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.

### An Introduction to Transformational Geometry by F.M. Eccles

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