ON THE COMPOUND POISSON DISTRIBUTION

Transcription

1 Acta Sci. Math. Szeged) ), pp ON THE COMPOUND POISSON DISTRIBUTION András Préopa Budapest) Received: March 1, 1957 A probability distribution is called a compound Poisson distribution if its characteristic function can be represented in the form { } 1) ϕu) =exp iγu + e iux 1) dmx)+ e iux 1) dnx), where γ is a constant, Mx) andnx) are defined on the intervals, ) and, ), respectively, both are monotone non-decreasing, M) = N ) =, further the integrals x dmx), x dnx) 1 exist. We shall prove that under certain conditions we obtain 1) as a limit distribution of double sequences of independent and infinitesimal random variables and apply this theorem to stochastic processes with independent increments. Theorem 1 Let ξ n1,ξ n2,...,ξ nn n =1, 2,...) be a double sequence of random variables. Suppose that the random variables in each row are independent, they are infinitesimal, i.e. for every ε> lim max P ξ n >ε)=, n 1 n finally, there exists a finite-valued, non-negative random variable η such that n ξ n η n =1, 2,...) with probability 1. This last condition means that the sums of the absolute values of the sample summands are uniformly bounded.) Suppose, moreover, that the sequence of probability distributions of the variables ζ n = ξ n1 + ξ n2 + + ξ nn converges to a limiting distribution. Then this is a compound Poisson distribution. 1

2 Proof. Let us define the functions f + x), f x) as follows: f + x) = { x if x, if x<, f x) = { if x, x if x<. Clearly, f + x)f x) and ζ n + = n f + ξ n ) η n =1, 2,...), n ζn = f ξ n ) η n =1, 2,...), with probability 1. Hence it follows that for every K>therelations Pζ n + K) Pη K) n =1, 2,...), Pζn K) Pη K) n =1, 2,...) hold. This imply that the distributions of the sequences ζ n + and ζn are compact sets. Let F n + x) andfn x) denote the distribution functions of the variables ζ n + and ζn, respectively. Let us choose a sequence of integers n 1,n 2,... for which 2) lim F n + i x) =F + x), lim F n i x) =F x), where F + x) andf x) are distribution functions) at every point of continuity of the latters. Let τ be a positive number such that the functions F + x) andf x) are continuous at τ and τ, respectively. Since the random variables in the double sequences f + ξ n1 ),f + ξ n2 ),...,f + ξ nn ), f ξ n1 ),f ξ n2 ),...,f ξ nn ) are infinitesimal and independent in each row, moreover the relations 2) hold, we conclude that if F + n x) =Pf + ξ n ) <x), F n x) =Pf ξ n ) <x), then the sequences ni <x<τ x df + n i x), ni τ<x< x df n i x) are convergent see [1] 25, Theorem 4, Remar). This implies that ni lim <x<τ ) 2 x df + n i x) ni = lim ) 2 x df n i x) =. τ<x< 2

3 Thus if ϕ + n t) = e itx df + n x), then from the inequality e itx 1) dgx) t x <τ ϕ n t) = e itx df n x), x dgx)+2 x τ dgx), valid for every distribution function Gx) and every τ>, it follows using Theorem 4 of [1] 25) that ni lim ϕ + n i t) ni 1 2 = lim ϕ n i t) 1 2 =. Hence the conditions of Theorem 2 of [2] are fulfilled and thus the variables ζ n + i are asymptotically independent, i.e. and ζ n i 3) lim Pζ n + i <x, ζn i <y)=f + x)f y). Let F x) denote the limiting distribution of the random variables ζ n.sinceζ n = ζ n + + ζ n, we get from 3) 4) F x) =F + x) F x). The laws F x), F + x), F x) are infinitely divisible. In Lévy s formula iγu σ2 u 2 + e iux 1 iux ) 2 1+x 2 dmx)+ e iux 1 iux ) 1+x 2 dnx) there correspond to F x), F + x) andf x) constants and functions, which we denote by γ, γ 1, γ 2 ; σ 2, σ 2 1, σ2 2 ; Mx), M 1x), M 2 x); Nx), N 1 x), N 2 x), respectively. According to 4) γ = γ 1 + γ 2, σ 2 = σ1 2 + σ2, 2 Mx) =M 1 x)+m 2 x), Nx) =N 1 x)+n 2 x). If σ 2 >, then at least one of σ1 2 and σ2 2 is positive too. This is, however, impossible, since F + x) =ifx andf x) =1ifx>. We have therefore only to prove that the integrals 1 x dmx), x dnx) exist. We prove the existence of the second integral, the existence of the first one can be proved similarly. We now that if τ 1 is a point of continuity of Nx), then 5) τ1 ni x d F ni x) 3

4 converges [1], 25, Theorem 4) hence it is bounded. If x dnx) = then we can choose such a number τ <τ<τ 1 )that 6) τ1 τ x dnx) >L, where L is the upper bound of the terms in the sequence 5) and Nx) is continuous at the point τ. But we now from the limiting distribution theorems cf. [1] 25, Theorem 4) that ni lim F ni x) 1) = Nx) x>) at every point of continuity of Nx) whence τ1 7) lim τ ni x d F ni x) = Obviously 6) and 7) contain a contradiction. Let us separate in Lévy s formula the terms iu τ1 x dmx), iu 1+x2 τ x dnx). x 1+x 2 dnx) and unite them with iγ u, then we obtain the required form of the limiting distribution. Thus our theorem is completely proved. In the sequel we apply our result to the theory of stochastic processes with independent increments. We say that a stochastic process with independent increments ξ t is wealy continuous if for every ε> P ξ t+h ξ t >ε) when h, uniformly in t. We suppose that Pξ =)=1. Theorem 2 Let us suppose that the stochastic process with independent increments ξ t is wealy continuous and its sample functions are of bounded variation with probability 1 in every finite time interval. If ϕu, t) is the characteristic function of the random variable ξ t then it has the form { 8) ϕu, t) =exp iγt)u + e iux 1) dmx, t)+ } e iux 1) dnx, t), 4

5 where γt) is a continuous function of bounded variation in every finite time interval, Mx, t) and Nx, t) are continuous functions of the variable t and the integrals 1 x dmx, t), x dnx, t) exist for every t. Proof. According to our suppositions the double sequence of independent random variables ξ t,ξ2t ξ t,...,ξ t ξ n 1 n n n n t satisfies all the conditions of Theorem 1. Moreover, for every n n ) ξ t = ξ n t ξ 1 n t, hence we have only to prove the assertion regarding the functions γt), Mx, t), Nx, t). The continuity in t of these functions follows at once from the wea continuity of the process ξ t and the convergence theorems of infinitely divisible distributions see e.g. [1] Chapter 3). Now we show that for every T>γt) is of bounded variation in the interval t T. Let us consider the sequence of subdivisions I n) = [ 1 2 n T, 2 n T ] =1, 2,...,2 n ; n =1, 2,...) of the interval [,T] and let us denote the distribution function of the random variable ξ 2 n T ξ 1 2 n T by F x, In) ). We now from the limiting distribution theorems that ) ) 1 γ 2 n T γ 2 n T + x d M x, 2 ) τ n T M x, 1 )) 2 n T τ + x d N x, 2 ) n T N x, 1 )) 2 n T = lim x df x, I N) N j ) cf. [1] 25, Theorem 4), hence 9) 2 n ) γ 2 n T γ I N) j I n) x <τ ) 1 τ 2 n T x dnx, T ) x dmx, T ) τ + lim 2 N N x <τ x df x, I N) ). 5

6 The boundedness of the sequence on the right-hand side of 9) is a consequence of the fact that the non-decreasing sequence 2 N ξ 2 N T ξ 1 2 N T converges with probability 1, and of Theorem 4 of [1] 25. Since γt ) is continuous, this implies that it is of bounded variation. Thus Theorem 2 is proved. References [1] Gnedeno, B. V. and A. N. Kolmogorov 1949). Predel nüe raszpredelenija dlja szumm nezaviszimüh szlucsajnüh velicsin, Moszva Leningrad. [2] Préopa, A. and A. Rényi 1956). On the independence in the limit of sums depending on the same sequence of independent random variables, Acta Math. Acad. Sci. Hung., 7,