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?

  • 2
    $\begingroup$ Sum of n independent poisson rvs each with parameter h is poisson with parameter n*h. $\endgroup$
    – PeterR
    Commented Nov 5, 2014 at 17:53
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Nov 8, 2014 at 20:29

2 Answers 2


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 $\theta$.

  • $\begingroup$ 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 :) $\endgroup$ 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.


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