Support of the conditional distribution of a poisson process I am working on Problem 5.1.8 of this book. It states:

Let $\left\{X(t),t \geq 0 \right\}$ be a Poisson process of rate $\lambda$. For $s,t >0$, determine the conditional distribution of $X(t)$, given that $X(t+s) = n$.  

You do all the work and you get a very nice looking result:
$$\binom{n}{k}\dfrac{t^k s^{n-k}}{(t+s)^n}$$
My questions are:


*

*What type of distribution is this? I do not recognize it, but the wording of the question seems to imply it is identifiable.

*What is the support of this pdf? I usually sum over the support to make sure my answer is correct, but I am struggling to determine this.

 A: 
What type of distribution is this? I do not recognize it, but the wording of the question seems to imply it is identifiable.

Indeed it is!
Hint: The distribution puts weight $p_k$ on each integer $k$ between $0$ and $n$, where $$p_k=\binom{n}{k}a^k(1-a)^{n-k},\qquad a=\frac{t}{t+s}.$$ Does this ring a bell?
Second hint: The LaTeX encoding of the first factor in the formula for $p_k$ is \binom.
A: I don't know regarding the distribution, but support is easy to check: you know that $X_t$ can take any value in $\Bbb N = \{0,1,2,\dots\}$ unconditionally. You also know that $X_t$ is non-decreasing, so if $X_{t+s} = n$ then $X_t\in [0;n]$ almost surely. I'm pretty sure that you've used this fact while computing the conditional distribution. An educated guess is also that in a nice formula like that, binomial coefficient "implies" that $k\leq n$. It may not be the case in other situations, of course, but at least it can provide you a high-level hint to where to start. Anyways, $X_t\in [0;n]$ a.s. and clearly $X_t$ can be either $0$ or $n$ with positive probability since on $[t,t+s]$ you can have as many jumps as you want (with maybe little, but still positive probability). So, $[0;n]$ is your support. 
Actually, determining the support is a crucial part of finding the distribution, just the "formula" is not enough, since it's only part of the answer - the final formula is different. For example, in case of exponential distribution it's easy to remember that density is kinda $\lambda\mathrm e^{-\lambda x}$. This is not density: you can define infinitely many distributions of the shape $\lambda1_A(x)\mathrm e^{-\lambda x}$, and exponential distribution is the one for which $A = [0,\infty)$.
