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.


1 Answer 1


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.

  • $\begingroup$ 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 $n\lambda$). $\endgroup$ Commented Mar 24, 2020 at 20:45
  • $\begingroup$ Well, the link (and the second part of my answer) are justifications for the first part of the answer. There are many different ways to derive distributions of sums of random variables. I think using MGF's is the easiest one, so that's what I used (assuming you're familiar with MGF's by the point of discussing sufficient statistics), but the derivation in the link is fine too. $\endgroup$ Commented Mar 25, 2020 at 17:08

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