Suppose we have $n$ Poisson Processes $N_{i}(t)$ each with rate $\lambda_i,i=1,2,\ldots,n$. Combining $n$ Poisson Processes, we have a Poisson Process $N(t)$ with rate $\lambda:=\sum_{i=1}^n \lambda_i$. You can view each Poisson Process $N_{i}(t)$ as the arrival process of a customer type. Suppose we have a capacity of $X$, meaning we can only serve $X$ customers in total regardless of their customer types. In addition, the time horizon is [0,1]. Let $\tau \sim Gamma(X,\lambda)$, which denotes the time when the $X^{th}$ customer arrives. Let $T:=\min(1,\tau)$, which notes the time when the capacity is reached or we are at the end of the time horizon. I am interested in the number of customers served for each type: $N_{i}(T)$. Is it safe to say that $$N_{i}(T) \sim Poisson(\lambda_i T)?$$ Note that we actually have the dependence here: $\sum_{i=1}^n N_{i}(T)=X$. So I am not sure if what I have written is correct.
A little bit more explanation of the background: You can think of the problem as an inventory problem. We have $X$ units of products. There are $n$ types of customers seeking this product, with an arrival process of Poisson process with rate $\lambda_i$. The time horizon is [0,1], which you can think of as a day. I am interested in the number of customers of each type served by the end of day, and there are two possibilities: one is the inventory runs out, the other is it does not run out. That's why the terminal time is $T:=\min(1,\tau)$. And specifically, I am interested in the distribution of $N_{i}(T)$.
Thank you in advance.
By the thinning property of Poisson processes, you can start off with the combined process $N(t)$ and recover the individual processes $N_i(t)$ by taking each arrival in $N(t)$ and classifying them as type $i$ with probability $p_i := \lambda_i/\lambda$. Each arrival is classified independently.
Therefore conditional on $N(t)=n$, we have $N_i(t) \sim \text{Binomial}(n, p_i)$.
This still holds even at random times; so at time $\tau$, say, $N_i(\tau) \sim \text{Binomial}(X, p_i)$.
The case for $N(T)$ is slightly more difficult; $N(T)$ is distrubuted like the minimum of a $\text{Poisson}(\lambda)$ and $X$. We can find an expression for the probability generating function of $N_i(T)$ though: $$\mathbb{E}[t^{N_i(T)}] = \sum_{n=0}^\infty (1-p_i+p_it)^{n \wedge X} \frac{e^{-\lambda}\lambda^n}{n!}$$ where $n \wedge X := \text{min}(n, X)$.