TITLE: Poisson process basic problem
QUESTION [0 upvotes]: I've just started to learn stochastic and I'm stuck with these problems. Don't know how to start solving them.

Suppose that cars cross a certain point in the highway in accordance with a
Poisson process with rate λ = 3 per minute. What is the probability that the
time between the 10th and the 11th car crossing exceeds two minutes?

And, I guess, similiar (don't want to make another thread):  

For a Poisson process evaluate the probability Pr(Ns = k|Nt = n)
for 0 < s < t.

REPLY [0 votes]: 1) Let $\{N(t) : t\ge 0\}$ be a Poisson process with rate $\lambda = 3$. Recall that for a Poisson process, $N(t)-N(s)$ has the Poisson distribution with rate $\lambda(t-s)$ for any $0\le s < t$. Because of independent and stationary increments, the probability that the time between the 10th and the 11th car crossing exceeds two minutes is the same as the probability that there are no arrivals in the first two minutes. So the answer is $$\mathbb P(N(2)-N(0)=0) = \frac{e^{-3(2)}(2)^0}{0!} = e^{-6}.$$
2) Let $\lambda$ be the rate of the Poisson process. Then using the definition of conditional probability and properties of Poisson processes we see that
$$ \mathbb P(N(s)=k | N(t) = n) = \frac{\mathbb P(\{N(s)=k\}\cap \{N(t)=n\})}{\mathbb P(N(t)=n)} = \frac{\mathbb P(\{N(s)=k\}\cap \{N(t)-N(s)=n-k\})}{\mathbb P(N(t)=n)} = \frac{\mathbb P(N(s)=k)\mathbb P(N(t)-N(s)=n-k)}{\mathbb P(N(t)=n)} = \frac{(e^{-\lambda s}(\lambda s)^k/k!)(e^{-\lambda(t-s)}(\lambda(t-s))^{n-k}/(n-k)!)} {e^{-\lambda t}(\lambda t)^n / n!}.$$
That expression is a little ugly and I managed to simplify it to
$$\binom n k s^k(t-s)^{n-k}t^{-n}, $$
and further to
$$\binom n k \left(\frac s t\right)^k\left(1 - \frac s t\right)^{n-k}.$$
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