# Need help understanding Sufficient Statistics and using the formal definition

Let $$X = (X_1, X_2, . . . , X_n)$$ be a random sample from $$\rm Poisson(\theta)$$. 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 $$\sum_{i=1}^nx_i \sim \mathrm{Poisson}(n\theta) \implies f_{\sum_{i=1}^nx_i}(t) = \frac{e^{-n\theta}(n\theta)^t}{t!}$$, but I don't understand how they arrived to this point, nor do I understand how they found $$\sum_{i=1}^nx_i \sim \mathrm{Poisson}(n\theta)$$ either. Can someone help me?

• Sum of n independent poisson rvs each with parameter h is poisson with parameter n*h. Commented Nov 5, 2014 at 17:53
• Searching our site turns up many threads in which these relationships among Poisson variables are derived: visit stats.stackexchange.com/search?q=sum+poisson+random+variable.
– whuber
Commented Nov 8, 2014 at 20:29

First you need to show that $$\sum_{i=1}^nX_i \sim \mathrm{Poisson}(n\theta)$$. 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 $$X \sim \mathrm{Poisson}(\theta)$$, 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, $$T = \sum_{i=1}^n X_i \sim \mathrm{Poisson}(n\theta).$$

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 $$\theta$$.

• I really like the fact that you used the MGF to help us find the distribution for sum(xi). Hadn't thought of that before. Thank you very much for this :) Commented Nov 10, 2014 at 10:54

Also if the statistic is not sufficient, then we can always write it as a product with a conditional

$$f_\mathbf x(\mathbf x\mid\theta) = f_T(t,\theta) \cdot f_\mathbf x(\mathbf x\mid T=t, \theta).$$

In order to know whether there is independence of $$\theta$$ when conditioning on $$T$$

$$f_\mathbf x(\mathbf x\mid T=t, \theta) = f_\mathbf x(\mathbf x\mid T=t),$$ it is not necessary to find explicitly the two functions $$f_T(t,\theta)$$ and $$f_\mathbf x(\mathbf x\mid T=t, \theta)$$.

Instead, if you are able to factorise the function like below,

$$f_\mathbf x(\mathbf x\mid\theta) = g(t;\theta)\cdot h(\mathbf x),$$

then it is not necessary to explicitly obtain those two functions. There will exist another function $$d(t)$$ that makes the two equivalent

$$f_\mathbf x(\mathbf x\mid\theta) = \underbrace{\left[g(t;\theta)/d(t)\right]}_{f_T(t,\theta)}\cdot \underbrace{\left[d(t) h(\mathbf x)\right]}_{f_\mathbf x(\mathbf x\mid T=t)};$$

we don't need to know $$d(t)$$ explicitly -- just the existence is enough. Intuitively, for every value of $$t$$, the function $$d(t)$$ will be a normalisation parameter to turn $$h(\mathbf{x})$$ into a probability density.

• Commented Jun 15 at 11:16