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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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I reformatted the math, check if it's ok. – leonbloy Apr 30 '12 at 0:46
This is perfect. Thanks! – mary Apr 30 '12 at 0:48
up vote 1 down vote accepted

$$ P(N_t-N_s=0 \mid N_t>0) = \frac{P(N_t-N_s=0\ \&\ N_t>0)}{P(N_t>0)} = \frac{P(N_t-N_s=0\ \&\ N_s>0)}{P(N_t>0)}. $$ 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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So when independent I can say, P(nt- ns) * P(Ns)/P(Nt>0) but what cancells? – mary Apr 30 '12 at 2:04
You have to work out the three probabilities first and then worry about what cancels, if anything. At some points you might want to multiply a numerator and denominator by $e^{s\lambda}$ or $e^{t\lambda}$, thereby changing $1-e^{-s\lambda}$ to $e^{s\lambda}-1$, etc. – Michael Hardy Apr 30 '12 at 2:13
I'm not following the three probabilities. Can you elaborate please and provide details? Thx – mary Apr 30 '12 at 7:09
I added some more details above. – Michael Hardy May 1 '12 at 2:27

Because $0<s<t$ 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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I don't realy follow your argument, can you please elaborate on it – mary Apr 30 '12 at 2:04
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. – FiniteA Apr 30 '12 at 2:08

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