Two-sided bounds for L p -norms of combinations of products of independent random variables

Transcription

1 Two-sided bounds for L p -norms of combinations of products of independent random ariables Ewa Damek (based on the joint work with Rafał Latała, Piotr Nayar and Tomasz Tkocz) Wrocław Uniersity, Uniersity of Warsaw, Uniersity of Warwick Harmonic Analysis, Probability and Applications Orleans, June

2 Wojciechowski Question Let X, X 1, X 2,... be i.i.d. nonnegatie r.. s such that EX = 1 and P(X = 1) < 1. Define R 0 := 1 and k R k := X j for k = 1, 2,.... j=1 Obiously ER k = 1 and therefore for any a 0, a 1,..., a n, E a k R k a k. Question. (M. Wojciechowski) Is it true that for any i.i.d. sequence the aboe estimate may be reersed, i.e. there exists a constant c > 0 that depends only on the distribution of X such that E a k R k c a k for any a 0,..., a n?

3 Wojciechowski Question Let X, X 1, X 2,... be i.i.d. nonnegatie r.. s such that EX = 1 and P(X = 1) < 1. Define R 0 := 1 and k R k := X j for k = 1, 2,.... j=1 Obiously ER k = 1 and therefore for any a 0, a 1,..., a n, E a k R k a k. Question. (M. Wojciechowski) Is it true that for any i.i.d. sequence the aboe estimate may be reersed, i.e. there exists a constant c > 0 that depends only on the distribution of X such that E a k R k c a k for any a 0,..., a n?

4 Wojciechowski Question Let X, X 1, X 2,... be i.i.d. nonnegatie r.. s such that EX = 1 and P(X = 1) < 1. Define R 0 := 1 and k R k := X j for k = 1, 2,.... j=1 Obiously ER k = 1 and therefore for any a 0, a 1,..., a n, E a k R k a k. Question. (M. Wojciechowski) Is it true that for any i.i.d. sequence the aboe estimate may be reersed, i.e. there exists a constant c > 0 that depends only on the distribution of X such that E a k R k c a k for any a 0,..., a n?

5 L 1 bound for products of i.i.d. nonnegatie r. s R k := k j=1 X j The answer to Wojciechowski s question is positie een in the more general case of ector coefficients. Theorem (Latała 2013) Let X, X 1, X 2,... be an i.i.d. sequence of nonnegatie nondegenerate r. s such that EX = 1. Then there exists a constant c that depends only on the distribution of X such that for any 0, 1,..., n in a normed space (F, ), E k R k c k.

6 L 1 bound in the non i.i.d. case Consider sequence (X i ) satisfying the following assumptions: X 1, X 2,... are independent, nonnegatie mean one r. s, (1) E X l λ < 1 and E X l 1 µ > 0 for all l, (2) Theorem (Latała 2013) E X l 1 1 {Xl A} 1 µ for all l. (3) 4 Let X 1, X 2,... satisfy assumptions (1), (2) and (3). Then for any ectors 0, 1,..., n in a normed space (F, ), we hae E k R k 1 512r µ3 where r is a positie integer such that k, 2 17 (1 λ) 2 rλ2r 2 A µ 3. (4)

7 L 1 bound in the non i.i.d. case Consider sequence (X i ) satisfying the following assumptions: X 1, X 2,... are independent, nonnegatie mean one r. s, (1) E X l λ < 1 and E X l 1 µ > 0 for all l, (2) Theorem (Latała 2013) E X l 1 1 {Xl A} 1 µ for all l. (3) 4 Let X 1, X 2,... satisfy assumptions (1), (2) and (3). Then for any ectors 0, 1,..., n in a normed space (F, ), we hae E k R k 1 512r µ3 where r is a positie integer such that k, 2 17 (1 λ) 2 rλ2r 2 A µ 3. (4)

8 L p -bounds for products of i.i.d. r. s It turns out that L 1 -bounds may be extended to L p for p > 0. Positiity of X is not needed. Namely we hae Theorem (Latała, Nayar, Tkocz, Damek) Let p > 0 and X, X 1, X 2,... be an i.i.d. sequence of r.. s P( X = t) < 1 for all t. Then there exist constants 0 < c p,x C p,x < which depend only on p and the distribution of X such that for any ectors 0, 1,..., n in a normed space (F, ), c p,x i p E R i p p E i R i C p,x i p E R i p. R k := k j=1 X j, E X p not necessarily equal 1.

9 L p -bounds p > 0 R k := k j=1 X j Theorem (Latała, Nayar, Tkocz, Damek) Let p > 0 and X 1, X 2,... be independent r. s with finite p-th moments, X i non degenerate, satisfying some uniform behaior assumptions. Then for any ectors 0, 1,..., n in a normed space (F, ) we hae c(x 1, X 2,...) i p E R i p p E i R i C(X 1, X 2,...) i p E R i p, where c(x 1, X 2,...), C(X 1, X 2,...) are positie constants that depend only on the uniform behaior and they are quite explicit.

10 Example Assumption P( X = t) < 1 for all t is crucial since for any p > 0 by the Khintchine inequality, E k=1 l=1 k p p ε l = E ε k p (E k=1 2) p/2 ε k = n p/2. k=1 Here ε l are i.i.d symmetric random ariables taking alues 1, 1. E ε l p = 1, i = 1,. p nc p,x E R i nc p,x

11 Riesz products Let T = R/2πZ be the one-dimensional torus and m be the normalized Haar measure on T. Riesz products are defined on T by the formula i R i (t) = (1 + cos(n j t)), i = 1, 2,..., (5) j=1 where (n k ) k 1 is a lacunary increasing sequence of positie integers. The result of Y. Meyer gies that if n k+1 /n k 3 and k n k n k+1 < then a k R k Lp(T) (E a k R k p ) 1/p for p 1, where R k are products of independent random ariables distributed as R 1. Therefore main Theorem yields an estimate for n a i R i Lp(T).

12 Riesz products Let T = R/2πZ be the one-dimensional torus and m be the normalized Haar measure on T. Riesz products are defined on T by the formula i R i (t) = (1 + cos(n j t)), i = 1, 2,..., (5) j=1 where (n k ) k 1 is a lacunary increasing sequence of positie integers. The result of Y. Meyer gies that if n k+1 /n k 3 and k n k n k+1 < then a k R k Lp(T) (E a k R k p ) 1/p for p 1, where R k are products of independent random ariables distributed as R 1. Therefore main Theorem yields an estimate for n a i R i Lp(T).

13 Riesz products Corollary Suppose that (n k ) k 1 is an increasing sequence of positie integers such that n k+1 /n k 3 and n k k=1 n k+1 <. Then for any coefficients a 0, a 1,..., a n R and p 1, c p a k p R k (t) p p dm(t) a k Rk (t) dm(t) T T C p a k p T R k (t) p dm(t), where 0 < c p C p < are constants depending only on p and the sequence (n k ). Dechamps condition k=1 ( n k n k+1 ) 2 <. Anyway n k (k!) α

14 Riesz products We expect that the assumptions on the growths of n k may be weakened to n k+1 /n k C p, but we are able to show it only for p = 1. Theorem (Latała, Nayar, Tkocz) There exist constants C 1 < and c 1 > such that if n k+1 /n k C 1 then for any ectors 0, 1,..., n in a normed space (F, ), dm k k R k c1 k. T

15 Perpetuities Let us consider the random difference equation S d = XS + B, (6) where the equality is meant in law, (X, B) is a random ariable with alues in [0, ) R independent of S. Let (X i, B i ) be i.i.d. copies of (X, B). It is known that (under some mild integrability assumptions) the infinite series i 1 R i 1 B i = B 1 + X 1 R i 1 B i, R i 1 := X j i=2 j=1 conerges almost eerywhere to a solution of (6). It is called perpetuity. The conditions for conergence are E log X < 0 and E log + B <

16 Perpetuities Let us consider the random difference equation S d = XS + B, (6) where the equality is meant in law, (X, B) is a random ariable with alues in [0, ) R independent of S. Let (X i, B i ) be i.i.d. copies of (X, B). It is known that (under some mild integrability assumptions) the infinite series i 1 R i 1 B i = B 1 + X 1 R i 1 B i, R i 1 := X j i=2 j=1 conerges almost eerywhere to a solution of (6). It is called perpetuity. The conditions for conergence are E log X < 0 and E log + B <

17 Perpetuities Let us consider the random difference equation S d = XS + B, (6) where the equality is meant in law, (X, B) is a random ariable with alues in [0, ) R independent of S. Let (X i, B i ) be i.i.d. copies of (X, B). It is known that (under some mild integrability assumptions) the infinite series i 1 R i 1 B i = B 1 + X 1 R i 1 B i, R i 1 := X j i=2 j=1 conerges almost eerywhere to a solution of (6). It is called perpetuity. The conditions for conergence are E log X < 0 and E log + B <

18 Perpetuities Oer the last 40 years equation S d = XS + B, (7) and its arious modifications (in particular multidimensional analogues) hae attracted a lot of attention. It has a rather wide spectrum of applications including random walks in random enironment, branching processes, fractals, finance, insurance, telecommunications, arious physical and biological models. P. Bougerol, M. Babillot, S. Brofferio, L. Elie, Y. Guiarc h, E. Le Page, G. Letac, N. Picard, C. Sabot, B. Sapporta, O. Wintenberger G. Alsmeyer, S. Mentemeier, P. Diaconis, D. Freedman, Ch. Goldie, A. Grinceicius, R. Grübel, H. Furstenberg, H. Kesten, C. Klüppelberg, J. Collamore, T. Mikosch, W. Veraat, P. Hitchenko, J. Wesołowski D. Buraczewski, K. Kolesko, P. Dyszewski, J. Zienkiewicz.

19 Perpetuities Oer the last 40 years equation S d = XS + B, (7) and its arious modifications (in particular multidimensional analogues) hae attracted a lot of attention. It has a rather wide spectrum of applications including random walks in random enironment, branching processes, fractals, finance, insurance, telecommunications, arious physical and biological models. P. Bougerol, M. Babillot, S. Brofferio, L. Elie, Y. Guiarc h, E. Le Page, G. Letac, N. Picard, C. Sabot, B. Sapporta, O. Wintenberger G. Alsmeyer, S. Mentemeier, P. Diaconis, D. Freedman, Ch. Goldie, A. Grinceicius, R. Grübel, H. Furstenberg, H. Kesten, C. Klüppelberg, J. Collamore, T. Mikosch, W. Veraat, P. Hitchenko, J. Wesołowski D. Buraczewski, K. Kolesko, P. Dyszewski, J. Zienkiewicz.

20 Perpetuities More can be said if we assume additionally that for some p > 0, Then for eery q < p P(X = 1) < 1, EX p = 1, E B p <. (8) EX q < 1 because the function q EX q is conex and equal 1 at 0, p. In particular, for q < p No moment of order p. E R i 1 B i q <.

21 Perpetuities More can be said if we assume additionally that for some p > 0, Then for eery q < p P(X = 1) < 1, EX p = 1, E B p <. (8) EX q < 1 because the function q EX q is conex and equal 1 at 0, p. In particular, for q < p No moment of order p. E R i 1 B i q <.

22 Perpetuities More can be said if we assume additionally that for some p > 0, Then for eery q < p P(X = 1) < 1, EX p = 1, E B p <. (8) EX q < 1 because the function q EX q is conex and equal 1 at 0, p. In particular, for q < p No moment of order p. E R i 1 B i q <.

23 Perpetuities More can be said if we assume additionally that for some p > 0, Then for eery q < p P(X = 1) < 1, EX p = 1, E B p <. (8) EX q < 1 because the function q EX q is conex and equal 1 at 0, p. In particular, for q < p No moment of order p. E R i 1 B i q <.

24 Perpetuities But how fast does E R i 1 B i p grow with n? Moreoer if EX p = 1, EX p log X <, log X has nonlattice distribution, E B p < and P(X + B = ) < 1 for any then lim t tp P( R i 1 B i > t) = c (X, B) = 1 αρ E( S p S B p ) and c (X, B) is a finite positie constant. The latter was proed by Kesten, then the proof was simplified by Goldie. The proof goes ia the renewal theorem and there is no good expression for the constant c (X, B) called nowadays the Goldie constant or the Goldie-Kesten constant. Although Goldie proided a formula for c (X, B) but positiity of the latter could not be deried from it. A new idea was needed. It was proided by Grinceicius in the seenties and further deeloped by Goldie.

25 Perpetuities But how fast does E R i 1 B i p grow with n? Moreoer if EX p = 1, EX p log X <, log X has nonlattice distribution, E B p < and P(X + B = ) < 1 for any then lim t tp P( R i 1 B i > t) = c (X, B) = 1 αρ E( S p S B p ) and c (X, B) is a finite positie constant. The latter was proed by Kesten, then the proof was simplified by Goldie. The proof goes ia the renewal theorem and there is no good expression for the constant c (X, B) called nowadays the Goldie constant or the Goldie-Kesten constant. Although Goldie proided a formula for c (X, B) but positiity of the latter could not be deried from it. A new idea was needed. It was proided by Grinceicius in the seenties and further deeloped by Goldie.

26 Perpetuities But how fast does E R i 1 B i p grow with n? Moreoer if EX p = 1, EX p log X <, log X has nonlattice distribution, E B p < and P(X + B = ) < 1 for any then lim t tp P( R i 1 B i > t) = c (X, B) = 1 αρ E( S p S B p ) and c (X, B) is a finite positie constant. The latter was proed by Kesten, then the proof was simplified by Goldie. The proof goes ia the renewal theorem and there is no good expression for the constant c (X, B) called nowadays the Goldie constant or the Goldie-Kesten constant. Although Goldie proided a formula for c (X, B) but positiity of the latter could not be deried from it. A new idea was needed. It was proided by Grinceicius in the seenties and further deeloped by Goldie.

27 Perpetuities But how fast does E R i 1 B i p grow with n? Moreoer if EX p = 1, EX p log X <, log X has nonlattice distribution, E B p < and P(X + B = ) < 1 for any then lim t tp P( R i 1 B i > t) = c (X, B) = 1 αρ E( S p S B p ) and c (X, B) is a finite positie constant. The latter was proed by Kesten, then the proof was simplified by Goldie. The proof goes ia the renewal theorem and there is no good expression for the constant c (X, B) called nowadays the Goldie constant or the Goldie-Kesten constant. Although Goldie proided a formula for c (X, B) but positiity of the latter could not be deried from it. A new idea was needed. It was proided by Grinceicius in the seenties and further deeloped by Goldie.

28 Goldie-Kesten constant and finite sums of perpetuities There is an expression for c (X, B) in a paper by N.Enriquez, C.Sabot, O.Zindy 2009 but it is ery complicated and only for positie B independent of X. There is another one in a paper by J. Collamore, A. Vidyashankar 2013 again for positie B and the law of X being non singular. There is something simpler in a paper by K. Bartkowiak, A.Jakubowski, T.Mikosch and O.Wintenberger 2011 but only for B = 1 lim n 1 npρ E R i 1 p = c (X, 1) > 0. The assumptions are as in Goldie. In particular, ρ = EX p log X <. The latter was an inspiration for my team.

29 Goldie-Kesten constant and finite sums of perpetuities There is an expression for c (X, B) in a paper by N.Enriquez, C.Sabot, O.Zindy 2009 but it is ery complicated and only for positie B independent of X. There is another one in a paper by J. Collamore, A. Vidyashankar 2013 again for positie B and the law of X being non singular. There is something simpler in a paper by K. Bartkowiak, A.Jakubowski, T.Mikosch and O.Wintenberger 2011 but only for B = 1 lim n 1 npρ E R i 1 p = c (X, 1) > 0. The assumptions are as in Goldie. In particular, ρ = EX p log X <. The latter was an inspiration for my team.

30 Goldie-Kesten constant and finite sums of perpetuities There is an expression for c (X, B) in a paper by N.Enriquez, C.Sabot, O.Zindy 2009 but it is ery complicated and only for positie B independent of X. There is another one in a paper by J. Collamore, A. Vidyashankar 2013 again for positie B and the law of X being non singular. There is something simpler in a paper by K. Bartkowiak, A.Jakubowski, T.Mikosch and O.Wintenberger 2011 but only for B = 1 lim n 1 npρ E R i 1 p = c (X, 1) > 0. The assumptions are as in Goldie. In particular, ρ = EX p log X <. The latter was an inspiration for my team.

31 Goldie-Kesten constant and finite sums of perpetuities Recently Buraczewski, Damek and Zienkiewicz showed that if additionally E(X p+ε + B p+ε ) < for some ε > 0 then lim n 1 npρ E R i 1 B i p = c (X, B) > 0, where ρ := EX p log X, EX p = 1. The same for some multidimensional models. S = XS + B, B, S R d, or similarities in place of X The first obseration is that this finite sums grow like n. Secondly they gie an expression for the Goldie-Kesten constant. Is it good? It is simple, but not necessarily good for simulations.

32 Goldie-Kesten constant and finite sums of perpetuities Recently Buraczewski, Damek and Zienkiewicz showed that if additionally E(X p+ε + B p+ε ) < for some ε > 0 then lim n 1 npρ E R i 1 B i p = c (X, B) > 0, where ρ := EX p log X, EX p = 1. The same for some multidimensional models. S = XS + B, B, S R d, or similarities in place of X The first obseration is that this finite sums grow like n. Secondly they gie an expression for the Goldie-Kesten constant. Is it good? It is simple, but not necessarily good for simulations.

33 Goldie-Kesten constant and finite sums of perpetuities Recently Buraczewski, Damek and Zienkiewicz showed that if additionally E(X p+ε + B p+ε ) < for some ε > 0 then lim n 1 npρ E R i 1 B i p = c (X, B) > 0, where ρ := EX p log X, EX p = 1. The same for some multidimensional models. S = XS + B, B, S R d, or similarities in place of X The first obseration is that this finite sums grow like n. Secondly they gie an expression for the Goldie-Kesten constant. Is it good? It is simple, but not necessarily good for simulations.

34 Goldie-Kesten constant and finite sums of perpetuities c p,x i p p E i R i C p,x i p. Notice that our L p bounds on moments yield that if X, B are independent, EX p = 1 then for eery n, c p,x E B i p E R i 1 B i p C p,x E B i p c p,x E B p 1 n E R i 1 B i p C p,x E B p ia conditioning on B. There is no limit, only bounds but besides independence the assumptions are much weaker, een weaker then in the Goldie theorem. We need only EX p = 1, E B p <, P(X = 1) < 1. In fact we may get rid of the independence assumption.

35 Goldie-Kesten constant and finite sums of perpetuities c p,x i p p E i R i C p,x i p. Notice that our L p bounds on moments yield that if X, B are independent, EX p = 1 then for eery n, c p,x E B i p E R i 1 B i p C p,x E B i p c p,x E B p 1 n E R i 1 B i p C p,x E B p ia conditioning on B. There is no limit, only bounds but besides independence the assumptions are much weaker, een weaker then in the Goldie theorem. We need only EX p = 1, E B p <, P(X = 1) < 1. In fact we may get rid of the independence assumption.

36 A family of perpetuities Before we proceed further let us consider the family of random equations S (d) = XS (d) + B (d) B (d) being a random ector in R d. So S (d) R d. Suppose for a moment that X and B (d) are independent, S (d) = R i 1 B (d) i and if EX p = 1, E B (d) p <, P(X = 1) < 1 then c p,x E B (d) p 1 n E R i B (d) p i C p,x E B (d) p

37 A family of perpetuities c p,x E B (d) p 1 n E R i B (d) p i C p,x E B (d) p Together with Buraczewski, Zienkiewicz, Damek result (that requires more moments) lim n 1 npρ E R i B (d) i p = C (X, B (d) ) this gies uniform bounds for the Goldie constant C (X, B (d) ) = lim t p P( R i B (d) t i > t) (independent of the dimension) c p,x E B (d) p C (X, B (d) ) C p,x E B (d) p

38 L p bound for finite sums of perpetuities Theorem (Latała, Nayar, Tkocz, Damek) Suppose that F is a separable Banach space. Let p > 0 and let an i.i.d. sequence (X, B), (X 1, B 1 ),... with alues in [0, ) F be such that X is nondegenerate and E B p, EX p = 1. Assume additionally that P(X + B = ) < 1 for eery F. Then there are constants 0 < c p (X, B) C p (X) < which depend on p and the distribution of (X, B) such that for eery n, c p (X, B)nE B p p E R i 1 B i C p (X)nE B p.

39 Finite sums of perpetuities There are quite explicit formulae for the constants 0 < c p (X, B) C p (X) < and although they are not close it seems that in some particular cases of perpetuities one can elaborate. The sum R i 1 B i is much easier to study then partial sums n R i 1 B i due to the renewal theorem. Nobody looked at perpetuities in this way yet. Recently D.Buraczewski, J. Collamore, J.Zienkiewicz and myself hae deeloped methods to study tails partial sums i.e. P( R i 1 B i > t) 1 n t α(n,t) or t α(n,t) without using Latała, Nayar, Tkocz, Damek result. I hae a dream to combine both. In particular both approaches work for (X i, B i ) being not necessarily i.i.d just independent proided some uniform behaior is guaranteed.

40 Finite sums of perpetuities There are quite explicit formulae for the constants 0 < c p (X, B) C p (X) < and although they are not close it seems that in some particular cases of perpetuities one can elaborate. The sum R i 1 B i is much easier to study then partial sums n R i 1 B i due to the renewal theorem. Nobody looked at perpetuities in this way yet. Recently D.Buraczewski, J. Collamore, J.Zienkiewicz and myself hae deeloped methods to study tails partial sums i.e. P( R i 1 B i > t) 1 n t α(n,t) or t α(n,t) without using Latała, Nayar, Tkocz, Damek result. I hae a dream to combine both. In particular both approaches work for (X i, B i ) being not necessarily i.i.d just independent proided some uniform behaior is guaranteed.

41 Finite sums of perpetuities There are quite explicit formulae for the constants 0 < c p (X, B) C p (X) < and although they are not close it seems that in some particular cases of perpetuities one can elaborate. The sum R i 1 B i is much easier to study then partial sums n R i 1 B i due to the renewal theorem. Nobody looked at perpetuities in this way yet. Recently D.Buraczewski, J. Collamore, J.Zienkiewicz and myself hae deeloped methods to study tails partial sums i.e. P( R i 1 B i > t) 1 n t α(n,t) or t α(n,t) without using Latała, Nayar, Tkocz, Damek result. I hae a dream to combine both. In particular both approaches work for (X i, B i ) being not necessarily i.i.d just independent proided some uniform behaior is guaranteed.

42 Finite sums of perpetuities There are quite explicit formulae for the constants 0 < c p (X, B) C p (X) < and although they are not close it seems that in some particular cases of perpetuities one can elaborate. The sum R i 1 B i is much easier to study then partial sums n R i 1 B i due to the renewal theorem. Nobody looked at perpetuities in this way yet. Recently D.Buraczewski, J. Collamore, J.Zienkiewicz and myself hae deeloped methods to study tails partial sums i.e. P( R i 1 B i > t) 1 n t α(n,t) or t α(n,t) without using Latała, Nayar, Tkocz, Damek result. I hae a dream to combine both. In particular both approaches work for (X i, B i ) being not necessarily i.i.d just independent proided some uniform behaior is guaranteed.

43 Finite sums of perpetuities There are quite explicit formulae for the constants 0 < c p (X, B) C p (X) < and although they are not close it seems that in some particular cases of perpetuities one can elaborate. The sum R i 1 B i is much easier to study then partial sums n R i 1 B i due to the renewal theorem. Nobody looked at perpetuities in this way yet. Recently D.Buraczewski, J. Collamore, J.Zienkiewicz and myself hae deeloped methods to study tails partial sums i.e. P( R i 1 B i > t) 1 n t α(n,t) or t α(n,t) without using Latała, Nayar, Tkocz, Damek result. I hae a dream to combine both. In particular both approaches work for (X i, B i ) being not necessarily i.i.d just independent proided some uniform behaior is guaranteed.

44 A lemma Lemma Let p > 0. Suppose that P( X = 1) < 1 and E X p = 1. There is δ such that for any ectors in a normed space we hae Then one does induction E Xu + p δ( u p + p ) E i R i p =E 0 + X 1 i X 2... X i p? 0 p +?E i X 2... X i p

45 Another lemma Lemma Let 0 < p 1 and Y, Z be random ectors such that E Z p 1 { Y p 1 8 E Z p } 1 8 E Z p. Then E Y + Z p E Y p E Z p. Proof. We hae for any u, F, u + p u p u p p, therefore E Y + Z p E( Y p + Z p 2 Z p )1 { Y p 1 8 E Z p } + E( Y p + Z p 2 Y p )1 { Y p < 1 8 E Z p } E Y p + E Z p 2E Z p 1 { Y p 1 8 E Z p } 2E Y p 1 { Y p < 1 8 E Z p } E Y p + E Z p E Z p E Z p = E Y p E Z p.

46 Another lemma Lemma Let 0 < p 1 and Y, Z be random ectors such that E Z p 1 { Y p 1 8 E Z p } 1 8 E Z p. Then E Y + Z p E Y p E Z p. Proof. We hae for any u, F, u + p u p u p p, therefore E Y + Z p E( Y p + Z p 2 Z p )1 { Y p 1 8 E Z p } + E( Y p + Z p 2 Y p )1 { Y p < 1 8 E Z p } E Y p + E Z p 2E Z p 1 { Y p 1 8 E Z p } 2E Y p 1 { Y p < 1 8 E Z p } E Y p + E Z p E Z p E Z p = E Y p E Z p.

47 L p bound for finite sums of perpetuities R k := k j=1 X j Theorem (Latała, Nayar, Tkocz, Damek) Suppose that F is a separable Banach space. Let p > 0 and let an i.i.d. sequence (X, B), (X 1, B 1 ),... with alues in [0, ) F be such that X is nondegenerate and E B p, EX p <. Assume additionally that P(X + B = ) < 1 for eery F. Then there are constants 0 < c p (X, B) C p (X) < which depend on p and the distribution of (X, B) such that for eery n, p c p (X, B)E B p ER p i 1 E R i 1 B i C p (X)E B p ER p i 1.

48 Symmetric i.i.d sequences Corollary Let p > 0 and X, X 1, X 2,... be an i.i.d. sequence of symmetric r.. s such that E X p < and P( X = t) < 1 for all t. Then there exist constants 0 < c p,x C p,x < which depend only on p and the distribution of X such that for any ectors 0, 1,..., n in a normed space (F, ), c p,x i p E R i p p E i R i C p,x i p E R i p. Proof. Let (ε i ) be a sequence of independent symmetric ±1 r.. s, independent of (X i ). Then p p i i E i R i = E ε E X i ε k X k k=1 k=1 and it is enough to use preious Theorem for nonnegatie ariables ( X i ).

49 Symmetric i.i.d sequences Corollary Let p > 0 and X, X 1, X 2,... be an i.i.d. sequence of symmetric r.. s such that E X p < and P( X = t) < 1 for all t. Then there exist constants 0 < c p,x C p,x < which depend only on p and the distribution of X such that for any ectors 0, 1,..., n in a normed space (F, ), c p,x i p E R i p p E i R i C p,x i p E R i p. Proof. Let (ε i ) be a sequence of independent symmetric ±1 r.. s, independent of (X i ). Then p p i i E i R i = E ε E X i ε k X k k=1 k=1 and it is enough to use preious Theorem for nonnegatie ariables ( X i ).

50 Ideas of the proof - upper bound We hae n i R i n i R i, so it is enough to consider the case when F = R and k 0. Since it is only a matter of normalization we may also assume that EX p i = 1 for all i. We proceed by an induction on m = p. For m = 1 i.e. p 1 there is nothing to proe, since (x + y) p x p + y p for x, y > 0. For p > 1 we use the inequality (x + y) p x p + 2 p (yx p 1 + y p ) for x, y 0. (9) We hae by (??) p p E i R i E i R i + 2 p p 1 0 E i R i + p 0. Iterating this inequality we get p n 1 E i R i n p ERn p +2 p k ER k i=k+1 i R i p 1 n 1 + p i ER p i.

51 Ideas of the proof - upper bound We hae n i R i n i R i, so it is enough to consider the case when F = R and k 0. Since it is only a matter of normalization we may also assume that EX p i = 1 for all i. We proceed by an induction on m = p. For m = 1 i.e. p 1 there is nothing to proe, since (x + y) p x p + y p for x, y > 0. For p > 1 we use the inequality (x + y) p x p + 2 p (yx p 1 + y p ) for x, y 0. (9) We hae by (??) p p E i R i E i R i + 2 p p 1 0 E i R i + p 0. Iterating this inequality we get p n 1 E i R i n p ERn p +2 p k ER k i=k+1 i R i p 1 n 1 + p i ER p i.

52 Ideas of the proof - upper bound We hae n i R i n i R i, so it is enough to consider the case when F = R and k 0. Since it is only a matter of normalization we may also assume that EX p i = 1 for all i. We proceed by an induction on m = p. For m = 1 i.e. p 1 there is nothing to proe, since (x + y) p x p + y p for x, y > 0. For p > 1 we use the inequality (x + y) p x p + 2 p (yx p 1 + y p ) for x, y 0. (9) We hae by (??) p p E i R i E i R i + 2 p p 1 0 E i R i + p 0. Iterating this inequality we get p n 1 E i R i n p ERn p +2 p k ER k i=k+1 i R i p 1 n 1 + p i ER p i.

53 Ideas of the proof - upper bound We hae n i R i n i R i, so it is enough to consider the case when F = R and k 0. Since it is only a matter of normalization we may also assume that EX p i = 1 for all i. We proceed by an induction on m = p. For m = 1 i.e. p 1 there is nothing to proe, since (x + y) p x p + y p for x, y > 0. For p > 1 we use the inequality (x + y) p x p + 2 p (yx p 1 + y p ) for x, y 0. (9) We hae by (??) p p E i R i E i R i + 2 p p 1 0 E i R i + p 0. Iterating this inequality we get p n 1 E i R i n p ERn p +2 p k ER k i=k+1 i R i p 1 n 1 + p i ER p i.

54 Ideas of the proof - upper bound We hae n i R i n i R i, so it is enough to consider the case when F = R and k 0. Since it is only a matter of normalization we may also assume that EX p i = 1 for all i. We proceed by an induction on m = p. For m = 1 i.e. p 1 there is nothing to proe, since (x + y) p x p + y p for x, y > 0. For p > 1 we use the inequality (x + y) p x p + 2 p (yx p 1 + y p ) for x, y 0. (9) We hae by (??) p p E i R i E i R i + 2 p p 1 0 E i R i + p 0. Iterating this inequality we get p n 1 E i R i n p ERn p +2 p k ER k i=k+1 i R i p 1 n 1 + p i ER p i.

55 Howeer, ER k ( n i=k+1 i R i ) p 1 = ER p k E( n i=k+1 i R k+1,i ) p 1 and ER p k = k j=1 EX p j = 1. Hence p p 1 E i R i 2 p p n 1 i + 2 p k E i R k+1,i. The induction assumption yields p 1 E i R k+1,i C(p 1) i=k+1 = C(p 1) i=k+1 p 1 i i j=k+1 i=k+1 i=k+1 p 1 i ER p 1 k+1,i EX p 1 j C(p 1) To finish the proof we obsere that n 1 k p 1 i λ (p 1)(i k) 1 ( 1 p p i=k+1 0 k<i n p i i=k+1 k + p 1 p λ (p 1)j 1 = λp 1 1 j=1 1 p 1 i λ (p 1)(i k) 1. ) p i λ (p 1)(i k) 1 1 λ p 1 p i.

56 Howeer, ER k ( n i=k+1 i R i ) p 1 = ER p k E( n i=k+1 i R k+1,i ) p 1 and ER p k = k j=1 EX p j = 1. Hence p p 1 E i R i 2 p p n 1 i + 2 p k E i R k+1,i. The induction assumption yields p 1 E i R k+1,i C(p 1) i=k+1 = C(p 1) i=k+1 p 1 i i j=k+1 i=k+1 i=k+1 p 1 i ER p 1 k+1,i EX p 1 j C(p 1) To finish the proof we obsere that n 1 k p 1 i λ (p 1)(i k) 1 ( 1 p p i=k+1 0 k<i n p i i=k+1 k + p 1 p λ (p 1)j 1 = λp 1 1 j=1 1 p 1 i λ (p 1)(i k) 1. ) p i λ (p 1)(i k) 1 1 λ p 1 p i.

57 Howeer, ER k ( n i=k+1 i R i ) p 1 = ER p k E( n i=k+1 i R k+1,i ) p 1 and ER p k = k j=1 EX p j = 1. Hence p p 1 E i R i 2 p p n 1 i + 2 p k E i R k+1,i. The induction assumption yields p 1 E i R k+1,i C(p 1) i=k+1 = C(p 1) i=k+1 p 1 i i j=k+1 i=k+1 i=k+1 p 1 i ER p 1 k+1,i EX p 1 j C(p 1) To finish the proof we obsere that n 1 k p 1 i λ (p 1)(i k) 1 ( 1 p p i=k+1 0 k<i n p i i=k+1 k + p 1 p λ (p 1)j 1 = λp 1 1 j=1 1 p 1 i λ (p 1)(i k) 1. ) p i λ (p 1)(i k) 1 1 λ p 1 p i.

58 Ideas of the proof - lower bound Proofs of lower bounds are much more inoled. They are also based on some induction. For p 1 we hae Proposition Let 0 < p 1 and independent nonnegatie r.. s X 1, X 2,... satisfy EX p i = 1, EX p/2 i λ < 1 and E(X p i 1)1 {1 X p i A} δ. Then for any ectors 0, 1,..., n in a normed space (F, ) and any integer k 1 we hae p ( ) E i R i ε 0 0 p ε1 + k c i i p, where ε 0 = δ/8, ε 1 = δ 3 /8, c i = 0 for 1 i k 1, c i = Φ i j=k λ j for i k and Φ = 28 A 1 λ λk 2.

59 L p -bounds in the non iid case p 1 In the non iid case for p (0, 1] we assume that X 1, X 2,... are independent, nonnegatie r.. s, EX p i <, (10) 0<δ<1,A>1 i E(X p i Theorem λ<1 i EX p/2 i λ(ex p i ) 1/2, (11) EX p i )1 {EX p i X p i AEX p i } δex p i. (12) Let 0 < p 1 and X 1, X 2,... satisfy assumptions (??)-(??). Then for any ectors 0, 1,..., n in a normed space (F, ) we hae c(p, λ, δ, A) i p ER p i p E i R i i p ER p i, where c(p, λ, δ, A) is a constant which depends only on p, λ, δ and A.

60 L p -bounds in the non iid case p 1 In the non iid case for p (0, 1] we assume that X 1, X 2,... are independent, nonnegatie r.. s, EX p i <, (10) 0<δ<1,A>1 i E(X p i Theorem λ<1 i EX p/2 i λ(ex p i ) 1/2, (11) EX p i )1 {EX p i X p i AEX p i } δex p i. (12) Let 0 < p 1 and X 1, X 2,... satisfy assumptions (??)-(??). Then for any ectors 0, 1,..., n in a normed space (F, ) we hae c(p, λ, δ, A) i p ER p i p E i R i i p ER p i, where c(p, λ, δ, A) is a constant which depends only on p, λ, δ and A.

61 L p -bounds in the non iid case p > 1 For p > 1 to get the lower bound we assume µ>0,a< i E X i EX i µ(ex p i ) 1/p E X i EX i 1 {Xi >A(EX p i ) 1/p } 1 4 µ(ex p i ) 1/p, (13) q>max{p 1,1} λ<1 i (EX q i ) 1/q λ(ex p i ) 1/p. (14) To derie the upper L p -bounds we need k=1,2,..., p 1 λk <1 i (EX p k i ) 1/(p k) λ k (EX p k+1 i ) 1/(p k+1). (15)

62 L p -bounds in the non iid case p > 1 For p > 1 to get the lower bound we assume µ>0,a< i E X i EX i µ(ex p i ) 1/p E X i EX i 1 {Xi >A(EX p i ) 1/p } 1 4 µ(ex p i ) 1/p, (13) q>max{p 1,1} λ<1 i (EX q i ) 1/q λ(ex p i ) 1/p. (14) To derie the upper L p -bounds we need k=1,2,..., p 1 λk <1 i (EX p k i ) 1/(p k) λ k (EX p k+1 i ) 1/(p k+1). (15)