Given a Poisson Process with rate $\lambda$, by a fixed time $t$ we have observed $n$ arrivals at times $t_1 < \cdots < t_n$, with $t_0 = 0 < t_1$ and $t_n < t$
I'm trying to find a density $f(n, t_1, \cdots, t_n)$, taking $\lambda$ and $t$ as constant.
One approach I took was defining $f(n, t_1, \cdots, t_n) = (\prod_{i=1}^n{\lambda e^{-\lambda (t_i - t_{i-1})}}) \times e^{-\lambda (t - t_n)}$
That is, considering the times between arrivals as independent exponential random variables, and considering there was no arrival from $t_n$ to $t$.
Another approach would be $f(n, t_1, \cdots, t_n) = f(n) \times f(t_1, \cdots, t_n | n) = \frac{(\lambda t)^n e^{-\lambda t}}{n!} \times \frac{n!}{t^n}$
(I borrowed the relation $f(t_1, \cdots, t_n | n) = \frac{n!}{t^n}$ from http://www3.ul.ie/~mlc/_notes/book-MS4217.pdf, section 5.4.2)
In a couple of examples I've tried they appear to be the same function.
Now, I encounter a behaviour I don't understand. When $\lambda > 1$, as $n$ grows it seems like so does $f(n, t_1, \cdots, t_n)$
For example, with $\lambda=1.5$, $t = 10$, the density in its second formulation (just a function of $n$), has value 3.38 at $n=40$, 195 at $n=50$, 11247 at $n=60$. It does not seem to integrate to one. What I find most striking is that it seems to give higher probabilities to having a huge amount of arrivals, while the expected value of $n$ is 15.
I don't know if I'm making wrong assumptions or calculation errors, but I feel there's something fundamental I don't understand. Thanks for your help.
Indeed $f(n, t_1, \ldots, t_n) = \dfrac{\lambda^n \mathsf e^{-\lambda t}}{t^n}$ .
$$\begin{align}f(n, t_1, \cdots, t_n) =&~\Big(\prod_{i=1}^n{t^{-1}\lambda e^{-\lambda (t_i - t_{i-1})}}\Big) \times e^{-\lambda (t - t_n)}\\[1ex] =&~ t^{-n} \lambda^n\exp({-\lambda \sum_{i=1}^n(t_i-t_{i-1})}-\lambda (t-t_n)) \\[1ex] =&~ t^{-n} \lambda^n\exp(-\lambda t)\end{align}$$
Are you integrating enough times? Once for each $t_k$, with $k\in \{1,...,n\}$.
Are you integrating over the appropriate domains? Recall that $0\leq t_1\leq t_2\leq\ldots\leq t_n\leq t$ . The shape is a right-angled n-D hyper pyramid with area $t^n/n!$. (They are the order statistics for a sample from $n$ iid uniform distributions.)
Are you summing the integration result times the probability of $n$, for $n\in\Bbb N$? (The count of events is Poisson distributed, of course.)
You should find that:
$$\begin{align}&~ \sum_{n=0}^\infty \iiint_{0\leq t_1\leq \ldots\leq t_n\leq t} f(n, t_1, \ldots t_n) \operatorname d t_n\cdots\operatorname d t_1 \\[1ex] = &~ \sum_{n=0}^\infty \frac{(\lambda t)^n e^{-\lambda t}}{n!} \times \frac{n!}{t^n} \iiint_{0\leq t_1\leq \ldots\leq t_n\leq t}\operatorname d t_n\cdots\operatorname d t_1 \\[1ex] = &~ \sum_{n=0}^\infty \frac{(\lambda t)^n e^{-\lambda t}}{n!} \times \frac{n!}{t^n} \times \dfrac{t^n}{n!} \\[1ex] =&~ 1 \end{align}$$