# Poisson distribution marginal probability of sufficient statistic

I am self studying a theoretical statistics course I found online.

There is a question to show that for $$(X_1, ... X_n)$$ i.i.d Poisson variables with parameter $$\theta$$, the statistic $$T=\sum_{i=1}^N X_i$$ is sufficient. Now I know there are many answers online to show this, which I (think I) understand, but my question is around finding the marginal distribution $$\mathcal{P}_{\theta}(T=t)$$ and not actually about showing sufficiency.

I specifically want to know if there is a way to do this by marginalizing over $$x$$ WITHOUT using the fact that the poisson distribution of a sum of $$n$$ poisson variables with parameter $$\theta$$ is a distribution $$Po(n\theta)$$.

So I believe that:

\begin{align} \mathcal{P}_{\theta}(X=x, T=t) &= \mathcal{P}_{\theta}(X=x)I\{T(x)=t\} \\ &= I\{T(x)=t\}\prod_{i=1}^n \frac{\theta^{x_i}e^{-\theta}}{x_i!} \\ &= I\{T(x)=t\}\theta^t e^{-n\theta}\prod_{i=1}^n \frac{1}{x_i!} \end{align}

And have seen that \begin{align} \mathcal{P}_{\theta}(T=t) &= \frac{n^t\theta^t e^{-n\theta}}{t!} \end{align}

In the lecture notes I am following, they do something similar with a Bernoulli distribution and they marginalise over the possible outcomes of $$x$$ by multipliying the corresponding Bernoulli joint distribution by $$\begin{pmatrix}n \\ t \end{pmatrix}$$ to get the marginal distribution of $$\mathcal{P}_{\theta}(T=t)$$. My understanding of this is that there are this many ways to get $$t$$ successes in $$n$$ trials. (Sorry if including this is confusing but it is the rationale for why I am posing this q).

Hence, I feel like in the Poisson case it has something to do with the fact that the number of arrangements of the set of $$t$$ objects containing $$n$$ distinct elements $$a_i, a_2 ... a_n$$ with $$x_i$$ copies of element $$a_i$$ (s.t $$\sum_i^n x_i = t$$) is $$\frac{t!}{\prod_{i=1}^n x_i!}$$.

So if the total number of possible arrangements of the sample space is $$n^t$$ (as there are $$n$$ choices for $$t$$ total objects) then I see that the probability of such an event, $$\Omega$$ is:

$$\mathcal{P}_{\Omega} = \frac{t!}{n^t\prod_{i=1}^n x_i!}$$

Now I see that multiplying $$\mathcal{P}_{\theta}(X=x, T=t)$$ by $$\frac{1}{\mathcal{P}_{\Omega}}$$ gives the desired outcome however I don't understand why the inverse, and I don't really understand why you would divide by $$n^t$$ anyway as this wasn't done in the Bernoulli example (i.e it was not divided by $$2^t$$).

Can someone explain if my reasoning about the arrangements is correct, and if so where my logic is failing around taking the inverse/dividing by $$n^t$$?

If my reasoning is incorrect is there a way to marginalize over $$x$$ without using/showing the fact that the poisson distribution of a sum of $$n$$ i.i.d variables from $$Po(\theta)$$ is a distribution $$Po(n\theta)$$?

Thanks!

Since \begin{align} \mathcal{P}_{\theta}(X=x, T(X)=t) = \mathbb I\{T(x)=t\}\theta^t e^{-n\theta}\prod_{i=1}^n \frac{1}{x_i!} \end{align} by marginalisation \begin{align} \mathcal{P}_{\theta}(T(X)=t) &= \sum_{x;\,T(x)=t}\theta^t e^{-n\theta}\prod_{i=1}^n \frac{1}{x_i!}\\ &= \theta^t e^{-n\theta}\sum_{x;\,\sum_i x_i=t}\,\prod_{i=1}^n \frac{1}{x_i!}\\ &= \theta^t e^{-n\theta}\sum_{x;\,\sum_i x_i=t}\,\frac{t!}{t!}\prod_{i=1}^n \frac{1^{x_i}}{x_i!}\\ &= \frac{1}{t!}\theta^t e^{-n\theta}\sum_{x;\,\sum_i x_i=t}\,{t \choose x_1 \cdots x_n}\prod_{i=1}^n 1^{x_i}\\ &= \frac{1}{t!}\theta^t e^{-n\theta}(1+\cdots+1)^t\\ &= \frac{1}{t!}\theta^t e^{-n\theta}n^t \end{align}