Ratio of Poisson Distributions (Possible mistake on exams by instructor)

Question

I found the following on a practice final of one of my students and it does not seem to add up.

Let $X_1, X_2, \dots, X_n$ be a random sample from a Poisson distribution $f(k\vert \theta) = \frac{e^{-\theta}\theta^k}{k!}$ where $k \in \mathbb{N}$. Show that $S(X)= \frac{X_1}{\sum_{i=1}^n X_i}$ is an ancillary statistic (i.e. its distribution does not depend on $\theta$).

I think what they have in mind is that the above expression is scale invariant which it is. The issue is however twofold:

1. $S$ is not well defined when $X_1 = X_2 = \dots = X_n = 0$ which happens with non-zero probability.
2. The scale invariant property of the expression is not sufficient. One also needs to show that the $X_i$'s follow what is known as the scale model, namely there are Random Variables $W_i$ which do NOT depend on $\theta$ and a constant $c$ such that: $X_i = cW_i$. To my knowledge this is not the case for Poisson.

Is this a case of a simple oversight or am I missing something obvious?

(Note: This is a standard graduate course in statistics where students are expected to identify various kinds of statistics (sufficient, complete, ancillary etc) and apply these ideas to relevant theorems such as Basu's theorem.)

Answer 1

You’re right; this is wrong. The problem of well-definedness could perhaps be solved by treating “undefined” as a special value of the statistic or defining it in some way, but the distribution for the defined values depends on $\theta$ anyway. For instance, $S(X)=1$ iff all $X_i$ except $X_1$ are $0$, and the probability for this clearly depends on $\theta$.

What they might have meant:

• The same statement for the normal distribution (where the undefined case has probability $0$)
• The conditional distribution of $X_1$ given $\sum_iX_i$ doesn’t depend on $\theta$.