Let ${X_1},...,{X_n}$ be an iid sample from a Poisson distribution with mean $\lambda $ and let $T = \sum\limits_{i = 1}^n {{X_i}} $.
Show that the distribution of ${X_1},...,{X_n}$ given $T$ is independent of $\lambda $, and conclude that $T$ is sufficient for $\lambda $.
I actually have the solution for it, but I'm struggling following it.
It first shows the steps until it arrives at $P\left( {{X_1} = {x_1},...,{X_n} = {x_n}} \right)$, given by
Then it goes to show the conditional distribution of ${X_1},...,{X_n}$ given $T = t$:
Why does he stop at ${x_{n - 1}}$ in the above denominator? Why does he use $n\lambda $ everywhere there was supposed to be just a $\lambda$ in the whole denominator? I can't seem to grasp the mechanics behind just about everything in the second picture.
Why does he stop at $x_{n−1}$ in the above denominator?
Because conditioned on the sum being $t$, he uses $x_n= t-x_1-x_2-\cdots -x_{n-1}$
Why does he use $nλ$ everywhere there was supposed to be just a $λ$ in the whole denominator?
In the denominator, the sum of $n$ i.i.d. Poisson random variables with parameter $\lambda$ is a Poisson random variable with parameter $n\lambda$. In the numerator, you are multiplying $n$ cases of $e^{-\lambda}$ together to get $e^{-n\lambda}$