# Frequentist reasoning and conditioning on observations (example from Wagenmakers et al.)

I'm not an expert in statistics, but I gather there is disagreement whether a "frequentist" or "Bayesian" interpretation of probability is the "right" one. From Wagenmakers et. al p. 183:

Consider a uniform distribution with mean $\mu$ and width $1$. Draw two values randomly from this distribution, label the smallest one $s$ and the largest one $l$, and check whether the mean $\mu$ lies in between $s$ and $l$. If this procedure is repeated very many times, the mean $\mu$ will lie in between $s$ and $l$ in half of the cases. Thus, $(s, l)$ gives a 50% frequentist confidence interval for $\mu$. But suppose that for a particular draw, $s = 9.8$ and $l = 10.7$. The difference between these values is $0.9$, and this covers 9/10th of the range of the distribution. Hence, for these particular values of $s$ and $l$ we can be 100% confident that $s < \mu < l$, even though the frequentist confidence interval would have you believe you should only be 50% confident.

Are there really people who believe that there is only 50% confidence in this case or is it a straw man?

I guess more generally, the book seems to be saying that frequentists can't express a conditional claim like "Given $s = 9.8$ and $l = 10.7$, $s<\mu<l$ with probability 1". Is it true that conditioning implies Bayesian reasoning?

• All three of the current answers are very good. I would add only that Wagenmakers is making a strawman argument in the sense that no frequentist statistician would ever recommend this confidence interval -- it exists in the literature only as an example of a pathological confidence interval. From a frequentist point of view, it demonstrates that confidence coverage alone is not sufficient for good inference. (I'm a Bayesian.) – Cyan Apr 12 '12 at 1:35

There's some intricate cheating involved. The confidence interval $(s,l)$ does not use the information that the range of the uniform is 1, and is thus non-parametric, while the claim made about the sample with $l-s=0.9$ does, and is highly model-dependent. I am pretty sure one can improve either the coverage or the (expected) length of the confidence interval if this information is taken into account. For one thing, the end points of the distribution are at most $1-(l-s)$ away from either $s$ or $l$. Hence, a 100% confidence interval for $\mu$ is $(l-1/2, s+1/2)$.

This particular problem falls into the domain of inference for partially identified distributions studied in the last 10-15 years extensively in theoretical econometrics. Likelihood, and hence Bayesian, inference for the uniform distribution is ugly, since it constitutes an non-regular problem (the support of the distribution depends on the unknown parameter).

• I doubt you can bring the expected length down below $\frac{1}{3}$ for a 50% confidence interval on a sample of 2 items. – Henry Apr 13 '12 at 0:02

I'm hesitant to answer this. These Frequentist vs. Bayesian spats are generally unproductive, and can be nasty and juvenile. For what it's worth, Wagenmakers is kind of a big deal, whereas largely forgotten 3k+ year old Chinese philosophers on the other hand...

However, I would argue that the standard Frequentist interpretation of a 50% confidence interval is not that you should be 50% confident the true value lies within the interval, or that there is a 50% probability that it does. Rather, the idea is simply that, if the same process were repeated indefinitely, the percentage of CI's that included the true value would converge to 50%. For any given single interval, however, the probability that it includes the true value is either 0 or 1, but you don't know which.

I think it is a weak argument for a strong case.

$(s,l)$ may be a 50% confidence interval in the sense defined, but so too is $\left(\dfrac{3l+s-1}{4},\dfrac{3s+l+1}{4}\right)$, and I think the latter can be justified as being a better one in these circumstances, as it extends without further adjustment to larger sample sizes; note also that latter confidence interval is never wider than $\frac12$ and its expected width for a sample of size $n$ is $\frac{1}{n+1}$.

• In the quoted example of the sample $\{9.8, 10.7\}$, my suggested alternative would give the $50\%$ confidence interval $[10.225,10.275]$, which is clearly the middle half of the logical $100\%$ confidence interval $[10.2,10.3]$ – Henry Dec 7 '16 at 9:10