As I mentioned in other blogs, we can still use a classically derived test known as the generalized loglikelihood ratio as a way of simply ranking different AB combinations against each other according to how interesting they are. Even without being able to interpret the statistical score as a statistical test, we get useful results in practice.
The generalized loglikelihood ratio most commonly used in these situations is derived assuming we have two binomial observations. This test can be extended to compare two multinomial conditions for independence, but this is rarely done, if only because comparing two binomials is so darned useful.
With the binomial test, we look at the number of positive observations out of some total number of observations for each condition. In some situations, it is much more natural to talk about the number of positive observations not as a fraction of all observations, but as a fraction of the duration of the condition. For instance, we might talk about the number of times we noticed a particular kind of network error under different conditions. In such a case, we probably can say how long we looked for the errors under each condition, but it can be very hard to say how many observations there were without an error.
Count  $$\Delta t$$  
$$A$$

$$k_1$$

$$\,\,\,\, t_1 \,\,\,\,$$

$$\neg A$$

$$k_2$$

$$t_2$$

We can investigate whether the Poisson distribution the same under both conditions using the generalized loglikelihood ratio test. Such a test uses $\lambda$, the generalized loglikelihood ratio,
\[
\lambda = \frac{
\max_{\theta_1 = \theta_2 = \theta_0} p(k_1  \theta_0)p(k_2  \theta_0)
}{
\max_{\theta_1, \theta_2} p(k_1  \theta_1)p(k_2  \theta_2)
}
\]
According to Wilks\cite{wilks1938} and also later Chernoff\cite{Chernoff1954}, the quantity $2 \log \lambda$ is asymptotically $\chi^2$ distributed with one degree of freedom.
For the Poisson distribution,
\[
p(k  \theta, t) = \frac{(\theta t)^k e^{\theta t}}{k!} \\
\log p(k\theta, t) = k \log \theta t  \theta t  \log k!
\]
The maximum likelihood estimator $\hat\theta$ can be computed by maximizing the log probability
For the Poisson distribution,
\[
p(k  \theta, t) = \frac{(\theta t)^k e^{\theta t}}{k!} \\
\log p(k\theta, t) = k \log \theta t  \theta t  \log k!
\]
The maximum likelihood estimator $\hat\theta$ can be computed by maximizing the log probability
\[
\max_\theta \frac{(\theta t)^k e^{\theta t}}{k!} = \max_\theta \log \frac{(\theta t)^k e^{\theta t}}{k!} \\
\log \frac{(\theta t)^k e^{\theta t}}{k!} = k \log \theta t  \theta t  \log k! \\
\frac{\partial \log L(k  \theta, t)}{\partial \theta} = \frac{k}{ \theta}  t
=0 \\
\hat \theta = k
\]
Returning to the loglikelihood ratio test, after some cancellation we get
\max_\theta \frac{(\theta t)^k e^{\theta t}}{k!} = \max_\theta \log \frac{(\theta t)^k e^{\theta t}}{k!} \\
\log \frac{(\theta t)^k e^{\theta t}}{k!} = k \log \theta t  \theta t  \log k! \\
\frac{\partial \log L(k  \theta, t)}{\partial \theta} = \frac{k}{ \theta}  t
=0 \\
\hat \theta = k
\]
Returning to the loglikelihood ratio test, after some cancellation we get
\[
\log \lambda =
k_1 \log k_1 +
k_2 \log k_2
 k_1 \log \frac{k_1+k_2}{t_1+t_2} t_1  k_2 \log \frac{k_1+k_2}{t_1+t_2} t_2
\]
\log \lambda =
k_1 \log k_1 +
k_2 \log k_2
 k_1 \log \frac{k_1+k_2}{t_1+t_2} t_1  k_2 \log \frac{k_1+k_2}{t_1+t_2} t_2
\]
Some small rearrangement gives the following preferred form that is very reminiscent of the form most commonly to compute the loglikelihood ratio test for binomials and multinomials
\[
2 \log \lambda = 2 \left( k_1 \log \frac{k_1}{t_1} +
k_2 \log \frac{k_2}{t_2}  (k_1+k_2) \log \frac{k_1+k_2}{t_1+t_2}
\right)
\]
\[
2 \log \lambda = 2 \left( k_1 \log \frac{k_1}{t_1} +
k_2 \log \frac{k_2}{t_2}  (k_1+k_2) \log \frac{k_1+k_2}{t_1+t_2}
\right)
\]