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In my first post, I described the general logic of Bayes factors. I will continue discussing the general logic of Bayes factor, while introducing some of the basic functionality of the BayesFactor package.

### Recap: What is a Bayes factor?

The Bayes factor is the relative evidence provided by the data for comparing two statistical models. It has two equivalent definitions:
1. The Bayes factor is the relative predictive success between two hypotheses: it is the ratio of the probabilities of the observed data under each of the hypotheses. If the probability of the observed data is higher under one hypothesis than another, then that hypothesis is preferred.
2. The Bayes factor is evidence: it is the factor by which the relative beliefs must be multiplied after the data have been observed, providing new (posterior) relative beliefs. The stronger the evidence, the more beliefs must change.

### Student’s sleep data

Student (1908) describes a study by Cushny and Peebles (1905) on the effect of several medications on sleep. One of the drugs they studied was Laevo-hyoscine hydrobromide, which we’ll abbreviate as “LH”. LH was administered to 10 patients on several nights, and the average amount of sleep over the patients’ unmedicated sleep (measured on other nights) was recorded. For the purposes of this post, we will make the typical assumptions about the data that would be made if we were to apply the $$t$$ procedure: specifically, that the observation are independent and normally distributed.
We return to Carole and Paul, who have another disagreement. Carole does not believe that LH increases sleep at all, while Paul claims that it does. This would traditionally be handled by a t test. Under Carole’s hypothesis, the $$t$$ statistic has a Student’s $$t$$ distribution with $$N-1$$ degrees of freedom. That is, $t = \frac{\bar{y}}{s}\sqrt{N} \sim \mbox{Student’s }~t_{N-1}$
Another way of stating her prediction is in terms of the observed effect size, $$\hat{\delta}=\bar{y}/s$$, which is shown in the figure below. It is simply a Student’s $$t$$ distribution, scaled by $$1/\sqrt{N}$$.

Paul’s hypothesis — that LH causes an increase in sleep — is problematic. Typically, we require scientific hypotheses to constrain the data in some way. We require them to predict that some outcomes are more plausible than others. Paul’s hypothesis, however, predicts nothing. His hypothesis does not constrain the data.
This highlights a curious fact about traditional significance testing. The only hypothesis we can support using a classical significance test is the one that makes no predictions! If we applied the traditional t test to these data, we could only reject Carole’s hypothesis in favor of Paul’s — but never support Carole’s — even though Paul has advanced a decidedly unfalsifiable hypothesis.
To be testable, Paul’s hypothesis must make predictions, as Carole’s has. As a first step, let’s assume that Paul believes that LH will increase sleep by 1 standardized effect size unit. If $$\mu$$ is the mean increase in hours of sleep and $$\sigma$$ is the standard deviation of the increases, then the standardized increase is $\delta = \frac{\mu}{\sigma}.$ and Paul believes that $$\delta=1$$. Using the noncentral t distribution, we can obtain Paul’s predictions for the observed effect size $$\hat{\delta}$$, shown in the figure below:

As expected, the predictions are centered around $$\hat{\delta}=1$$. We now have two hypothesis: Carole believes that $$\delta=0$$, and Paul believes that $$\delta=1$$. These hypotheses imply two different predictions for the observed effect size, as shown above. We can now compare the predictions with the results presented by Student to see whether Paul or Carole is favored by the evidence.
The sleep data set is built into R; the following R code will extract the data for LH (which is labelled group 2):
improvements = with(sleep, extra[group == 2])
improvements

##  [1]  1.9  0.8  1.1  0.1 -0.1  4.4  5.5  1.6  4.6  3.4

The improvements in hours of sleep range from -0.1 hours (the patient slept less when medicated by LH) and an improvement of 5.5 hours. The mean improvement was 2.33 hours. The figure below shows the hours of improvement for all 10 patients.

We begin our analysis by computing $$N$$ and the $$t$$ statistic from the data:
N = length(improvements)
t = mean(improvements)/sd(improvements) * sqrt(N)
t

## [1] 3.68

Typically, at this point we would look up the $$p$$ value for this $$t$$ statistic; but for a Bayes factor analysis, we use it to compute the relative evidence for the two hypotheses. Of interest is the effect size:
deltaHat = t/sqrt(N)
deltaHat

## [1] 1.164

The effect size $$\hat{\delta}=1.1637$$ seems to favor Paul, since Paul’s hypothesis ($$\delta=1$$) was actually quite close to the observed result . In order to know by how much the result favors Paul, we need to look at the predictions and see where the observation falls.
The figure below shows both Carole’s and Paul’s predictions for the observed effect size as overlapping gray distributions. The observed effect size is denoted by the vertical line segment. In order to compute the Bayes factor, we must compare the relative heights of the two distributions for the observed data. The probability density of the observed data is 67.9451 times higher under Paul’s hypothesis than under Carole’s. The Bayes factor favoring Paul is thus 67.9451.

We can easily compute this using R’s built in functions. Under Carole’s hypothesis, the $$t$$ statistic has a central $$t$$ distribution with 9 degrees of freedom; Under Paul’s hypothesis, the $$t$$ statistic has a noncentral $$t$$ distribution with 9 degrees of freedom, and a noncentrality parameter of $$\delta\sqrt{N}=1\times\sqrt{10}$$. The ratio of the two densities, at the observed $$t$$ statistic, yields the Bayes factor.
dt(t, df = 9, ncp = 1 * sqrt(10))/dt(t, df = 9)

## [1] 67.95


### Uncertain hypotheses, and a first look at BayesFactor

In the section above Carole’s and Paul’s original hypotheses, when stated in terms of effect sizes, were \begin{align*} {\cal H}_c : \delta &= 0\\ {\cal H}_p : \delta &> 0; \end{align*} that is, Paul hypothesized that the drug increases sleep on average, and Carole hypothesized that it had no effect. The problem with this was that Paul’s hypothesis was untestable, because it does not lead to any proper constraint on the data.
The solution is for Paul to propose a hypothesis that does constrain data. We chose the hypothesis $$\delta=1$$ for Paul, and showed how that can be used to compute a Bayes factor. The resulting Bayes factor, however, does not seem to capture very well the spirit of Paul’s original hypothesis. Paul was not specific about what the exact effect size was. But $$\delta>0$$ won’t do either; it is so unspecific as to be completely unfalsifiable.
What we need is a middle ground between the two. In the previous post I showed how a Bayesian analysis can spread uncertainty out over a range of parameter values using a probability distribution.
The figure below shows two hypotheses: Carole’s null hypothesis ($$\delta=0$$) whose predictions I’ve already presented, and a plausible hypothesis for Paul ($$\delta>0$$). Under Paul’s hypothesis, plausible effect sizes are not spread evenly across the positive effect sizes, because that would not lead to any constraint on the data. Instead, small effect sizes are preferred to large ones. It is implausible for the effect size to be too large, so the plausibility of the effects diminish as the effect sizes get larger.

The shape of this prior distribution is the positive half of a Cauchy distribution, or a Student’s $$t$$ distribution with 1 degree of freedom. The Cauchy prior was suggested by Rouder et al. (2009), and is the one implemented in the BayesFactor package.
Of course, this half-Cauchy distribution is not the only way we could implement Paul’s hypothesis. We could choose a different shape of distribution; we could make it wider or narrower around 0; we could even restrict the range further. We must choose some valid probability distribution in order to create a hypothesis that constrains the data, and we should do our best to make the test meaningful by choosing a defensible distribution.
Recall from the previous post that the way we obtain predictions from an uncertain hypothesis like Paul’s, we weight the predictions for each possible $$\delta$$ by their plausibility and average them. This is represented by the integral: $P(\hat{\delta}\mid{\cal H}_p) = \int P(\hat{\delta}\mid \delta)p(\delta)\,d\delta,$ where $$P(\hat{\delta}\mid \delta)$$ gives the predictions for a specific $$\delta$$, and $$p(\delta)$$ is the distribution in the figure above, giving the weightings for all possible values of $$\delta$$.
The figure below shows Carole’s predictions (which are the same as before) and Paul’s new predictions for the observed effect size. Notice that Paul’s predictions are now substantially more spread out than before. The uncertainty Paul has about the true value of $$\delta$$, added to the uncertainty that is inherent in the sampling given a particular true value, has caused his predictions to be less certain. This spreading out is the penalty for a having a less-specific hypothesis.

Now that we have predictions from both Carole’s and Paul’s hypothesis, we can compute a Bayes factor. The Bayes factor is the ratio of the heights at the observed $$\hat{\delta}$$ value, shown in the figure below by the vertical line segment. The Bayes factor is 21.3275 in favor of Paul, because the probability density of the observed data is 21.3275 times greater under Paul’s hypothesis than under Carole’s. Note that this is substantially lower than the Bayes factor in favor of Paul’s hypothesis when it was more specific. More flexible hypotheses, like Paul’s, are automatically penalized by the Bayes factor.

As previously mentioned, the distribution used as Paul’s hypothesis happens to be the one implemented in the BayesFactor package. We can thus use the function ttestBF, part of the package’s suite of functions, to compute the Bayes factor for the Student sleep data.
If you have not installed the BayesFactor package yet, install the package first (help here). If you define the improvements vector as above, you can run the following two lines of code to load the package and perform the a Bayes factor $$t$$ test. The nullInterval argument tells ttestBF that you want to consider that range as a hypothesis. Paul’s hypothesis ranged from 0 to $$\infty$$, so we specify that range using the nullInterval argument:
library(BayesFactor)
ttestBF(improvements, nullInterval = c(0, Inf))

## Bayes factor analysis
## --------------
## [1] Alt., r=0.707 0<d<Inf    : 21.33  ±0.01%
## [2] Alt., r=0.707 !(0<d<Inf) : 0.1036 ±1.4%
##
## Against denominator:
##   Null, mu = 0
## ---
## Bayes factor type: BFoneSample, JZS