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For a Poisson process having rate $\lambda$. Given the number of events by time $T$ the set of event times are iid Uniform $(0,T)$ random variables. Suppose that each event are independently counted with probability $\lambda(t)/\lambda$. The times of these counted event have a common distribution, $F(s)$, where

$F(s) = P\{time \leq s | counted \} =\frac{\int P(time \leq s, counted,time=x)dx}{P(counted)} =\frac{\int P(time \leq s, counted|time=x)P(time=x)dx }{P(counted)} = \frac{\int_0^T P\{time \leq s, counted |time=x \}dx/T }{P\{counted\} } = \frac{\int_0^s \lambda(x) dx}{\int_0^T \lambda(x)dx }$.

How does 1/T get cancelled out in the last equality?

  • The integrand $P\{time \leq s, counted |time=x \} = \lambda(t)/\lambda$ ?
  • $P\{counted\} = \int_0^T \lambda(x)/ \lambda dx$ ?
J.doe
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  • In your first edit, you somehow lost the division by T in the numerator. Just revert back to your original question.(it should even earn you a new badge doing so ;-)) – Ute Jul 17 '23 at 09:03
  • You are computing the conditional event $$P(time \leq s, time = x|counted)$$ what does that mean? – Sextus Empiricus Jul 18 '23 at 08:15
  • @SextusEmpiricus, OP is trying to follow an argument in a textbook that uses that informal notation. (Some extra confusion got introduced in an edit of the OP, therefore I asked to revert. I can't see the very formula you quote in the post, though. )I added a clip from the textbook to my answer – Ute Jul 18 '23 at 08:27
  • @Ute $$\frac{\int P(time \leq s, counted,time=x)dx}{P(counted)}= \int\frac{ P(time \leq s, counted,time=x)}{P(counted)} dx = \int P(time \leq s, time=x|counted) dx$$ – Sextus Empiricus Jul 18 '23 at 12:23
  • @SextusEmpiricus, that was not the original line of argumentation. You are continuing on the two extra equations that the author added in their first edit (I suggested to revert that edit because these lines messed up). The original post only quoted the textbook, which is basically correct. – Ute Jul 18 '23 at 12:58
  • @Ute the quote from the textbook doesn't make it much clearer. It mentions $$P(time \leq s| counted)$$ and $$P(time \leq s, counted | time =x) = \begin{cases} 1 & \text{if $x \leq s$} \ 0 & \text{if $x > s$}\end{cases}$$ but the motivation is not clear. It is that many bits of the notation is unclear, and 'not the original line of argumentation' doesn't help. – Sextus Empiricus Jul 18 '23 at 22:35
  • I also find the textbooks notation hard to interpret, and do not wonder why the question's author struggled to formalize it. But I read $$P(time≤s,counted|time=x)\ \ \text{as}\ \ =\frac{\lambda(x)}{\lambda}\begin{cases} 1 & \text{if $x \leq s$} \ 0 & \text{if $x > s$}\end{cases}.$$ It would have taken up quite some space to define independent thinning of point processes rigorously - the textbook tried to avoid that. – Ute Jul 18 '23 at 22:56
  • @Ute your reading seems better, and using that reading I can reverse engineer the meaning of $counted$ which is a boolean variable, but this should be made clear in the question. Also the meaning of $P(counted)$ is not clear to me. Should this have been instead $P(counted|time = x)$? – Sextus Empiricus Jul 19 '23 at 20:32
  • Should this have been instead $P(counted|time=x)$ - no, it is the average. I also reverse engineered. What are you going for - amending the question? I feel it was actually a question provoked by that handwavy not closer explained notation, so if you repair the question (up to reverting the edit that introduced errors), it does no longer make sense. But you could of course try to translate Ross' heuristic argument into rigorous mathematics. After all, it is an introductory text book, no time for formalism and probably the goal to teach intuitive understanding. – Ute Jul 19 '23 at 20:48
  • @Ute "> what are you going for?" I am hoping that the person that asks the question will explain their question. ”> no, it is the average" I don't understand what 'the average' relates to in this context. It is already unclear what the event 'counted' means. – Sextus Empiricus Jul 19 '23 at 20:55

1 Answers1

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Inhomogeneity by independent thinning of a homogeneous (Poisson) point process

Your question concerns a method to generate an inhomogeneous point process with intensity function $\lambda(x)$ from a homogeneous one with intensity $\lambda$, by independent thinning, that is by counting points in $x$ with probability $\lambda(x)/\lambda$. The original point process need not necessarily be a Poisson point process, it would work with any other point process, too, and also the formulae hold.

Let us look at denominator and numerator in the second last fraction, $$\frac{\int_0^T P\{\text{time}\leq s, \text{counted}\mid\text{time}=x\} d x\,/T}{P\{\text{counted\}}} $$

Denominator:

$P\{\text{counted}\}$ denotes the mean probability of a randomly chosen event in the base interval $[0,T]$ to be counted. So, no,

  • $P\{counted\} = \int_0^T \lambda(x)/ \lambda dx$

is not totally right. You would need to calculate the integral average of the $\lambda(x)/\lambda$, which is $$ \begin{aligned} P\{\text{counted}\} &= \frac{1}{T}\int_0^T P\{\text{counted}\mid\text{time}=x\} d x \\&=\frac{1}{T}\int_0^T \lambda(x)/\lambda\ dx \\&=\frac{1}{\lambda T}\int_0^T \lambda(x)\ dx . \end{aligned} $$

Numerator:

the integrand $P\{time \leq s, counted |time=x \} = \lambda(t)/\lambda$ ?

Well, the numerator stands for the mean probability that a randomly chosen event lies in the shorter interval $[0,s]$ and is counted, $$\begin{aligned} P(\text{time}\leq s \wedge \text{counted}) &= \frac{1}{T}\int_0^T P\{\text{time}\leq s, \text{counted}\mid\text{time}=x\} d x \\&=\frac{1}{T}\int_0^T \mathbf{1}_{[0,s]}(x) \lambda(x)/\lambda\ dx \\&=\frac{1}{T}\int_0^s \lambda(x)/\lambda\ dx. \\&=\frac{1}{\lambda T}\int_0^s \lambda(x)\ dx. \end{aligned} $$

Final (and now easy) step:

When taking the quotient of this numerator and denominator, the factor $1/(\lambda T)$ cancels out.

Additional remark:

You get the formula for the denominator as a special case of the formula for the numerator, by setting $s=T$.

From the comments: The source is an argument in Sheldon Ross, "Introduction to Probability Models" page 675. The informal notation is taken from there, here is the original: textbook paragraph that provoked the question

Ute
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  • I am not seeing how the numerator and denominator winds up to the last equation. You will still have the factor $\frac{1}{T} = p(time)$ left . please check out the edit on the question @ute – J.doe Jul 16 '23 at 21:19
  • This is some quite informal notation you are using - what is the source of these formulae? Would it help you to do it more formally? In the last equation, you need to divide Numerator by denominator. Both are integrals with a factor $1/T$ in front. This one cancels out. In your own calculation, the factor 1/T is missing in the denominator. – Ute Jul 16 '23 at 21:23
  • @J.doe, I have edited my question, hope it helps. I started out with the original formulation, because it seems you have introduced something in your edit that leads to your comment question "$\frac{1}{T} = p(time)$" - hope it helps! – Ute Jul 16 '23 at 21:52
  • Sorry,There might be something missing in my understanding but In your answer, the second equality: $F(s) =P(time \leq s|counted) = \frac{\int P(time \leq s, counted|time=x)dx }{P(counted)}$ you are missing that it should be $F(s) =P(time \leq s|counted) = \frac{\int P(time \leq s, counted|time=x)dx/\textbf{T} }{P(counted)}$. Check whatever is in bold (T) @ute . You can find the derivation in "sheldon ross introduction to probability models" page 675 – J.doe Jul 17 '23 at 04:18
  • ah, thanks, overlooked it when copying from your question, sorry. it is fixed - is the rest understandable? – Ute Jul 17 '23 at 07:13
  • Ok if i replace: $\frac{ \int_0^T P(time \leq s, counted|time=x )dx/T}{P(counted)} = \dfrac{ 1/\lambda T \int_0^s \lambda(x) dx /T }{1/\lambda T \int_0^T \lambda(x) dx } $ you see that in the numerator we still have $T$ left. Maybe I am not understanding something that leads to that confusion. appreciate the help @ute – J.doe Jul 17 '23 at 08:42
  • No, look at my derivation of the numerator. It includes that division by T right from the start and does not end in the result you are proposing here – Ute Jul 17 '23 at 08:50
  • ok then it makes sense – J.doe Jul 17 '23 at 08:56
  • "a randomly chosen event in the base interval [0,T]" What does this mean? – Sextus Empiricus Jul 19 '23 at 20:38
  • From all events in $[0,T]$ chosen with equal probability. This is doubly stochastic and to be understood as conditional on the realization of the original process. A formal definition can e.g. be found in Last & Penrose (2017) p.43, Def. 5.7 – Ute Jul 19 '23 at 21:01