MATH68181: EXTREME VALUES FIRST SEMESTER ANSWERS TO IN CLASS TEST

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1 MATH688: EXTREME VALUES FIRST SEMESTER ANSWERS TO IN CLASS TEST ANSWER TO QUESTION 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, ie Mn b n Pr a n x = F n a n x + b n Gx 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 <, if x for some α > The necessary and sufficient conditions for the three extreme value distributions are: F t + xγt I : γt > st = exp x, x R, t wf F t F tx II : wf = and t F t = x α, x >, III : wf < and t F wf tx F wf t = xα, x > 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 G t + xht = exp x t wg Gt for every x, But, using L Hopital s rule, we note that F t + xht t wf F t t wf t wg + xh t + xh t f t + xht ft log G t + xht] g t + xht log Gt]

2 t wg t wg t wg t wg + xh t g t + xht + xh t + xh t + xh t log G t + xht] log Gt] g t + xht G t + xh t g t + xht G t + xht g t + xht g t + xh t g t + xht + xh t g t + xht g t + xht G t + xht t wg Gt = exp x for every x, So, it follows that F also belongs to the max domain of attraction of the Gumbel extreme value distribution with Pr a n M n b n x = exp exp x n for some suitable norming constants a n > and b n 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 Gtx Gt = x β for every x > But, using L Hopital s rule, we note that F tx F t xftx ft x log Gtx] gtx log Gt] xgtx log Gtx] log Gt] xgtx xgtx xgtx Gtx Gt = x β Gt Gtx xgtx xgtx xgtx for every x > So, it follows that F also belongs to the max domain of attraction of the Fréchet extreme value distribution with Pr a n M n b n x = exp n x β

3 for some suitable norming constants a n > and b n Third, suppose that G belongs to the max domain of attraction of the Weibull extreme value distribution Then, there must exist a α >, such that GwG tx t GwG t = xα for every x > But, using L Hopital s rule, we note that F wf tx t F wf t t xfwf tx fwf t x log GwF tx] a gwf tx t log GwF tx] gwf t xgwf tx log GwF tx] t gwf t log GwF t] t xgwf tx gwf t t xgwf tx gwf t t xgwf tx gwf t t GwF tx GwF t = x α GwF t GwF tx xgwf tx gwf t gwf t xgwf tx xgwf tx gwf t So, it follows that F also belongs to the max domain of attraction of the Weibull extreme value distribution with Pr a n M n b n x = exp x α n for some suitable norming constants a n > and b n ANSWER TO QUESTION i Note that wf = and take γt = Then F t + x exp t x] exp t x t F t t exp t] = exp x t exp t So, the exponentiated exponential cdf F x = exp x] belongs to the Gumbel domain of attraction ii Note that wf = Then F t + xγt F t exp t + xγt exp t + xγt exp t exp t 3

4 exp t + xγt exp t exp t + xγt exp t exp t + xγt exp t exp t + xγt exp t exp xγt = exp x if γt = 5 So, the exponentiated exponential geometric cdf F x = to the Gumbel domain of attraction iii Note that wf = Then F t + xγt F t exp x 5+5 exp x belongs 5 exp 4 t + xγt 5 exp t + xγt] 5 exp 4t 5 exp t] 5 exp 4 t + xγt 5 exp t + xγt] 5 exp 4t 5 exp t] exp 4 t + xγt 5 exp t + xγt] exp 4t 5 exp t] exp 4xγt 5 exp t + xγt] 5 exp t] exp 4xγt ] ] exp 4xγt = exp x if γt = /4 So, the exponential-negative binomial distribution cdf F x = 5 exp 4x 5 exp x] belongs to the Gumbel domain of attraction 4

5 iv For the degenerate distribution, pk =, if k =,, if k So, and F k = PrX = k F k =, if k,, if k <, /, if k =, /, if < k < 3, /, if k 3, /, if k < Hence, there can be no sequences a n > and b n such that M n b n /a n has a non-degenerate iting distribution v For the Poisson distribution, j= PrX = k F k = = k /k! j=k j /j! + j=k+ k! j k /j! The term in the denominator of the last term can be rewritten as j k + k + k + j j = /k k /k when k > and the bound tends to as k and so it follows that p k / F k Hence, there can be no sequences a n > and b n such that M n b n /a n has a non-degenerate iting distribution ANSWER TO QUESTION 3 i The cdf of M = maxx, Y, Z is F M m = PrX < m, Y < m, Z < m = PrX > m PrY > m PrZ > m + PrX > m, Y > m j= + PrX > m, Z > m + PrY > m, Z > m PrX > m, Y > m, Z > m = + m ] + m ] + m ] + + m] + + m] + + m] + 3m ] ii Differentiating F M m with respect to m gives the pdf of M as f M m = + m ] m ] m ] 3 + m] 3 + m] 3 + m] m ] 3 5

6 iii The nth moment of M can be calculated as E M n = m n + m ] 3 dm m n + m m n + m] 3 dm m n + m] 3 dm m n + 3m m n + m ] 3 dm ] 3 dm m n + m] 3 dm ] 3 dm = n+ x n x n dx + n+ x n x n dx + n+ x n x n dx +3 x n x n dx x n x n dx 3 n x n x n dx = n B n, n + + n B n, n + + n B n, n + B n, n + B n, n + B n, n + 3 n + B n, n + x n x n dx iv The cdf of L = minx, Y, Z is F L l = PrL > l = PrX > l, Y > l, Z > l = + 3l ] v Differentiating F L l with respect to l gives the pdf of L as 3 f L l = + 3l ] 3 vi The nth moment of L can be calculated as 3 E L n = l n + 3l ] 3 dl 3 n = B n, n + 6