I am looking at a multinomial Poisson distribution. Failing to obtain the denominator.

We have that $X_1, ..., X_n \sim \mathbb{P}(\lambda)$, where $\theta = \exp^{-\lambda}$

We know that $$\mathbb{P}(X=x) = \frac{\lambda^x \theta}{x!}$$

Since the variables are IID, we know that:

$$\mathbb{P}(\mathbf{X}) = \prod_{i=1}^n \frac{\lambda^{x_i} \theta}{x_i!} $$

this when simplified, gives me:

$$\mathbb{P}(\mathbf{X}) = \left( \theta^n \lambda^{\sum_i x_i} \right)\left( \frac{1}{\prod_{i=1}^n x_i!} \right) $$

If denote the sufficient statistic $T$ as $\sum_i x_i = t$, then we have:

$$\mathbb{P}(\mathbf{X}) = \left( \theta^n \lambda^t \right) \left( \frac{1}{\prod_{i=1}^n x_i!} \right)$$

However, Cambridge stats notes, page 15 example 3.4 (b) last equation line, implies that $\prod_{i=1}^n x_i! = t!$. Actually, I think I am mixing something up... Isn't the distribution $\mathbb{P}(\sum_i X_i = t)$ the one that I have been deriving above?

  • $\begingroup$ $\frac{1}{\prod_{i=1}^n x_i!}$ is in the expression for $\mathbb P(\mathbf{X}=\mathbf{x})$ while $\frac{1}{t!}$ in in the expression for the rather larger $\mathbb P\left(\sum X_i =t\right)$ $\endgroup$ – Henry Dec 19 '18 at 16:50
  • $\begingroup$ thanks! Can you give me a tip on how to derive the equation for $\mathbb{P} \left( \sum_i X_i = t \right)$ $\endgroup$ – i squared - Keep it Real Dec 19 '18 at 16:53

The key point is that the $X_i$´s are poisson distributed. Then the sum of poisson distributed random variables can be derived. For instance here (two variables). For $n$ variables we get

$$P\left( \sum_{i=1}^{n} X_i=t \right)=\frac{\left( n\cdot \lambda\right)^t}{t!}\cdot e^{-n\cdot \lambda}$$

Now it is straigtforward to caclculate

$$\frac{\mathbb P(X_1=0 \cap \sum_{i=2}^n X_i=t)}{\mathbb P(\sum_{i=1}^n X_i=t)}=\frac{\mathbb P(X_1=0) \cdot \mathbb P( \sum_{i=2}^n X_i=t)}{\mathbb P(\sum_{i=1}^n X_i=t)}$$


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