Heavy Tailed Analysis Eurandom Summer 2005
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HEAVY TAILED ANALYSIS EURANDOM SUMMER 2005 SIDNEY RESNICK School of Operations Research and Industrial Engineering Cornell University Ithaca, NY 14853 USA [email protected] http://www.orie.cornell.edu/»sid & Eurandom http://www.eurandom.tue.nl/people/EURANDOM_chair/eurandom_chair.htm 1 80 SIDNEY RESNICK 11. The Laplace functional. The Laplace functional is a transform technique which is useful for manipulating distri- butions of random measures and point processes. When applied to Poisson processes and empirical measures, algebraic manipulations become familiar to ones used with either char- acteristic functions or Laplace transforms applied to sums of iid random variables. Continue to assume the state space of the random measures or point processes is the nice space E. For a non-negative, bounded measureable function f : E 7! R+ and for ¹ 2 M+(E), we continue to use the notation Z ¹(f) = f(x)d¹(x): x2E P For a point measure m = i ²xi , this is Z X m(f) = f(x)m(dx) = f(xi): x2E i If we think of f ranging over all the non-negative bounded functions, m(f) yields all the information contained in m; certainly we learn about the value of m on each set A 2 E since we can always set f = 1A. So integrals of measures with respect to arbitrary test functions contain as much information as evaluating the measures on arbitrary sets. De¯nition 3 (Laplace functional). Suppose B+ are the non-negative, bounded, measurable functions from E 7! R+ and let » : (; A; P) 7! (M+(E); M+(E)) be a random measure. The Laplace functional of » is the non-negative function on B+ given by Z Z ¡1 ª»(f) = E expf¡»(f)g = expf¡N(!; f)dP(!) = expf¡m(f)P ± N (dm): M+(E) Note if P is a probability measure on Mp(E), its Laplace functional is Z expf¡m(f)P (dm); f 2 B+: Mp(E) Proposition 10. The Laplace functional of » uniquely determines the distribution of », ¡1 namely P ± » on M+(E)). Proof. The topology (and hence the Borel σ-algebra on M+(E) is generated by basic open sets of the form f¹ 2 M+(E): ¹(fi) 2 Ii; i = 1; : : : ; kg; + where fi 2 CK (E) and Ii are bounded open intervals. (Review page 33. ) This class of sets is closed under ¯nite intersection and hence is a ¦-system generating the σ-algebra. So from (Resnick, 1998, page 38) is su±ces to show that P ± »¡1 is uniquely determined on this class by the Laplace functional. HEAVY TAIL ANALYSIS 81 Pk For f = i=1 ¸ifi, with ¸i > 0, we have Pk ¡»(f) ¡ ¸i»(fi) û(f) = Ee = Ee i=1 : So û determines the joint distribution of the random vector (»(f1);:::;»(fk)) and hence ¡1 P ± » f¹ : ¹(fi) 2 Ii; i = 1; : : : ; kg = P[(»(f1);:::;»(fk)) 2 ¢] is determined, as required. ¤ 11.1. The Laplace functional of the Poisson process. Recall De¯nition 2 of the Poisson process and the two parts 1 and 2 of the de¯nition. The Poisson process can be identi¯ed by the characteristic form of its Laplace functional as discussed next. Theorem 8 (Laplace functional of PRM). The distribution of PRM(¹) is uniquely deter- mined by 1 and 2 in De¯nition 2. Furthermore, the point process N is PRM(¹) i® its Laplace functional is of the form Z ¡f(x) (11.1) ªN (f) = expf¡ (1 ¡ e )¹(dx)g; f 2 B+: E So PRM(¹) can be identi¯ed by the characteristic form of its Laplace functional. Proof. We ¯rst show (1) and (2) imply (11.1). If f = ¸1A where ¸ > 0 then because N(f) = ¸N(A) and N(A) is Poisson with parameter ¹(A) we get ³ ´ ¡¸N(A) ¡¸ ªN (f) =E e = expf(e ¡ 1)¹(A)g Z = expf¡ (1 ¡ e¡f(x))¹(dx)g E which is the correct form given in (11.1). Next suppose f has a somewhat more complex form Xk f = ¸i1Ai i=1 where ¸i ¸ 0;Ai 2 E; 1 · i · k and A1;:::;Ak are disjoint. Then à ( )! Xk ªN (f) =E exp ¡ ¸iN(Ai) i=1 Yk ³ ´ = E expf¡¸iN(Ai)g from independence i=1 Yk ½ Z ¾ ¡¸ 1 (x) = exp ¡ (1 ¡ e i Ai )¹(dx) from the previous step i=1 E 82 SIDNEY RESNICK ( ) Z Xk ¡¸ 1 (x) = exp (1 ¡ e i Ai )¹(dx) E ½Z i=1 ¾ P ¡ k ¸ 1 (x) = exp (1 ¡ e i=1 i Ai )¹(dx) ½ZE ¾ = exp (1 ¡ e¡f(x))¹(dx) E which again veri¯es (11.1). Now the last step is to take general f 2 B+ and verify (11.1) for such f. We may approximate f from below by simple fn of the form just considered. We may take for instance n Xn2 i ¡ 1 fn(x) = 1[ i¡1 ; i )(f(x)) + n1[n;1)(f(x)) 2n 2n 2n i=1 so that 0 · fn(x) " f(x): ¡f By monotone convergence N(fn) " N(f) and since e · 1 we get by dominated convergence that ªN (f) = lim ªN (fn): n!1 We have from the previous step that ½ Z ¾ ¡fn(x) ªN (fn) = exp ¡ (1 ¡ e )¹(dx) : E Since 1 ¡ e¡fn " 1 ¡ e¡f we conclude by monotone convergence that Z Z (1 ¡ e¡fn(x)¹(dx) " (1 ¡ e¡f(x)¹(dx) E E and thus we conclude (11.1) holds for any f 2 B+. Since the distribution of N is uniquely determined by ªN we have shown that 1 and 2 in De¯nition 2 determine the distribution of N. Conversely if the Laplace functional of N is given by (11.1) then N(A) must be Poisson distributed with parameter ¹(A) for any A 2 E which is readily checked by substituting f = ¸1A in (11.1) to get a Laplace transform of a Poisson distribution. Furthermore if Pk A1;:::;Ak are disjoint sets in E and f = i=1 ¸i1Ai then substituting in (11.1) gives ½ Z ¾ P P ¡ k ¸ N(A ) ¡ k ¸ 1 Ee i=1 i i = exp ¡ (1 ¡ e i=1 i Ai )d¹ ( E ) Z Xk ¡¸ 1 = exp ¡ (1 ¡ e i Ai )d¹ E i=1 HEAVY TAIL ANALYSIS 83 Yk ¡¸i = expf¡(1 ¡ e )¹(Ai)g i=1 Yk = Ee¡¸iN(Ai) i=1 and so the joint Laplace transform of (N(Ai); 1 · i · k) factors into a product of Laplace transforms and this shows independence. ¤ 11.1.1. Application: A general construction of the Poisson process. When E = [0; 1), we know the renewal theory construction yields a Poisson process with Lebesgue measure as the mean measure. Here is a general scheme for constructing a Poisson process with mean measure ¹. Start by supposing that ¹(E) < 1. De¯ne the probability measure F F (dx) = ¹(dx)=¹(E) on E. Let fXn; n ¸ 1g be iid random elements of E with common distribution F and let º be independent of fXng with a Poisson distribution with parameter ¹(E). De¯ne (P º ² ; if º ¸ 1 N = i=1 Xi 0; if º = 0: We claim N is PRM(¹). Verify this by computing the Laplace functional of N: Pº ¡ f(Xi) ªN (f) =Ee i=1 1 X P ¡ j f(X ) = Ee i=1 i P [º = j] j=0 1 X ¡ ¢j = Ee¡f(X1) P [º = j] j=0 © ª = exp ¹(E)(Ee¡f(X1) ¡ 1) ½ µ Z ¶¾ ¹(dx) = exp ¡¹(E) 1 ¡ e¡f(x) ¹(E) ½ Z E ¾ ¡ ¢ = exp ¡ 1 ¡ e¡f(x) ¹(dx) : E The Laplace functional has the correct form and hence N is indeed PRM(¹). Note that what the construction does is to toss points at random into E according to distribution ¹(dx)=¹(E); the number of points tossed is Poisson with parameter ¹(E). When the condition ¹(E) < 1 fails, we proceed as follows to make a minor modi¯cation in the foregoing construction: Decompose E into disjoint bits E1; E2;::: so that E = [iEi and ¹(Ei) < 1 for each i. Let ¹i(dx) = ¹(dx)1Ei (x) and let Ni be PRM(¹i) on E (do the 84 SIDNEY RESNICK constructionP just outlined) and arrange things so the collection fNig is independent. De¯ne N := i Ni and N is PRM(¹) since Y ªN (f) = ªNi (f) i Y ½ Z ¾ ¡ ¡f(x)¢ = exp ¡ 1 ¡ e ¹i(dx) Ei i ( ) X Z ¡ ¡f(x)¢ = exp ¡ 1 ¡ e ¹i(dx) E ( i ) Z X ¡ ¡f(x)¢ = exp ¡ 1 ¡ e ¹i(dx) E ½ Z i ¾ ¡ ¢ = exp ¡ 1 ¡ e¡f(x) ¹(dx) E P since i ¹i = ¹: This completes the construction. 11.2. Weak convergence of point processes and random measures. We now discuss weak convergence in M+(E) and specialize to give criterion for weak convergence to a Poisson process. 11.2.1. Basic criterion via Laplace functionals. Since M+(E) can be metrized as a complete separable metric space, the theory of weak convergence applies. A useful criterion for our purposes is in terms of Laplace functionals. Theorem 9 (Convergence criterion). Let f»n; n ¸ 0g be random elements of M+(E). Then »n ) »0 in Mp(E); i® ¡»n(f) ¡»0(f) + (11.2) ûn (f) = Ee ! Ee = û0 (f); 8f 2 CK (E): + So weak convergence is characterized by convergence of Laplace functionals on CK (E). Proof. Here is the proof of the easy half of the equivalence. Suppose »n ) »0 in M+(E). The map M+(E) 7! [0; 1) de¯ned by ¹ 7! ¹(f) is continuous, so the continuous mapping theorem gives »n(f) ) »0(f) in R.