Need help understanding Sufficient Statistics and using the formal definition Let $X = (X_1, X_2, . . . , X_n)$ be a random sample from $Poisson(\theta)$. Use the factorization theorem to find a sufficient statistic $T(X)$ and then use the formal
definition of sufficiency to confirm that $T(X)$ is sufficient.
So the factorization theorem states that if our $ f_\mathbf x(\mathbf x\mid\theta) = g(T;\theta)h(\mathbf x) $ then our sufficient statistic can be found through $T(X)$.
In this particular case, the joint pdf : $f_\mathbf x(\mathbf x\mid\theta) = \prod_{i=1}^{n} \frac{e^{-\theta}\theta^{x_i}}{x_i!} = e^{-n\theta} \theta^{\sum_{i=1}^nx_i} \prod_{i=1}^{n}\frac{1}{x_i!}$, with $T(X) =  \sum_{i=1}^nx_i$ being our sufficient statistic.
What I don't understand is the second part where we use the formal definition to prove the statistic is sufficient ie to show $f_\mathbf x(\mathbf x\mid T=t)$ is independent of $\theta$.
We use $f_\mathbf x(\mathbf x\mid \sum_{i=1}^nx_i=t)$ = $\frac{f_\mathbf x,_{\sum_{i=1}^nx_i}( x, t)}{f_{\sum_{i=1}^nx_i}(t)}$, but I don't understand how to obtain $f_{\sum_{i=1}^nx_i}(t)$.
In the solutions, they use the fact that $\sum_{i=1}^nx_i \sim Poisson(n\theta) \implies f_{\sum_{i=1}^nx_i}(t) =  \frac{e^{-n\theta}(n\theta)^t}{t!}$, but I don't understand how they arrived to this point, nor do I understand how they found $\sum_{i=1}^nx_i \sim Poisson(n\theta) $ either. Can someone help me?
I'm not used to posting on here, nor have I ever used LaTeX before, so feedback would be appreciated.
 A: First you need to show that $\sum_{i=1}^nX_i \sim Poisson(n\theta)$. This can be done by showing that the sum of the $X_i$'s has the same MGF of a poisson random variable.
Recall that for a poisson random variable $X \sim Poisson(\theta)$, its MGF is
$M_x(t) = exp\bigl(\theta(e^t - 1) \bigl)$.Since we have iid random variables, the MGF of $T = \sum_{i=1}^n X_i$ is thus
$$
M_T(t) = M_{\sum_{i=1}^n X_i} (t) = M_{X_1}(t)M_{X_2}(t)\dots M_{X_n}(t) = exp\bigl(n\theta(e^t - 1) \bigl).
$$
This is exactly the MGF of a poisson random variable with parameter $n\theta$. Therefore, $$T = \sum_{i=1}^n X_i \sim Poisson(n\theta).$$
Knowing the above result makes it easy to show $f_\mathbf x(\mathbf x\mid T=t)$ is independent of $\theta$, as following.
\begin{align}
f_\mathbf x(\mathbf x\mid T=t) &= \frac{f_\mathbf x(\mathbf x, \,T=t)}{f_\mathbf x(T=t)} \\
&= \frac{e^{-n\theta} \theta^t \prod_{i=1}^{n}\frac{1}{x_i!}}{e^{-n\theta} (n\theta)^t \frac{1}{t!}} \\ &= \frac{\prod_{i=1}^{n}\frac{1}{x_i!}}{n^t \frac{1}{t!}}.
\end{align}
And this is independent of $\theta$.
Hope it helps!
