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Consider a homogeneous Poisson process $N$ with rate $\lambda$. For For $0 < s < t$, I'm trying to show that:

$P(N_t-N_s=0\mid N_t>0)= \frac{e^{\lambda s} - 1}{e^{\lambda t} - 1}$

I'm mainly thinking of using some sort of conditioning and/or rewriting the expression

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    This is perfect. Thanks!2012-04-30

2 Answers 2

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$ P(N_t-N_s=0 \mid N_t>0) = \frac{P(N_t-N_s=0\ \&\ N_t>0)}{P(N_t>0)} = \frac{P(N_t-N_s=0\ \&\ N_s>0)}{P(N_t>0)}. $ First figure out what the above is true. Think about what it means.

Then exploit the fact that the two events with "$\&$" between them are independent. Make sure you understand why they're independent.

Later note in response to comments below: What was done above was for the purpose of writing the expression in terms of events that are independent. $ \Pr(N_t-N_s=0) = e^{-(t-s)\lambda} $ $ \Pr(N_s>0) = 1 - e^{-s\lambda} $ $ \Pr(N_t>0) = 1 - e^{-t\lambda} $ So we have $ \frac{e^{-(t-s)\lambda}(1 - e^{-s\lambda})}{1 - e^{-t\lambda}}. $ Multiplying out the numerator, we get: $ \frac{e^{-(t-s)\lambda} - e^{-t\lambda}}{1 - e^{-t\lambda}}. $ Then multiply both the numerator and the denominator by $e^{t\lambda}$, and we get: $ \frac{e^{s\lambda}-1}{e^{t\lambda}-1}. $

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    I added some more details above.2012-05-01
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Because $0 and the probability is conditioned on $t$ instead of $s$, I would use Bayes' rule so that the two condition becomes reversed. That is,

$P(N_t-N_s=0\mid N_t>0)= \frac{P(N_t-N_s=0)}{P(N_t>0)}P(N_t>0|N_t=N_s) = \frac{P(N_t-N_s=0)}{P(N_t>0)}P(N_s>0)$

Then you should get it after some algebra.

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    Intuitively, I want the condition to be based on past, not future. So I apply Bayes' rule in the first equality. The second equality is just simplifying the second term.2012-04-30