# Poisson sufficient statistics problem

I have the following problem:

Let $$Y_1, \dots, Y_n$$ be a random sample from a Poisson distribution $$\text{Pois}(\lambda)$$. Recall, the $$\text{Pois}(\lambda)$$ distribution has the probability function $$f_{\lambda}(y) = e^{-\lambda} \dfrac{\lambda^y}{y!}$$, if $$y = 0, 1, 2, 3, \dots$$, and $$\lambda > 0$$.

(a) Show that $$T(\mathbf{Y}) = \sum_{i = 1}^n Y_i$$ is a sufficient statistic for $$\lambda$$ using the Fisher-Neyman factorisation theorem.

(b) What is the distribution of $$T(\mathbf{Y})$$? Obtain this result directly using the definition of a sufficient statistic.

For (a), we have that $$L(\lambda, \mathbf{y}) = \prod_{i = 1}^n e^{-\lambda}\dfrac{\lambda^{y_i}}{y_i!} = e^{-n \lambda} \dfrac{\lambda^{\sum_{i = 1}^n y_i}}{\prod_{i = 1}^n y_i!}$$. So $$T(\mathbf{y}) = \sum_{i = 1}^n y_i$$, $$g(t, \lambda) = e^{-n \lambda} \lambda^t$$ and $$h(\mathbf{y}) = \dfrac{1}{\prod_{i = 1}^n y_i!}$$.

For (b), the solution is given as follows:

$$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$

$$P(\mathbf{Y} \mid T(\mathbf{Y})) = \dfrac{P(\mathbf{Y}, T(\mathbf{Y}))}{P(T(\mathbf{Y}))} = \dfrac{\prod_{i = 1}^n e^{-\lambda} \dfrac{\lambda^{Y_i}}{Y_i!}}{e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}} = \dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$$

I don't understand the solution for (b). In particular, I don't understand how the author concluded that $$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$ and $$P(T(\mathbf{Y})) = e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}$$. I would greatly appreciate it if someone would please take the time to clarify this.

If $$Y_1, ..., Y_n$$ are iid $$\text{Pois}(\lambda)$$, then $$T(Y) = \sum\limits_{i=1}^n Y_i \sim \text{Pois}(n\lambda)$$
This is probably most easily seen by noting that the MGF for each $$Y_i$$ is $$M_{y_i}(t) = e^{\lambda(e^t - 1)}$$, so then the MGF for $$T(Y)$$ is $$(M_{y_i}(t))^n = (e^{\lambda(e^t - 1)})^n = e^{(n\lambda)(e^t - 1)}$$, which is the MGF for a $$\text{Pois}(n\lambda)$$ random variable.
• The first part of your answer is fine, but I don't think the second part really explains how the author came to that conclusion. For the second part, I think this llc.stat.purdue.edu/2014/41600/notes/prob1805.pdf explains it well, combined with the knowledge of the first part of your answer (that is, that the parameter is now $n\lambda$). Commented Mar 24, 2020 at 20:45