Question

Let N be a point process on IRk. For any Borel set A c RK, let E[N(A)] = A(A). Assume that A(A) < co if A is a bounded set. Define a rectangle to be a set of the form (x = (x1, ..,XK) ERK: -co < a; < x; S bj,i = 1,.., k} Let me assume two properties. The first property is that for any collection of disjoint rectangles /1, ...,Jn, we have P{N(1) = 0, .., N(n) = 0} = e-(4(1)+.+A(In)) The second property is that any rectangle can be split into two disjoint rectangles with the same expected number of points. Please note that N() has a Poisson distribution. The other property that we need to show is that under the above assumptions, N(1), ..., N(n) are independent when /1, ..../n are disjoint rectangles. Please prove this.

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Answer 1

Under the assumptions of a Poisson point process with properties of disjoint rectangles having joint probabilities expressed as a product of exponential terms, and the ability to split rectangles while maintaining the same expected points, it follows that the number of points in disjoint rectangles,
`\(N(1), \ldots, N(n)\)`, is independent. This independence arises from the product of independent exponential terms in the joint probability expression, demonstrating the desired result.

Step-by-step explanation:

The properties you've outlined are reminiscent of the Poisson point process and its properties. Let's prove the independence property for disjoint rectangles. We'll use the given assumptions and properties.

Let's consider the notation:

- N(A) is the number of points in a Borel set A under the point process N.

- P(A) is the probability of event A.

The first property states that the joint probability of no points in disjoint rectangles is the product of the probabilities of no points in each rectangle. Mathematically, it is given by:

`\[ P(N(1) = 0, \ldots, N(n) = 0) = e^(-(A(1) + \ldots + A(n))) \]`

The second property states that any rectangle can be split into two disjoint rectangles with the same expected number of points.

Now, let's prove the independence property for disjoint rectangles:

1. Start with the definition of joint probability:

`\[ P(N(1) = n_1, \ldots, N(n) = n_n) = (e^(-(A(1) + \ldots + A(n))) \cdot (A(1) + \ldots + A(n))^(n_1 + \ldots + n_n))/((n_1! \cdot \ldots \cdot n_n!)) \]`

This is the joint probability of having n₁ points in rectangle 1, n₂ points in rectangle 2, and so on.

2. Use the second property to split each rectangle:

According to the second property, each rectangle i can be split into two disjoint rectangles with the same expected number of points:

`\[ P(N(i) = n_i) = P(N(i,1) = n_i,1 \cap N(i,2) = n_i,2) \]`

where
`\( n_i = n_i,1 + n_i,2 \)`.

3. Apply the first property to each split rectangle:

`\[ P(N(i,1) = n_i,1, N(i,2) = n_i,2) = e^(-(A(i,1) + A(i,2))) \]`

Therefore,

`\[ P(N(i) = n_i) = e^(-(A(i,1) + A(i,2))) \]`

4. Independence of disjoint rectangles:

Now, consider the joint probability of
`\( N(1), \ldots, N(n) \)`:

`\[ P(N(1) = n_1, \ldots, N(n) = n_n) = \prod_(i=1)^(n) e^(-(A(i,1) + A(i,2))) \]`

Since each term is a product of independent terms, it implies that
`\( N(1), \ldots, N(n) \)` are independent for disjoint rectangles.

Therefore, under the given assumptions and properties,
`\( N(1), \ldots, N(n) \)` are independent when
`\( 1, \ldots, n \)` are disjoint rectangles.