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 ?
