Homogenous Poisson process 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 
 A: $$
P(N_t-N_s=0 \mid N_t>0) = \frac{P(N_t-N_s=0\  \&\  N_t>0)}{P(N_t>0)} = \frac{P(N_t-N_s=0\  \&\  N_s>0)}{P(N_t>0)}.
$$
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:
$$
\frac{e^{s\lambda}-1}{e^{t\lambda}-1}.
$$
A: 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.
