Sufficient statistic for the distribution of a random sample of Poisson distribution Let $X_1,...,X_n$ be a random sample from a Poisson distribution with mean $\lambda$ and $T = \sum_{i=1}^n X_i $ . Show that the distribution of $X_1,...,X_n$ given T is independant of $\lambda$ so that $T$ is a sufficient statistic for $\lambda$.
By definition of sufficient statistic :
$$ P(X_1 = x_1,...,X_n=x_n | T =t  ) = \frac{P(X_1=x1,...,X_n=x_n,T=t)}{P(T=t)}$$
According to the teaching assistant, this is equal to :
$$ \frac{P(X_1=x1)...P(X_n=t-\sum_{i=1}^{n-1}x_i)}{P(T=t)}$$
I understand that because $X_1,...X_n$ are iid, probability of their intersection is product of their probabilities. I also understand that $ x_n=t-\sum_{i=1}^{n-1}x_i) $. What I cannot figure out is why the factor $P(T=t)$ disappears of the numerator.

The answer to my own question would be that :
$$ 
P(X_1=x_1,...X_n = x_n,T=t) = P(X_1=x_1,..,X_n=x_n \cap T = t) $$
And
$$ P(X_1=x_1,..,X_n=x_n \cap T = t) = P(X_1,..,X_n=X)  $$
Because here the realisation of the event T has probability one if $ X_1=x_1...X_n=x_n $.
In other words, the probability that : "n random variables equal n observed values and sum of this n random variables is equal to the sum of the n observed values" is equal to the probability that n random variables equal n observed values.
Now I wonder if this generalize to any statistic computed on these random variables ?
Is $P(X_1 = x_1,...X_n=x_n, T =t)$ with T a statistic which only depends on the data equals to $ P(X_1=x_1,...X_n=x_n) $ ?
Also, the reverse is not true, knowing a statistic about a sample of random variables does not tell us anything about $P(X_1=x_1)$. The data determines the sample mean/variance/kurtosis... But not the other way around. Am I correct ?
 A: As the sum of $n$ i.i.d. samples from a Poisson distribution with mean $\lambda$, $T$ is itself Poission-distributed with mean $n\lambda$. We can simplify your expression:
$$
\frac{\prod_i \mathbb{P}(X_i = x_i)}{\mathbb{P}(T=t)} =
\frac{\prod_i \frac{\lambda^{x_i}e^{-\lambda}}{x_i!}}{\frac{(n\lambda)^{t}e^{-n\lambda}}{t!}} =
\frac{\lambda^t e^{-n\lambda}\prod_i\frac{1}{x_i!}}{\frac{n^t\lambda^t e^{-n\lambda}}{t!}} =
\frac{t!}{n^t \prod_i x_i!}
$$
Note that this expression is independent of $\lambda$, which is what you needed to show. (In fact, it is the probability mass function of a multinomial distribution.)

Is $P(X_1 = x_1,...X_n=x_n, T =t)$ with $T$ a statistic which only depends on the data equal to $ P(X_1=x_1,...X_n=x_n) $ ?

This is correct.

Also, the reverse is not true, knowing a statistic about a sample of random variables does not tell us anything about $P(X_1=x_1)$. The data determines the sample mean/variance/kurtosis... But not the other way around. Am I correct ?

This isn't quite true. For example, if you know that the variance is 0, that tells you that the random variable is equal to its mean with probability 1, so $\mathbb{P}(X=x)$ must be either 1 (if $x$ is the mean) or 0.
