0
$\begingroup$

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:

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

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

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.

$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $n\lambda$). $\endgroup$ – The Pointer Mar 24 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. $\endgroup$ – Nick Koprowicz Mar 25 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.