I am self studying a theoretical statistics course I found online.

There is a question to show that for $(X_1, ... X_n)$ i.i.d Poisson variables with parameter $\theta$, the statistic $T=\sum_{i=1}^N X_i$ is sufficient. Now I know there are many answers online to show this, which I (think I) understand, but my question is around finding the marginal distribution $\mathcal{P}_{\theta}(T=t)$ and not actually about showing sufficiency.

I specifically want to know if there is a way to do this by marginalizing over $x$ WITHOUT using the fact that the poisson distribution of a sum of $n$ poisson variables with parameter $\theta$ is a distribution $Po(n\theta)$.

So I believe that:

\begin{align} \mathcal{P}_{\theta}(X=x, T=t) &= \mathcal{P}_{\theta}(X=x)I\{T(x)=t\} \\ &= I\{T(x)=t\}\prod_{i=1}^n \frac{\theta^{x_i}e^{-\theta}}{x_i!} \\ &= I\{T(x)=t\}\theta^t e^{-n\theta}\prod_{i=1}^n \frac{1}{x_i!} \end{align}

And have seen that \begin{align} \mathcal{P}_{\theta}(T=t) &= \frac{n^t\theta^t e^{-n\theta}}{t!} \end{align}

In the lecture notes I am following, they do something similar with a Bernoulli distribution and they marginalise over the possible outcomes of $x$ by multipliying the corresponding Bernoulli joint distribution by $\begin{pmatrix}n \\ t \end{pmatrix}$ to get the marginal distribution of $\mathcal{P}_{\theta}(T=t)$. My understanding of this is that there are this many ways to get $t$ successes in $n$ trials. (Sorry if including this is confusing but it is the rationale for why I am posing this q).

Hence, I feel like in the Poisson case it has something to do with the fact that the number of arrangements of the set of $t$ objects containing $n$ distinct elements $a_i, a_2 ... a_n$ with $x_i$ copies of element $a_i$ (s.t $\sum_i^n x_i = t$) is $\frac{t!}{\prod_{i=1}^n x_i!}$.

So if the total number of possible arrangements of the sample space is $n^t$ (as there are $n$ choices for $t$ total objects) then I see that the probability of such an event, $\Omega$ is:

$$ \mathcal{P}_{\Omega} = \frac{t!}{n^t\prod_{i=1}^n x_i!} $$

Now I see that multiplying $\mathcal{P}_{\theta}(X=x, T=t)$ by $\frac{1}{\mathcal{P}_{\Omega}}$ gives the desired outcome however I don't understand why the inverse, and I don't really understand why you would divide by $n^t$ anyway as this wasn't done in the Bernoulli example (i.e it was not divided by $2^t$).

Can someone explain if my reasoning about the arrangements is correct, and if so where my logic is failing around taking the inverse/dividing by $n^t$?

If my reasoning is incorrect is there a way to marginalize over $x$ without using/showing the fact that the poisson distribution of a sum of $n$ i.i.d variables from $Po(\theta)$ is a distribution $Po(n\theta)$?



1 Answer 1


Since \begin{align} \mathcal{P}_{\theta}(X=x, T(X)=t) = \mathbb I\{T(x)=t\}\theta^t e^{-n\theta}\prod_{i=1}^n \frac{1}{x_i!} \end{align} by marginalisation \begin{align} \mathcal{P}_{\theta}(T(X)=t) &= \sum_{x;\,T(x)=t}\theta^t e^{-n\theta}\prod_{i=1}^n \frac{1}{x_i!}\\ &= \theta^t e^{-n\theta}\sum_{x;\,\sum_i x_i=t}\,\prod_{i=1}^n \frac{1}{x_i!}\\ &= \theta^t e^{-n\theta}\sum_{x;\,\sum_i x_i=t}\,\frac{t!}{t!}\prod_{i=1}^n \frac{1^{x_i}}{x_i!}\\ &= \frac{1}{t!}\theta^t e^{-n\theta}\sum_{x;\,\sum_i x_i=t}\,{t \choose x_1 \cdots x_n}\prod_{i=1}^n 1^{x_i}\\ &= \frac{1}{t!}\theta^t e^{-n\theta}(1+\cdots+1)^t\\ &= \frac{1}{t!}\theta^t e^{-n\theta}n^t \end{align}

  • 1
    $\begingroup$ Awesome thank you, just fyi for anyone else the lines from 4-5 use the multinomial theorem $\endgroup$
    – emv
    Sep 14, 2020 at 7:18

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