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By F.M. Eccles

ISBN-10: 0201018519

ISBN-13: 9780201018516

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Then N (A(n) ) ↓ N ({x0 }), and by monotone convergence, P0 ({x0 }) = limn→∞ P0 (A(n) ) = 0. Equivalently, Pr{N {x0 } > 0} = 1, so that x0 is a fixed atom of the process, contradicting (i). IV. II that we have a Poisson process without fixed atoms. Thus, the following theorem, due to Prekopa (1957a, b), is true. V. s. boundedly finite and without fixed atoms. III. To extend this result to the nonorderly case, consider for fixed real z in 0 ≤ z ≤ 1 the set function Qz (A) ≡ − log E(z N (A) ) ≡ − log Pz (A) defined over the Borel sets A.

Then, for N to be a stationary Poisson process it is necessary and sufficient that for all sets A that can be represented as the union of a finite number of finite intervals, P0 (A) = e−λ (A) . 2) It is as easy to prove a more general result for a Poisson process that is not necessarily stationary. 1)] r Pr{N (Ai ) = ki (i = 1, . . , r)} = [µ(Ai )]ki −µ(Ai ) e ki ! i=1 (r = 1, 2, . ) for some nonatomic measure µ(·) that is bounded on bounded sets. II). 6). 32 2. II. Let µ be a nonatomic measure on Rd , finite on bounded sets, and suppose that the simple point process N is such that for any set A that is a finite union of rectangles, Pr{N (A) = 0} = e−µ(A) .

X(n) ) has the same distribution as the vector whose kth component is Xn Xn−k+1 Xn−1 + ··· + . + n−1 n−k+1 n 24 2. Basic Properties of the Poisson Process (b) Write Y = X1 + · · · + Xn and set Y(k) = (X1 + · · · + Xk )/Y . Then Y(1) , . . d. s uniformly distributed on (0, 1). s have no memory. v. v. v. f. has as its tail R(z) ≡ Pr{XY > z} = Pr{X > Y + z | X > Y }. f. 1) has Pr{N (t − x − ∆, t − ∆] = 0, N (t − ∆, t] = 1, N (t, t + y] = 0 | N (t − ∆, t] > 0} → e−λx e−λy (∆ → 0), showing the stochastic independence of successive intervals between points of the process.

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An Introduction to Transformational Geometry by F.M. Eccles


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