Robust Estimation and Inference for Extremal Dependence in Time Series. Appendix C: Omitted Proofs and Supporting Lemmata

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1 Robust Estimation and Inference for Extremal Dependence in Time Series Appendix C: Omitted Proofs and Supporting Lemmata Jonathan B. Hill Dept. of Economics University of North Carolina - Chapel Hill January 24, 29 This appendix contains the omitted proof of Lemma A.3 (Section C.), and supporting Lemmas B-B4 (Section C.2). Recall and take values on [) with probability one, denotes either or and ( ) = () as!, slowly varying (C.) and ( )! (C.2) C. LEMMA A.3 We present an expanded version of Lemma A.3. The proofs of Lemmas A.2 and A.6 use Lemma A.3.iii,iv which is proved by Lemma A.3.i,ii. Recall ¹ () := ( ) ( ), 2 R + LEMMA A.3 Let Assumption A hold. f ¹ () 2 g and f 2 g are 4 -NED on fz g with constants ½ Z ¾ ~ () ~ () = ( ) uniformly in 2 for arbitrary 2, and common coe cients = (( ) 2 2 ). For any, f¹ ()¹ () 2 g and f 2 g are 2 -NED on fz g with constants n () () ~ () o = 2 ( ) uniformly in Dept. of Economics, University of North Carolina-Chapel Hill, jbhill@ .unc.edu.

2 Proof. Claim i. and common coe cients ª = (( ) 2 2 ). For any 2 R, =, f P = ¹ () ¹ () 2 g and f P = 2 g are 2 -NED on fz g with constants n o ½ n o n o ¾ () () () = max () () max ~ () = ( ) uniformly in 2 and common coe cients ª () = (( ) 2 2 ). Under the conditions of ( ), f P = ¹ () ¹ () 2 z g and f P = 2 z g form 2 -mixingale sequences with size 2 and constants f g = ( 2 ) uniformly in. We tackle ¹ () and in two steps. Step ( ¹ ()): By the de nition of E-NED and the statement of Assumption A, it follows for some () = (( ) ) and = (( ) 2 2 ), where 2, ¹ h i () 2 ¹ () 2 jz+ 4 2 () = ~ () where ~ () = () 2 = 2 ( ) is Lebesgue integrable by Assumption A. Therefore f ¹ () 2 g is 4 -NED on fz g with constants ~ () and coe cients with size 2. Step 2 ( ): De ne () := ( ) jz + 2

3 and note [ jz + ] 4 h = (ln ( )) + " µz = = = = " " Z " Z Z Z Ã Z (ln( )) + jz + i 4 h 4 # 4 ( ) jz + i Z Z Z Z Z Z Z Z Z k ()k 4!# 4 4Y ( ) = " 4 # 4 Y ( ) # = # 4 4Y k ( 2 )k = = ( ) jz + 4 µz () The second equality follows from (ln( )) + = s ( )and the Fubini- Tonelli Theorem: [(ln( )) + jz + ] = [s ( )jz + ] = s ( jz + ). The fourth equality follows from the Fubini-Tonelli Theorem. The rst inequality is Cauchy-Schwartz s. The last inequality follows from Assumption A. This proves f g is 4 -NED on fz g with constants s ()and coe cients, where s ()= (( ) ) uniformly in under Assumption A. Finally, the proof that f 2 g is 4 -NED on fz g with constants s ~ () = ( 2 ( ) ) uniformly in and coe cients follows instantly. Claim ii. Use the 4 -NED properties of f ¹ ()g and f ¹ () 2 g by Claim () to deduce the convolution ¹ ¹ () := ¹ ( ) ¹ ( 2 ) 2 satis es (see Davidson 994: Theorem 7.9) ¹ ¹ i () h ¹ ¹ ()jz ¹ ( ) 4 ~ ( 2 ) + ¹ ( 2 ) 2 ( ) 4 = 3 ¹ ( ) 4 ~ ( 2 ) + ¹ ( 2 ) 4 ~ ( ) Now use Lemma B. to deduce ¹ ¹ i () h ¹ ¹ ()jz + 2 n ( ) 4 max ~ ( ) ~ o ( 2 ) max ª = () () ª 3

4 say, where by Claim () n o () () = ( ) 4 max ~ ( )~ ( 2 ) = ( ) 4 2 = ( ) uniformly in 2 and ª = max ª = ( ) 2 2 Therefore f¹ ( )¹ ( 2 ) 2 g is 2 -NED on fz g with constants () () = ( 2 ( ) ) and coe cients ª = (( ) 2 2 ). The proof that f 2 g is 2 -NED follows from an argument similar to Step 2 of Claim (), where the constants are ~ () ½Z = ( ) 4 max = ( ) uniformly in 2 and the coe cients are also ª. Claim iii. X = Use Minkowski s inequality to deduce " ¹ () ¹ X () 2 X = Z ¾ ~ () ~ () = n () () ª max = () () ª() ¹ () ¹ () 2 jz + # 2 () () o ( ª ) say, where () () = ( 2 ( ) ) uniformly in andª () = (( ) 2 2 ) under Claim ( ). An identical argument applies to P = 2. Claim iv. The nal claim follows from Claim ( ) and arguments in Davidson (994: p ). Consider ¹ ¹ () := P = ¹ () ¹ () 2, the proof for P = 2 being similar. If the base f g is strong mixing with coe cients then (see Davidson 994: eq. 7.8) ¹¹ ()jz 2 max ¹¹ () () ª + ª () 2 The Minkowski andthe Cauchy-Schwartzinequalities, =, and jj¹ ()jj = (( ) ) uniformly in by Lemma B., below, imply for any 2 ¹ ¹ () 2 ¹ 2 ¹ 2 = 2 ( ) Since () () = ( 2 ( ) ) uniformly in and ª () = (( ) 2 2 ) by Claim ( ), and 2 = (( ) 2 (2 ) ( 2) ) = (( ) 2 4

5 2 ) by Assumption A, it follows ¹ ¹ ()jz 2 2 ( ) 2 + ª () = 2 ( ) ) (( ) 2 2 ) + (( ) 2 2 ) = 2 ( ) 2 (( ) 2 2 ) + (( ) 2 2 = 2 ( 2 ) = ( 2 ) say. A similar argument applies to the remaining mixingale inequality jj ¹ ¹ () [¹¹ ()jz + ]jj 2 ( 2 ), and in the uniform mixing case, cf. Davidson (994: eqs. (7.9)-(7.2)). C.2 SUPPORTING LEMMAS B.-B.4 Throughout = (( ) ). LEMMA B. Under (C.)-(C.2), jj jj = (( ) ) and jj ¹ jj = (( ) ) uniformly in for any. LEMMA B.2 Under Assumptions A, B and E, for all 2 2 N = X j j ^ ( 2 ) = () LEMMA B.3 Under Assumptions A, B and E: P = ^ ^ = ( 2 );. P = ( [ ]) = ( 2 ). LEMMA B.4 Under Assumptions A, B and E Ã! X ^ ^ 2^ ^ () = ^ = Proof of Lemma B.. Use Minkowski s and Liapunov s inequalities and arguments in Hsing (99: p. 548), cf. (C.)-(C.2), to deduce for all k k 2 (ln( )) +» 2 ( )! ª = (( ) ) which does not depend on. Similarly, the construction of implies for any ¹ ( ) + ( )» ( ) + ( ) = (( ) ) 5

6 Proof of Lemma B.2. Write ^ := ^ and (C.3) ^ ( 2 ) = ^ ^ ^ 2 ^ 2 = ^ ^ 2 ^ + ^ ^ 2 ^ + ^ ^ 2 ^ + ^ ^ ^ 2 + ^ 2 ^ + + ^ ^ ^ 2 ^ We will prove X j j 2 ^ = () = Nearly identical arguments su ce to a show kernel weighted sum of each remaining term in (C.3) is (). Exploiting ^ = + ( 2 ) by Theorem 5. of Hill (29b) and Assumption B we may write (cf. Hsing 99: p. 554) ^ = ln ( +) + (ln( )) + ( ) ( ª )(ln( )) + ( ) ^ ª = ln ( +) + (ln( )) + +( 2 ) By cases it is straightforward to show ln ( +) + (ln ( )) + ln ( +) Note = X 2 ^ = X 2 = n ln o ( +) + (ln( )) + + X 2 (( ) 2 ) 2 = = X 2 = n ln o (+) + (ln( )) + + () 6

7 The last line follows from applications of the Minkowski and Cauchy-Schwartz inequalities, and Lemma B.: use P = j j = ( 2 ) under Assumption E to get X = 2 = X j j k k 2 k k 4 2 µ = 2 ( ) 2 ( ) 42 = 2 and ( 2 ( ) 2 2 ) = (( ) 2 ) = (). Therefore X ln 2 ( +) = = + (ln( )) + X j j k 2 k 2 ln ( +) 2 + X 2 j j k 2 k 2 (( ) 2 ) = 2 ln ( +) 2 X 2 j j k k 8 k k 8 k k 4 = + (( ) 2 ) = () X 2 j j + (( ) 2 ) = () = Similarly, because ln( ( +) ) = ( 2 ) by Theorem 5. Hill (29b), X ^ ^ = ^ 2 2 ^ ln 2 X (+) ln (+) 2 = ( ) X j j = = ( 2 ) 2 = X j j = () = j j Proof of Lemma B.3. Claim i: Since the kernel symmetrically and negligibly trims over =, and P = j j = ( 2 34 ) = () under Assumption E and ( ) = (), 7

8 the claim follows by an argument identical to the proof of Lemma A.. Claim ii: Simply augment the argument used to prove Lemma A.2. Since P = j j = ( 2 34 ) = () it is easy to prove under Assumptions A and B ( 34 X 2 = ) is 2 -NED on fz g with constants () = ( 2 ) uniformly in and coe cients ª () = (( ) 2 2 ), cf. Lemma A.3. Now apply Corollary 3.3 of Hill (29b) to deduce à X 34 " X 34 #! X = = 2 = = ( 2 ) 2 Multiply both sides by and 2 34 and compress the summation to conclude Ã! X 2 ( [ ]) = 34 given = (). = 2 = 2 Proof of Lemma B.4. Write := (2) and ^ := (2)^ ^. We rst show that everywhere ^ and ^ ^ can be replaced with and X = and then prove X µ ^^ ^ ^ () = = X = µ () + () µ () = 2 32 Step : The arguments in the proof of Lemma A. carry over entirely if we replace ^ ^ with ^ ^. In particular it is straightforward to show à X = =! X ^ ^ ^ à = X = =! X ^ + 2 8

9 Since ^ = + ( 2 ) is implied by Theorem 5. of Hill (29b) it follows Ã! X X ^^ ^ = = = à X = =! X + 2 à X = = + 2 X! Since jj jj 2 = (( ) 2 ) uniformly in by Lemma B., use Assumption E and the Cauchy-Schwartz inequality to deduce à X X! = = X j j = = 2 Therefore Ã! X X ^ ^ ^ = = = Ã! X X + = = = Ã! X X + () = = 2 Now observe Lemma A., Theorem 3. and ^ = + ( 2 ) imply ^ () = X ^ ^ ^ = + ( ) 2 = () + X ^ ^ = + ^ = () + 2 X ^ ^ = and P = j j = ( 2 ) under Assumption E, hence X = = = µ ^ ^ ^ ^ () X = µ () + X = 2 X µ () + ( 2 ) = 9

10 Step 2: By construction X = µ () = X = = X ( [ ]) so we need only prove X X = = ( [ ]) = 2 since 2 = ( 32 ) follows trivially from = (). The rate follows instantly from Lemma B.3 and = ().