Extreme Values by Resnick

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1 1 Extreme Values by Resnick 1 Preliminaries 1.1 Uniform Convergence We will evelop the iea of something calle continuous convergence which will be useful to us later on. Denition 1. Let X an Y be metric spaces an suppose we have a sequence of functions f n : X Y. We say f n converges continuously to f if whenever x n x 0 in X we have f n (x n ) f 0 (x 0 ) in Y. Lemma 2. Suppose X is compact, an f 0 is continuous. Then f n f 0 continuously if an only if f n f 0 uniformly on X, Proof. ( ) Suppose f n f 0 uniformly on X. Take any sequence x n x 0 in X. Use to enote the metric on Y. Have: (f n (x n ), f 0 (x 0 )) (f n (x n ), f 0 (x n )) + (f 0 (x n ), f 0 (x 0 )) sup (f n (z), f 0 (z)) + (f 0 (x n ), f 0 (x 0 )) z X The rst term goes to 0 since f n f 0 uniformly, while the secon term goes to zero since f 0 is continuous an x n x 0. ( ) Suppose by contraiction that f n oes not converge to f 0 uniformly. Then there is a subsequence n k an ɛ > 0 so that: sup (f nk (z), f 0 (z)) > ɛ z X Hence, by enition of sup, for each n k we can n a x k X so that (f nk (x k ), f 0 (x k )) > ɛ/2. Since the space X is compact now, there is a limit point x 0 for the sequence ( x k an a subsubseuqnece ) ( so that x kl ) x 0. But this is a contraiction in the continuous convergence hypthesis, as ɛ/2 < f nkl (x kl ), f 0 (x kl ) f nkl (x kl ), f 0 (x 0 ) + (f 0 (x kl ), f 0 (x 0 )) so the latter is boune away from zero. Corollary 3. Suppose U n, n 0 are non-ecreasing real value functions on R. Suppose also that U 0 is continuous. If U n U 0 point wise as n then U n U 0 locally uniformly. i.e. for any a < b we have: sup U n (x) U 0 (x) 0 x [a,b] Proof. We will check that U n U 0 continuously on any [a, b]. (Rmk: the reason we nee to restric our attention to [a, b] is that the lemma only works on compact spaces X ) Suppose that x n x 0. We want to show that U n (x n ) U 0 (x 0 ). It suces to consier two cases: a) x n > x 0 an b) x n < x 0 (If necessary we can always partition x n into two subsequences that have this property. Our argument will hol for both subsequences). We will look at the case a), an the case b) is analogous. Given any ɛ > 0, use the continuity of U 0 to n η > 0 such that: U 0 (x 0 + η) U 0 (x 0 ) < ɛ Take n 0 so large now so that for each n n 0 we have x n x 0 < η (ok since x n x) an U n (x 0 + η) U 0 (x 0 + η) < ɛ an U n (x 0 ) U 0 (x 0 ) < ɛ. (Ok since U n U 0 pointwise) Then we have on one han that: an on the other han: U n (x n ) U n (x 0 + η) since Uis monotone an x n < x 0 + η U 0 (x 0 + η) + ɛ U 0 (x 0 ) + ɛ + ɛ U n (x n ) U n (x 0 ) since x n > x 0 an Umonotone U 0 (x 0 ) ɛ

2 Remark 4. Say X n are ranom variables converging X n X 0 in the weak sense, an suppose X 0 has no atoms. Then we know that the istribution functions F n converge pointwise to F 0 pointwise an that F 0 is continuous. The corrolary tells us that F n F 0 uniformly on compact subsets. We can actually strengthen this a bit to uniform convergence everywhere because of the aitional fact that F n 0/1 as x / +. Corollary 5. Suppose X n are ranom variables an X n X 0. Suppose further that X 0 has no atoms. Then the istribution functions converge uniformly on all of R: sup P(X n x) P(X 0 x) 0 x R Proof. Fix any ɛ > 0. Choose [a, b] so large now so that P (X 0 [a, b]) > 1 ɛ. By the corollary/remark we know that F n F uniformly in [a, b] so we know that for n large enough sup x [a,b] F n (x) F 0 (x) < ɛ. Outsie of [a, b], consier as follows. By pointwise convergence at b, we know that for n large enough F n (b) F 0 (b) < ɛ. Have then for any x > ɛ an such n suciently large: So then: sup x>b F n (x) F n (b) 1 F n (b) 1 F 0 b) + F 0 (b) F n (x) 2ɛ F n (x) F n (b) sup { F n (x) F n (b) + F n (b) F 0 (b) + F 0 (b) F 0 (x) } x>b 2ɛ + ɛ + ɛ The same iea works for x < a. Combining the convergence in the 3 pieces of R = (, a) [a, b] (b, ) we get sup x R F n (x) F n (b) < 5ɛ. Since ɛ arbitary, we have uniform convergence. 1.2 Inverses of Monotone Functions Denition 6. If H is a non-ecreasing function on R, we ene the left-continupus-inverse of H as: H (y) = inf {s : H(s) y} We take the convention that the inmum over an empty set is +. Proposition 7. i) H is non-ecreasing ii) H (y) < z = H(z) y iii) H is left continuous at points y R Proof. i) For y 1 < y 2 we have {s : H(s) y 2 } {s : H(s) y 1 } since H is non-ecreasing. Taking inf's gives: ii) H (y 1 ) = inf {s : H(s) y 1 } inf {s : H(s) y 2 } = H (y 2 ) H (y) z = inf {s : H(s) y} < z = H(z) y since H is non-ecreasing ii) Suppose by contraiction that x n x but H (x n ) H (x ) < H (x). Then there exists δ > 0 so that: H (x n ) < y < H (x) δ By the enition of H, the left inequality tells us H(y) x n for every n. Taking n then gives H(y) x. By enition of H again now, H(y) x gives H (x) y. But this is impossible since H (x) δ > y! Proposition 8. If H is right continuous (i.e. x n x 0 = H(x n ) H(x 0 )) then: A(y) := {s : H(s) y} is close H(H (y)) y H (y) t y H(t) t < H (y) y > H(t)

3 3 Proof. i) Since H is non-ecreasing, the set A(y) is unboune on the right (i.e. s A(y) = [s, ) A(y)) so we only have to check that the left enpoint is a close enpoint. Inee, if s n A(y) an s n s then y H(s n ) for each s n by ef'n of A(y). Since H is right continuous, we have H(s n ) H(s) an so we conclue y H(s), which shows s A(y) ii) Since A(y) is close, H (y) = inf A(y) is achieve at some point s A(y). Hence H(H (y)) = H(s ) y since s A(y). iii) As in ii), let s = H (y) = inf A(y) be the points in the close set A(y) that achieves the minimum. We use the fact that the entire interval [s, ) A(y) because H is non-ecreasing. Consier now: The strict sie is: H (y) t inf A(y) t t A(y) since [s, ) A(y) H(t) y H (y) > t inf A(y) > t t / A(y) since [s, ) A(y) H(t) < y Proposition 9. Let (Ω, F, P) = ([0, 1], B[0, 1], m) be the Lebesgue measure on [0, 1] with the usual Borel sigma algebra. Let U be the ientity function on Ω i.e. U Unif([0, 1]) is a uniformly istribute ranom variable. If F is any istribution function on R with law f, then F (U) is a ranom variable on [0, 1] with law f. Proof. Such F are always right continuous, so by the last proposition, we have: m [F (U) t] = m [U F (t)] = F (t) Denition 10. For any function H enote C (H) = {x R : H is nite an continuous at x} (the set of H-continuity points) We ene weak convergence of non-ecreasing functions by H n (x) H 0 (x) for all x C (H). Proposition 11. If H n are non-ecreasing functions an H n H 0 then H n H 0. Proof. These proof is along the same lines of the last few...just play with the enition of until you get it...i'm going to skip it. Theorem 12. (Skoroho's Representation Theorem) Suppose X n on (Ω n, B n, P n ) is a sequence of ranom variables on a sequence of probability spaces so that X n X 0. Then there exists ranom variables X n on the Lebesgue probability space ([0, 1], B[0, 1], m) so that: X n = Xn for each n N X n X 0 almost surely with respect to m Remark 13. The X n 's ont respect possibly epenencies between the X ns. For example E(X 1 X 2 ) is not the same as E( X 1 X2 ); the joint istributions are nor respecte. Proof. Let U be the ientity function on [0, 1] so that U is uniformly istribute. Suppose that the istribution function of X n is F n. Dene: X n = F n (U) As we have alreay seen, Xn = Xn with this enition. To see the convergence is almost sure now, consier as follows. Firstly, X n X 0 means that F n F 0 (in the sense of weak convergence). By the last proposition, we know that Fn F0 too then! Therefore: { 1 m 0 u 1 : Xn (u) X } 0 (u) = m {u : F n (u) F 0 (u)} m {C (F 0 )} = 1 The last equality follows by the famous result that a non-ecreasing function can only have countably many iscontinuities (so in particular the set of iscontinuous points, C (F0 ) c, is a null set)

4 4 1.3 Convergence to Types Theorem an Limit Distribution of Maxima Denition 14. We say two istribution functions U(x) an V (x) are of the same type if for some A > 0, B R we have: V (x) = U(Ax + B) Example 15. The family of Gaussian istributions scales like this, so they are the same type. Proposition 16. Suppose U(x) an V (x) are two istributions, neither of which concentrates at a point. (i.e. they are not the heavisie step function) a) Suppose F n is a sequence of istribution functions, a n 0 an b n R, α n > 0 an β n R are seqeunces so that: F n (a n x + b n ) U(x) an F n (α n x + β n ) V (x) (where convergence is the weak convergence of istribution functions). Then: an moreover: α n /a n A > 0 an (β n b n ) /a n B R V (x) = U(Ax + B) An equivalent way to state this in terms of ranom variables is as follows: a') Let X n be a sequence of ranom variables an suppose U, V are ranom variables (neither of which is almost surely constant). If: (X n b n ) /a n U an (X n β n ) /α n V Then: an: α n /a n A > 0 an (β n b n ) /a n B R V = (U B) /A Remark 17. This proposition is saying that if you start with some ranom variables X n, an then you center an scale somehow to get convergence to SOME ranom variable, say (X n b n ) /a n U, then the ONLY ranom variable you coul hope to converge to are things of the same TYPE as U. Inee, once you have one convergence (X n b n ) /a n U its obvious how changing a n by a constant factor (or an o(1) factor) an changing b n by aing a constant (or an o(1) aition) will shift the istribtuion of U aroun. The theorem says this is the ONLY thing you can o. Its really a nice uniqueness result. If you knew before han that U an V were of the same type, you coul play aroun with the sequences to extract the constant A an B. This is how the proof goes. We play aroun with the convergence until we n A an B. Once this is one, we use Skohor to see the convergence in istribution as almost sure convergence in a certain light, an nally conclue that V = (U B) /A. Proof. By using the fact that H n H 0 = H n H 0, an other properties we have seen of, we can take as our hypothesis that: (F n (y) b n ) /a n U (y) an (F n (y) β n ) /α n V (y) weakly. Since neither U(x) nor V (x) are a point masses at a single point, we can n 0 < y 1 < y 2 < 1 so that y 1, y 2 are continuity points for both U an V so that: U (y 1 ) < U (y 2 ) an V (y 1 ) < V (y 2 ) (Roughly speaking: this just says that the values y 1 an y 2 are not both in the same atom of U or V ). Pluggin in y = y 1 an y = y 2 into our hypothesis, we have (since these are continuity points) four statements of convergence, one to each of U (y 1 ), U (y 2 ), V (y 1 ), V (y 2 ). By subtracting each pair, we get: Diviing the rst relation by the secon now, we get: (F n (y 2 ) F n (y 1 )) /a n U (y 2 ) U (y 1 ) > 0 (F n (y 2 ) F n (y 1 )) /α n V (y 2 ) V (y 1 ) > 0 α n U (y 2 ) U (y 1 ) a n V (y 2 ) V (y 1 ) =: A > 0 Now, replacing α n = Aa n + o(1) we compare our convergences at y = y 1 : (Fn (y 1 ) b n ) /a n U (y 1 ) an (Fn (y 1 ) β n ) /a n V (y 1 ) A So subtracting these two limits gives: β n b n U (y 1 ) V (y 1 ) =: B a n A So we inee have esire connection between the a s an b s an the α s an β s. The fact that V (x) = U(Ax + B) follows by Skohor's theorem now. We put this in the next proposition.

5 5 Proposition 18. If α n /a n A > 0 an (β n b n ) /a n B R, then either of the two conitions: (X n b n ) /a n U or (X n β n ) /α n V will imply the other, an the conclusion of the last proposition will hol. Proof. The main iea is to use Skohoro's representation theroem to replace the covnergence with almost sure convergence, at which point the statement is obvious. Let Y n = (X n b n ) /a n an suppose Y n U. By Skoroho's Representation Theorem we can n Ỹn, Ũ on the probability space ([0, 1], B[0, 1], m) such that Ỹn = Y n an Ỹn a.s Ũ where the convergence is now almost sure. Dene X n := a n Ỹ n + b n so that X n = Xn. Then: So we see that (X n β n ) /α n (U B) /A ( ) (X n β n ) /α n = Xn β n /α n ( ) an = Ỹ n + (b n β n ) /α n a.s α n 1 AŨ 1 A B = (U B) /A Remark 19. A nice byprouct of this proof is that if we are given the istribution functions Fn try are: a n = F n (y 2 ) F n (y 1 ) b n = F n (y 1 ) then the natural constants to (Of course there are many ierent values of y 1 an y 2 one coul choose...so the choice of constants is still a bit free. The proposition shows that actually all the choices lea to the same type so all choices are equally vali in some sense) We will now focus our attention on the theory of the maximum of n i.i.. ranom variables. Proposition 20. Let X 1,... be ii ranom variables. Let M n := max 1in X i. Notice M n is non-ecreasing as a function of n. a) The istribution function for M n is F n (x) where F (x) is the istribution function for X i. b) Let x 0 = sup {x : F (x) < 1}. Then: lim n M n = x 0 a.s. Proof. a) This is the stanar calculation: ( n ) P (M n x) = P {X i x} = i=1 n P (X i x) = F n (x) b) For any x < x 0, we have F (x) < 1. Hence lim n F n (x) = 0. Hence, since the M n are non-ecreasing, we have: ( ( ) ) P lim M n < x = P {M n < x} n i=1 n=1 = lim n P (M n < x) = 0 Hence for all x < x 0 we know that a.s M n > x eventually. On the other han, by the enition of x 0 we know that M n x 0 always hols. Hence we have that M n x 0 in probability. Finally, since M n are non-ecreasing convergence in probability implies convergence almost surely (since the ba sets: { M n x 0 > ɛ} = {M n < x 0 ɛ} an a ecreasing family of sets as ɛ 0 because M n is non-increasing. Convergence in probability is saying the measure of the sets goes to zero as n while almost sure convergence is saying that their lim-sup (i.e. their intersection) is a null set. Since these sets are ecreasing, this is the same thing!) Theorem 21. (The THREE TYPES THEOREM). Let X 1,... be ii ranom variables anm n := max 1in X i. Suppose there exist sequences a n > 0, b n R so that: P ((M n b n ) /a n x) = F n (a n x + b n ) G(x) Where the convergence is weakly. Assume that G is non-egenerate. Then G is of the type of one of the following three classes: { 0 x < 0 i) Φ α (x) = exp ( x α for some α > 0 ) x 0 { exp ( x α ) x < 0 ii) Ψ α (x) = for some α > 0 1 x 0 iii) Λ α (x) = exp ( exp ( x)) for x R

6 6 Proof. For t R ene the oor: t = greatest integer less than or equal to t For xe t, the sequence nt is a subsequence of the naturals going to innity. Hence by the convergence in the hypothesis we have for any t that: F nt ( a nt x + b nt ) G(x) On the other han, since nt /n t, we also have: F nt (a n x + b n ) = (F n (a n x + b n )) nt n G t (x) If we think of α n = a nt an β n = b nt we see that we have two ierent limits for the centere/scale istribution function F nt, namely: F nt (α n x + β n ) G(x) an F nt (a n x + b n ) G t (x). By the convergence of types proposition, we know that it must be the case that G(x) an G t (x) are the same type an that there are constants A, B that relate the a's an b's to the α's an β's. We can apply this argument for any t, so we will actually get functions A(t) an B(t) so that: (I'm going to actually call them α an β to match the notation in the book...ont get mixe up with the α n an β n we just ha...we are one with those now!) lim n a n a nt = α(t) b n b nt lim = β(t) n a nt G t (x) = G (α(t)x + β(t)) The functions α, β are measurable with respect to t. This follows because the function t an a nt is measurable for every n (its piecewise constant) an limits of measurable functions are still measurable! (Same iea for β). We can now apply a sneaky trick: for any t > 0 an s > 0 we will evaluate G ts (x) for s, t R + an we will write it in two ways, namely as st once an t s once. Have: G st (x) = G (α(ts)x + β(ts)) G st (x) = (G s (x)) t = (G(α(s)x + β(s)) t = G (α(t) [α(s)x + β(s)] + β(t)) = G (α(t)α(s)x + α(t)β(s) + β(t)) Since G is non-egenerate, by a technical lemma which follows, the arguments must be equal an it must be that α(ts)x + β(ts) = α(t)α(s)x + α(t)β(s) + β(t) for every x. Hence: α(ts) = α(t)α(s) β(ts) = α(t)β(s) + β(t) = α(s)β(t) + β(s) (The last line follows by the symmetry s t). The equation α(ts) = α(t)α(s) has a one parameter family of solutions: α(t) = t θ for some constant θ R. (After oing the change of variable ϕ(x) = log (α(e x )), this equation says that ϕ(x+y) = ϕ(x)+ϕ(y), which is the famous Hamel equation. The pathological cases are exclue because α is measurable.) There are now three cases that give rise to the three types, either θ < 0 or θ = 0 or θ > 0. Each case is hanle iniviually: Case 1 (θ = 0; this gives rise to the type Λ(x) = exp ( e x )) In this case α(t) 1 is constant, so the equation for β is: β(ts) = β(t) + β(s) After changing variables again, this is again the Hamel equation, so we have: This yiles the following equaltion for G: β(t) = c log(t) for some c R G t (x) = G(x c log t) We will now argue c > 0. If c = 0, we have G t (x) = G(x) for every t, which only happens if G is a egenerate istribution (a point mass at some point...so it takes only values in {0, 1}). c cannot be negative, as G t (x) must be a non-increasing function of t (as G(x) 1 always), an G(x c log t) is a ecreasing function when c < 0. So it must be the case that c > 0.

7 We will now argue that G(x) < 1 for every x R. Otherwise, if G(x 0 ) = 1 for some x 0 plugging in this value woul yiel 1 = G(x 0 c log t) for every t. But then G(x) = 1 for every x since the range of x 0 c log t is all of R. Contraiction! So it must be that G(x) < 1 for all x R. A similar argument shows that G(x) > 0 for all x. Dene p := log ( log(g(0))) so that G(0) = exp ( e p ). Let u = c log(t) (the range of u is (, ) as t ranges in (0, ) ). Changing variables in the expression G t (0) = G( c log t) gives: G(u) = exp ( e p t ) = exp ( e ) (c 1 u+p) = Λ(c 1 u + p) So inee, G is in the same type class as Λ in this case. Case 2 (θ > 0; gives rise to the the type Φ α = exp ( x α ) supporte on x > 0) Rearranging the equation for β gives: β(s) 1 α(s) = β(t) 1 α(t) Hence this value is constant for every value of t. Writing c for this constant, we have then: So the equation for G is then: Let y = x c now to get: β(t) β(t) = (1 α(t)) 1 α(t) = c (1 α(t)) = c ( 1 t θ) G t (x) = G(t θ x + c ( 1 t θ) ) = G ( t θ (x c) + c ) G t (y + c) = G ( t θ y + c ) If we let H(x) = G(x + c), this is in the same type-class as G, so it suces to work with H which satises H t (y) = H(t θ y). We claim now that H(0) = 0. Setting y = 0 in the above gives H t (0) = H(0) so it is clear that H(0) can only be 1 or 0. However, H(0) = 1 is impossible however by the following argument by contraiction. Suppose that H(0) = 1, then since H is non-egenrate, then we have an x 0 < 0 with 0 < H(x 0 ) < 1 an so H t (x 0 )is a ecreasing function of t. On the other han, H(t θ x 0 ) is an increasing function of t (since t θ x 0 is increasing as x 0 < 0), which is a contraiction since these are equal. Hence H(x) = 0 for x < 0 since H is a istribution function. Now, plugging in y = 1 we have H t (1) = H(t θ ). It must be that 0 < H(1) < 1 or else H is ientically 0 or 1 everywhere. Let α := θ 1, an ene p so that H(1) = exp ( p α ). Uner the change of variable u = t θ so that t = u θ 1 = u α. Notice that u ranges in (0, ) as t > 0. We have: H(u) = H t (1) = exp ( tp α) = exp ( (pu) α) = Φ α (pu) So inee H an therefore G are in the typeclass of Φ. Case 3 (θ < 0; gives rise to the the type Ψ α = exp ( x α ) supporte on x < 0) This case is very similar to the previous case. As before, we get to (to clarify things I write θ as θ to remin myself θ < 0 here) : (1 t θ ) β(t) = c G t (y + c) = G ( ) t θ y + c H t (y) = H(t θ y) We claim now that H(0) = 1. (This is analagous to the argument that showe H(0) = 0 in Case 2) Setting y = 0 in the above gives H t (0) = H(0) so it is clear that H(0) can only be 1 or 0. However, H(0) = 0 is impossible however by the following argument by contraiction. Suppose H(0) = 0, then since H is non-egenrate, then we have an x 0 > 0 with 0 < H(x 0 ) < 1 an so H t (x 0 )is a ecreasing function of t. On the other han, H(t θ x 0 ) is an increasing function of t (since t θ x 0 is increasing), which is a contraiction since these are equal. Hence H(x) = 1 for x > 0 since H is a istribution function.

8 8 Now, plugging in y = 1 we have H t ( 1) = H( t θ ). It must be that 0 < H(1) < 1 or else H is ientically 0 or 1 everywhere. Let α := θ 1, an ene p > 0 so that H( 1) = exp ( p α ). Uner the change of variable u = t θ so that t = ( u) θ 1 = u α. Notice that u ranges in (, 0) as t > 0. We have: So inee H an therefore G are in the typeclass of Ψ α. H(u) = H t ( 1) = exp ( tp α ) = exp ( (p u ) α ) = Ψ α (pu) Lemma 22. (This was an exercise in Resnick) If F is a non-egenerate istribution an a > 0, c > 0, b R an R an if: Then a = c an b =. F (ax + b) = F (cx + ) Proof. Uner the change of variables y = cx +, x = c 1 (y ) we are left with the hypothesis F (ac 1 y + b ) = F (y). Relabeling these as A an B we have F (Ay + B) = F (y) an we esire to prove that A = 1 an B = 0. Firstly notice that if we let T (x) = Ax + B then we can iterate We rst claim that B = 0. Otherwise, plugging in y = 0 gives F (B) = F (0) which shows F is constant on the interval [0, B] (or if B is negative, on [ B, 0]. From now on we will only consier the case B > 0, the other case is very similar). Then for any B A < y < 0 we have that 0 < Ay + B < B so Ay + B [0, B] an hence F (Ay + B) = F (0) = F (B) is the same constant. On the other han, F (Ay + B) = F (y) so F must achieve the same constant value in the range [ B A, 0] [0, B]. The same argument can be repeate to further exten the interval to the left a bit further. For any B ( 1 A + ) 1 A 2