# Are random variables sampled upon stopping rules exchangeable?

the author uses an example to introduce some limits of p-values in frequentists analyses.

He takes an example where two different sampling schemes lead to different inferences even if the same data are collected:

1. A sampling of 10 Bernoulli, resulting in 8 sucesses and 2 failures
2. A sampling of Bernoulli until 8 sucesses are obtained, resulting in 10 draws with 8 sucesses and 2 failures

Letting $$\theta$$ be the probability of success the likelihood are the same for both experiments : $$\theta^8(1-\theta)^2$$

When testing $$\theta = \frac{1}{2}$$ versus $$\theta > \frac{1}{2}$$ those sampling schemes lead to different p-values. Since p-values can be stated as $$P(T(X) \ \text{"more extreme than"} \ T(X_{obs}) \mid H_0)$$ this lead to different definitions of "more extreme observations" base on the sampling scheme used and to different p-values.

However, even if this shows a limit of usual frenquentist analyses regarding the Likelihood principle my question is: are Bayesian analyses still relevant under the second sampling scheme? It seems to me that those observations are not exchangeable in this case since observing 8 sucesses in a row will prevent from observing the 2 remaining values.

• In the second case, the number of failures is the random variable, so you can observe $0, 1, 2, \dots$. If you observe $2$, this is what goes into the likelihood function that is used in the Bayesian analysis. If you observe $0$, this is what goes into the likelihood function that is used in the Bayesian analysis. In either case, or any of the others, you can still do the Bayesian analysis. Mar 23, 2019 at 3:38