# 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 ?

• I believe you mis-stated the question: don't you want to show the $X_i$ are independent of $T$, not $\lambda$? – whuber Dec 12 '20 at 16:46
• actually, it is that $P(X_1,...,X_n|T=t)$ is independent of $\lambda$ and thus $T$ is sufficient statistic of $\lambda$ – Jean de Léry Dec 12 '20 at 20:38

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.