Median Concenraion and Flucuaions for Lévy Processes Chrisian Houdré and Philippe Marchal Augus 23 2006 Absrac We esimae a median of f(x ) where f is a Lipschiz funcion X is a Lévy process and is an arbirary ime. This leads o concenraion inequaliies for f(x ). In urn corresponding flucuaion esimaes are obained under assumpions ypically saisfied if he process has a regular behavior in small ime and a possibly differen regular behavior in large ime. Key words and phrases: Lévy processes median flucuaions concenraion AMS Subjec Classificaion (2000): 60E07 60F10 60G51 60G52. 1 Inroducion In R d le X = (X 0) be a Lévy process wihou Gaussian componen. Is characerisic exponen ψ X is given for all u R d by where E exp(i u X ) = exp(ψ X (u)) ( ) ψ X (u) = i u b + e i uy 1 i u y 1 y 1 ν(dy) (1) R d School of Mahemaics Georgia Insiue of Technology Alana GA 30332 USA houdre@mah.gaech.edu. This research was done in par while I visied L École Normale Supérieure whose hospialiy and suppor is graefully acknowledged. DMA École Normale Supérieure 75005 Paris France Philippe.Marchal@ens.fr 1
b R d and ν 0 (he Lévy measure) is a posiive Borel measure wihou aom a he origin and such ha R d ( y 2 1)ν(dy) < + (hroughou and are respecively he Euclidean inner produc and norm in R d ). While he asympoic behavior of X in small or large ime can be deduced from he asympoic behavior of ψ near he origin or a infiniy i is more difficul o ge precise esimaes for he law of X a some given ime. However when X has finie mean Marcus and Rosiński [MR] (see he nex secion for a precise saemen) provide a fine esimaion of E X involving he funcions V (R) = x 2 ν(dx) and M(R) = x R x >R x ν(dx) R > 0. If one removes he assumpion of finie mean in which case M(R) becomes infinie he naural way o express he order of magniude of X is o consider one of is medians. One may hen wan o esimae his median and o furher know how X is concenraed around i. More generally one may ask he same quesion for f(x ) where f is a Lipschiz funcion wih respec o he Euclidean norm. The aim of his paper is o invesigae hese quesions and relaed ones. In essence he main resul of he presen paper is ha under some raher general hypoheses if f is a Lipschiz funcion wih Lipschiz consan 1 (a 1-Lipschiz funcion) he order of magniude of he median and of he flucuaions of f(x ) is given by funcions of he form h c () = inf { x > 0 : V (x) x 2 = c } (2) where c is some posiive real. More precisely denoe by ν he ail of ν i.e. le ν(r) = ν(dx) hen we have: x >R 2
Theorem 1 Le X be a Lévy process wih characerisic exponen (1). Le f be a 1-Lipschiz funcion le c > 0 le > 0 and le h c be given by (2). Then for every > 0 such ha any median mf(x ) of f(x ) saisfies: ν(h c ()) 1/4 (3) mf(x ) f(0) h c () [1 + 3g c (1/4)] + E c () (4) where g c (x) is he soluion in y of he equaion ( y (y + c) log 1 + y ) = log(x) c and where E c () = d ( 2 e k b e k y ν(dy) + e k y ν(dy)) h c()< y 1 1< y h c() k=1 e 1... e d being he canonical basis of R d. Remark 1 (i) Noe ha if X is symmeric i.e. if X d = X hen E c () = 0. (ii) The proof of he above heorem acually shows ha 3g c (1/4) can be replaced by g c (1/4)+2g c (1/2 ν(h c ())) whenever he condiion ν(h c ()) 1/4 is weakened o ν(h c ()) < 1/2. (iii) Noe also ha he main assumpion of Theorem 1 namely (3) is saisfied as soon as here exiss a consan A > 0 such ha for every R > 0 ν(r) A V (R) R 2. (5) Indeed when (5) holds choosing c = 1/4A ensures ha mf(x ) is of order a mos h c () + E c (). This is in paricular rue if X is a sable vecor in which case A = (2 α)/α will do. In fac in he sable case for any c > 0 h c () = (σ(s d 1 )/(2 α)c) 1/α where σ is he spherical componen of he corresponding sable Lévy measure. In he nex secion a furher naural class of examples saisfying (3) is presened. Our nex sep is o sudy he deviaions from he median. 3
Theorem 2 Under he assumpions of Theorem 1 for all c > 0 such ha R = h c () saisfies (5) here exiss m(c ) R such ha for all reals x > x > 0 he quaniies P(f(X ) m(c ) x) and P(f(X ) m(c ) x) are upper bounded by Ac + exp ( x x h c () ( ) x x h c () + c log (1 + x )) x. (6) ch c () In paricular if q > 0 hen for every > 0 such ha R = h q/2a () saisfies (5) and every x > 0 such ha here exiss a real m() such ha x [ 1 + g q/2a (q/2) ] h q/2a () (7) P(f(X ) m() x) q (8) and P(f(X ) m() x) q. (9) Remark 2 (i) From he proof of he above heorem i can be seen ha m(c ) = Ef(Y (hc()) ) where he Lévy process Y is obained by runcaing he Lévy measure of he process X a R = h c () will do. Then aking c = q/2a in m(c ) gives m() in (8) and (9). Remark also ha since f(x ) is concenraed around some value his value is necessarily close o he median and so mf(x ) is necessarily close o m(c ). (ii) In view of (4) when X is symmeric as well as (8) and (9) he median and he flucuaions of f(x ) are of order h 1/4A (). (iii) I is easily seen ha when q 0 g q/2a (q/2) 1. So for q small enough P(f(X ) m() x) q and P(f(X ) m() x) q as soon as x (2 + ε)h q/2a () ε > 0. Le us now reurn o he mean and le us precisely recall he resul of Marcus and Rosiński. Le X have finie expecaion and be cenered i.e. such ha E(X ) = 0 (10) 4
le > 0 and le x 0 () be he soluion in x of he equaion: V (x) x 2 + M(x) x = 1. (11) Then 1 4 x 0() E( X ) 17 8 x 0() (12) and he facor 17/8 can be replaced by 5/4 when X is symmeric. The inequaliy (12) suggess ha one should have flucuaions of order x 0 () a ime. We shall acually prove his under he following addiional assumpion: There exiss a consan K such ha for every R > 0 Under his las hypohesis (11) enails M(R) K V (R) R. (13) h 1/(1+K) () x 0 () h 1 () and so E X h c () where means ha he raio of he wo quaniies is bounded above and below by wo posiive consans. We can now sae: Theorem 3 Using he noaion of Theorem 1 assume also ha (10) and (13) hold. Then for all b > 0 all c > 0 such ha R = h c () saisfies (5) and for every 1-Lipschiz funcion f [ ( P(f(X ) Ef(X ) (b + ck)h c ()) Ac + exp b (b + c) log 1 + b )]. c In paricular if q > 0 hen for every > 0 such ha R = h q/2a () saisfies (5) and for every x such ha [ ] qk x 2A + g q/2a(q/2) h q/2a () (14) we have P(f(X ) Ef(X ) x) q. 5
Remark 3 (i) Of course if X has finie mean bu is no cenered one obains a similar resul by considering he Lévy process X E(X ). (ii) Here again for q small enough one has P(f(X ) Ef(X ) x) q as soon as x (1 + ε)h q/2a () ε > 0. (iii) Above i is clear ha lef ails inequaliies also hold rue. For example for all x saisfying (14) we have: P(f(X ) m() x) q. (iv) The resuls on norm esimaes of infiniely divisible vecors derived in [MR] were used o obain relaed esimaes for sochasic inegrals of deerminisic and possibly random predicable inegrands wih respec o infiniely divisible random measures. Similar applicaions and exensions will also carry over o our seings. 2 Examples: symmeric runcaed sable processes In many imporan siuaions ha have been considered in he lieraure he assumpion (13) is saisfied. This is he case in paricular of Lévy processes for which ν(dx) = g(x/ x )ρ( x )dx where ρ is a funcion such ha say ρ(r) cr α 1 for r small enough while ρ(r) cr β 1 for r large enough 0 < α β < 2 or such ha r 2 ρ(dr) <. Processes of his ype 1 have been inroduced in physics and are also of use in mahemaical finance where hey provide models of asse prices differen from he usual modeling via diffusions. Le us examine more precisely he runcaed sable case. Le X be he real symmeric Lévy process wihou Gaussian componen and Lévy measure ν(dx) dx = K x 1+α 1 { x M} wih K M > 0 and 0 < α < 2. Then for every R > 0 V (R) = 2K inf(r M)2 α 2 α and for any c > 0 we have for 0 (2 α)cm α /2K h c () = ( ) 1/α 2K (2 α)c 6
while for (2 α)cm α /2K h c () = ( ) 2KM 2 α 1/2. (2 α)c Taking say c = α/4(2 α) se for 0 αm α /(8K) while for αm α /(8K) se ( ) 1/α 8K H α () = α ( ) 8KM 2 α 1/2 H α () =. α Moreover since ν(r) = 2K α (5) holds wih Thus furher seing ( 1 R 1 ) 1 α M α {R M} A = 2 α α. K(α) = 1 + 3g α/4(2 α) (1/4) i follows from our firs heorem ha for every 1-Lipschiz funcion f mf(x ) f(0) K(α)H α (). So we recovered he fac ha in small ime X behaves like a sable process of index α while in large ime X behaves like a Brownian moion. Bu furhermore we see ha he ransiion occurs around a ime of order αm α /K and we have precise bounds esimaing how his ransiion happens. Our second and hird heorems also apply and give upper bounds for he flucuaions around he median and around he mean. For insance choose q > 0. I is hen easily seen ha 1 + g qα/2(2 α) (q/2) c α 7
wih ( c α = 1 + max 1 ) (1 + 2e)α. 2(2 α) Therefore Theorem 2 says ha if qαm α /2K hen here exiss some m() R such ha P(f(X ) m() x) q as soon as ( ) 1/α 2K x c α. qα On he oher hand if qαm α /2K hen here exiss some m () R such ha P(f(X ) m () x) q as soon as ( ) 2KM 2 α 1/2 x c α. qα Moreover if ones akes R = M hen (5) is auomaically saisfied. This amouns o aking A = (2 α)qm α /2K and so we also have as soon as P(f(X ) m() x) q x [1 + g K/(2 α)m α(q/2)]m. To sum up here exiss some real m() such ha if one of hese wo condiions holds: qαm α /2K and { x min qαm α /2K and { x min c α ( 2K qα c α ( 2KM 2 α qα ) 1/α [1 + g K/(2 α)m α(q/2)]m} ) 1/2 [1 + g K/(2 α)m α(q/2)]m} 8
hen P(f(X ) m() x) q. Suppose for insance ha qαm α /2K. For q no oo small he minimum is aained for he firs erm and so he condiion is x c(/q) 1/α. On he oher hand for very small q he condiion is x G (q) where G can be expressed in erms of he funcion g. Alernaively one can wrie for x Mc α { } C P(f(X ) m() x) min x G (x) 2 and for x Mc α wih P(f(X ) m() x) min C = 2Kcα α α { } C x G (x) α C = 2KM 2 α c 2 α α and [ ( x ) ( ) ( x G (x) = exp M 1 M 1+ K log 1+ (2 α)m )] α 1 (x M). (2 α)m α K 3 Proofs 3.1 Proof of Theorem 1 Fix > 0 and as in [HM] decompose X by runcaing he Lévy measure ν a R (o be chosen laer). Wrie X = Y (R) + Z (R) where Y (R) = (Y (R) 0) and Z (R) = (Z (R) 0) are wo independen Lévy processes. Their respecive characerisic exponen ψ (R) Y and ψ (R) Z are given for u R d by: ψ (R) Z (u) = ( e i uy 1 ) ν(dy) y >R ψ (R) Y (u) = i u b ( ) + e i uy 1 i u y 1 y 1 ν(dy) y R 9
wih b = b y >R y1 y 1 ν(dy) where he las inegral is undersood coordinae-wise (and so is he above difference). Nex our global sraegy is o bound mf(x ) f(0) using: mf(x ) f(0) mf(x ) mf(y (R) ) + mf(y (R) ) Ef(Y (R) ) + Ef(Y (R) ) f(0). Le us sar by bounding mf(x ) mf(y (R) ). To do so i is easy o check (see for insance [HM] p.1498) ha On he oher hand [H] ells us ha P(f(Y (R) P(Z (R) 0) ν(r). (15) ) mf(y (R) ) x) H (R) (x) where ( ( ) ( x x H (R) (x) = exp 2R V (R) + log 1 + Rx )). 2R R 2 2V (R) Nex le and le Then we have (see [HM] p. 1500) I (R) (y) = sup{x 0 : H (R) (x) y} P m = P(f(X ) mf(x )) 1/2. mf(y (R) ) mf(x ) I (R) (P m P(Z (R) 0)) I (R) (1/2 ν(r)) (16) provided ha ν(r) < 1/2. To bound Ef(Y (R) ) mf(y (R) ) we use he concenraion inequaliy [H] ( ( ) )) P( f(y (R) ) Ef(Y (R) x ) x x ) 2 exp R V (R) + log (1 + Rx. R R 2 V (R) (17) 10
By he very definiion of a median and aking x = Ef(Y (R) ) mf(y (R) ) we see ha (17) lead o our second esimae: Finally we bound Ef(Y (R) ) f(0). Ef(Y (R) ) f(0) E Y (R) ( = d 2 Ef(Y (R) ) mf(y (R) ) I (R) (1/4). (18) = E Y (R) 2 k=1 y R ( yk 2ν(dy) + 2 bk + y R ) ) 2 y k 1 y >1 ν(dy) V (R) + EY (R) 2. (19) Combining (16) (18) and (19) gives for any > 0 and R > 0 such ha ν(r) < 1/2 mf(x ) f(0) I (R) (1/2 ν(h c ()))+2 1 I (R) (1/4)+ V (R) + E(Y (R) ) 2. Now choosing R = h c () gives mf(x ) f(0) I (hc()) (1/2 ν(h c ()))+2 1 I (hc()) (1/4)+h c ()+ E(Y (hc()) ) where E(Y (hc()) ) is equal o E() given in he saemen. Finally noe ha I (R) (x) 2R ( I (R) (x) 2R ) ( ) V (R) + log 1 + RI(R) (x) = log(x) R 2 2V (R) and so I (hc()) (x) = 2h c ()g c (x) wih he definiion of g c given in he saemen of Theorem 1. This concludes he proof. 3.2 Proof of Theorem 2 Recall he assumpions and noaion as he previous subsecion: > 0 is fixed and c > 0 is such ha R = h c () saisfies (5). Pu m(c ) = Ef(Y (R) ) 11
Since f is 1-Lipschiz we have f(x ) f(y (R) ) Z (R). Therefore for every x < x P(f(X ) m(c ) x) P(f(Y (R) ) m(c ) x x ) + P( Z (R) x ). (20) The firs erm of he above righ-hand side is bounded as in (17). On he oher hand recall ha Z (R) can be seen as he value a ime of a compound Poisson process (Z s (R) s 0). Therefore if Z (R) x he process s 0) has a leas a jump before ime. This implies ha (Z (R) s Using (5) gives P( Z (R) x ) 1 e ν(r) ν(r). P( Z (R) x ) Ac (21) which when combined wih (20) lead o he inequaliy (6) giving he firs par of he heorem. The second par of he heorem follows by aking x = h c () and c = q/2a. This choice provides firs he upper bound q/2 on (21) and moreover enails ha he condiion (7) becomes (x x )/h c () g c (q/2) leading o anoher upper bound q/2 on he righmos erm in (6). 3.3 Proof of Theorem 3 Le c > 0. Decompose X by runcaing he measure ν a R = h c (). As above wrie X = Y (R) + Z (R) where Y (R) Z (R) are wo independen infiniely divisible random vecors whose Lévy measures are respecively ν(dx)1 x R and ν(dx)1 x >R. Then for every a > ck P(f(X ) Ef(X ) ah c ()) P(f(Y (R) ) Ef(X ) ah c ()) + P(Z (R) 0). (22) Since Z (R) is a compound Poisson process we have as seen in he proof of Theorem 2 On he oher hand P(f(Y (R) P(Z (R) 0) ν(r). (23) ) Ef(X ) ah c ()) P(f(Y (R) ) Ef(Y (R) ) x ) wih x = ah c () Ef(X ) Ef(Y (R) ). 12
To bound x from below remark ha Ef(X ) Ef(Y (R) = E ) ( f(y (R) E Z (R) x >R = M(R) ckr ) + Z (R) ) f(y (R) ) 1 (R) {Z x ν(dx) using boh (13) and (2) for he las inequaliy. Hence x (a ck)r. 0} Moreover [H] ells us ha P(f(Y (R) ) Ef(Y (R) ) x ) exp ( ( x x R R ) )) V (R) + log (1 + Rx. R 2 V (R) Using he fac ha R = h c () and x (a ck)r we ge [ P(f(Y (R) ) Ef(Y (R) ) x ) exp b (b + c) log ( 1 + b c )] wih b = a ck. Togeher wih (23) his yields he firs par of Theorem 3. The second par follows by aking c = q/2a and b = g q/2a (q/2). References [H] C. Houdré Remarks on deviaion inequaliies for funcions of infiniely divisible random vecors. Ann. Probab. 30 (2002) 1223 1237. [HM] C. Houdré P. Marchal On he Concenraion of Measure Phenomenon for Sable and Relaed Random Vecors. Ann. Probab. 32 (2004) 1496 1508. [MR] M.B. Marcus J. Rosiński L 1 -norms of infiniely divisible random vecors and cerain sochasic inegrals. Elecron. Comm. Probab. 6 (2001) 15 29. 13