m n dm + d + m + ) b + ) b + ) +d + b + ) m n + m + m ] dm b m n + m + m ] dm m n + m b + m ] dm m n + m + m b + m ] dm 9

11 d m n + m ] dm dx dx set x + m ] x, so, m nd dm x dx ] x 2 + d m n + m ] dm b b dx dx set x + m ] x, so, m b nd dm b x dx b ] x 2 + d m n + m ] dm dx dx set x + m ] x, so, m nd dm x dx ] x ( 2 + ) m n + m b + m ] dm b dx dx set x + m + m ] (, so, m b + ) x b x + ) m n + m + m ] dm dx dx set x + m + m ] (, so, m + ) x x ) m n + m b + m ] dm dx dx set x + m b + m ] (, so, m b + ) x x + b + ) m n + m + m b + m ] dm dx dx + +d set x + m + m b + m ], so, m + b + nd dm dx + ) ] b x 2 nd dm dx + ) ] x 2 nd dm dx b + ) ] x 2 ) x x nd dm dx + b + ) ] x 2

12 d d b n ( x)n x d+ x dx 2 x n n ( x)n b x d+ b x dx 2 x n d n ( x)n x d+ x n x dx 2 +d + ) b + ) n ( ( x) n x d+ b x n + ) b x dx 2 +d + ) + ) n ( ( x) n x d+ x n + ) x dx 2 +d b + ) b + ) n ( ( x) n x d+ x n b + ) x dx 2 + b + ) + b + ) n ( x) n x d+ x n + b + d n+ x d n ( x) n dx + d b bn+ x d n ( x) n dx + d n+ x d n ( x) n dx d + b + ) + ) n b + ) b + ) n +d + b + ) + b + ) n x d n ( x) n dx x d n ( x) n dx ) x 2 dx ) + ) n x d n ( x) n dx b x d n ( x) n dx d n B(d n, n + ) + db n B(d n, n + ) +d n B(d n, n + ) d + ) n B(d n, n + ) b + ) n B(d n, n + ) b + ) n B(d n, n + ) +d + b + ) n B(d n, n + ).

13 d) The df of L min(x, Y, Z) is F L (l) Pr(L > l) Pr(X > l, Y > l, Z > l) + l + l b ] + l. e) Differentiting F L (l) with respet to l gives the pdf of L s f L (l) d + b + ) + l + l b ] + l. f) The nth moment of L n be lulted s E (L n ) d d ) l n + l + l b + l ] dl + b + + b + ) n B(d n, n + ). 2

14 Solutions to Question 5 (i) The df of X is (ii) Differentiting F Y (y) Pr(Y y) with respet to y gives the pdf of Y s Pr (mx (X,..., X α ) y) Pr (X x,..., X α y) Pr (X y) Pr (X α y) exp( λy)] exp( λy)] exp( λy)] α. F Y (y) exp( λy)] α f Y (y) αλ exp( λy) exp( λy)] α. (iii) The nth moment of Y n be lulted s E (Y n ) αλ x n exp( λx) exp( λx)] α dx ( ) n αλ n (log y) n y] α dy ( ) n n n αλ y β y] α dy β n ( ) n n n αλ B(β +, α) βn. β β gives (iv) Setting exp( λy)] α p VR p (Y ) λ log p /α). (v) The expeted shortfll is ES p (Y ) λp λp λp λp p p i i 3 log v /α) dv v i/α i i p v i/α i dv dv p i/α+ i (i/α + ).

15 (vi) The likelihood funtion is ( L(α, λ) α n λ n exp λ The log-likelihood funtion is ) { n n α y i exp ( λy i )]}. i n n log L n log(αλ) λ y i + (α ) log exp ( λy i )]. i i i The prtil derivtives with respet to α nd λ re log L α n n α + log exp ( λy i )] i nd log L λ n n λ y i + (α ) i n i y i exp ( λy i ) exp ( λy i ). Setting these to zero, we obtin the mle of α s α n. n log exp ( λy i )] i The mle of λ is the root of the eqution n n λ y i i n + n log exp ( λy i )] i n i y i exp ( λy i ) exp ( λy i ). 4

16 Solutions to Question 6 ) If X i Exp(λ i ), i, 2 re independent rndom vribles then Pr (X < X 2 ) exp ( λ x 2 )] λ 2 exp ( λ 2 x 2 ) dx 2 λ 2 λ + λ 2 λ. λ + λ 2 λ 2 exp ( λ x 2 λ 2 x 2 ) dx 2 b) If X i Exp(λ i ), i, 2, 3 re independent rndom vribles then Pr (X < X 2 < X 3 ) x x 2 λ λ 2 λ 3 exp ( λ x λ 2 x 2 λ 3 x 3 ) dx 3 dx 2 dx λ λ 2 exp ( λ x λ 2 x 2 λ 3 x 2 ) dx 2 dx x λ λ 2 exp ( λ x λ 2 x λ 3 x ) dx λ 2 + λ 3 λ λ 2 (λ 2 + λ 3 ) (λ + λ 2 + λ 3 ). ) If X i Exp(λ i ), i, 2, 3, 4 re independent rndom vribles then Pr (X < X 2 < X 3 < X 4 ) x x 2 x x 3 λ λ 2 λ 3 λ 4 exp ( λ x λ 2 x 2 λ 3 x 3 λ 4 x 4 ) dx 4 dx 3 dx 2 dx x 2 λ λ 2 λ 3 exp ( λ x λ 2 x 2 λ 3 x 3 λ 4 x 3 ) dx 3 dx 2 dx λ λ 2 λ 3 exp ( λ x λ 2 x 2 λ 3 x 2 λ 4 x 2 ) dx 2 dx (λ 3 + λ 4 ) x λ λ 2 λ 3 exp ( λ x λ 2 x λ 3 x λ 4 x ) dx 2 dx (λ 3 + λ 4 ) (λ 2 + λ 3 + λ 4 ) λ λ 2 λ 3 (λ 3 + λ 4 ) (λ 2 + λ 3 + λ 4 ) (λ + λ 2 + λ 3 + λ 4 ). If X i Exp(λ i ), i, 2,..., k re independent rndom vribles then the generl formul will be Pr (X < X 2 < < X k ) This n be proved by indution. λ λ 2 λ k (λ k + λ k ) (λ k + λ k + λ k 2 ) (λ k + λ k + + λ ). 5

17 If λ i λ then Pr (X < X 2 ) 2, Pr (X < X 2 < X 3 ) 2 2 3, nd Pr (X < X 2 < X 3 < X 4 ) Pr (X < X 2 < < X k ) k k!. 6