Negative Wilks' theorem statistic & zero degrees of freedom

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In one of my text examples, we are to use Wilks' theorem to test $H_0 : y \sim \textrm{Bernoulli}(p)$ versus $H_1 : y \sim \textrm{Poisson}(\lambda)$ given $n$ samples and $t = \sum_{i=1}^n y_i$. The likelihood ratio $\Lambda$ is reasonably simple to compute $$ \Lambda = e^t\left(1-\frac{t}{n}\right)^{n-t}.$$ Thus the relevant statistic for Wilks' theorem must be $$ -2\log\Lambda = -2\left[t + (n-t)\log\left(1-\frac{t}{n}\right)\right].$$ The values given are $n = 30$ and $\sum_{i=1}^n y_i = 4$, but this just gives $-2\log\Lambda = -0.558756$. The book actually just leaves it at $\Lambda$ and makes some argument of it probably being too small to be statistically significant rather than applying Wilks' theorem as the question asks. I just have a few issues with this

  • How can we have a negative value for the Wilks' statistic if it is supposed to be distributed as a $\chi^2(\nu)$ random variable for $\nu = \dim \Theta_1 - \dim \Theta_0$, where $\Theta_j$ is the parameter space of $H_j$, $j=0,1$.
  • Furthermore, wouldn't this test give $\nu = 0$ since we are estimating 1 parameter in both $H_0$ and $H_1$ (the MLE $t/n$)? To me it seems like this question is not suitable for Wilks' theorem, in which case I'm not sure how to test these hypotheses.