I've posted a fair amount about confidence intervals for various quantities. All of the ones I've posted so far - central limit based theorem intervals, Wald theory intervals, and likelihood intervals - are based on a frequentist understanding of probability - that is to say, probability is defined as the limiting result of the proportion of times an event (say, a hit) happens as the number of trials (say, at-bats) goes to infinity.
The term "95% confidence" refers to the construction of the interval itself - that is, if we were to calculate a 95% confidence interval for the true batting average $\theta$ for each of our millions and millions of trials, then 95% of them will contain the true batting average $\theta$.
Statisticians tend to avoid making probability statements about confidence intervals. The statement that "There is a 95% chance that $\theta$ is in the interval." is incorrect because $\theta$ is conceptualized as a fixed quantity. Furthermore, the statement "There is a 95% chance that the interval contains $\theta$." is awkward because once an interval has been calculated, there's no more randomness anymore - it either contains $\theta$, or it doesn't. This is why statisticians prefer to use confidence rather than probability to describe intervals.
But what if you are working in a Bayesian framework? The end result of a Bayesian analysis is a distribution $p(\theta | x)$ that represents the distribution of believe in $\theta$ after the data has been accounted for - so it makes perfect sense to write, for example, $P(0.250 \le \theta \le 0.300)$.
All the code used to generate the images in this article may be found on my github.
Credible Intervals
Instead of confidence intervals, Bayesian statisticians instead calculate "Credible" intervals - these are intervals from the distribution of $p(\theta | x)$ that contain the desired amount of probability. Say, for example, that $P(0.250 \le \theta \le 0.300) = 0.95$ - then the interval $(0.250, 0.300)$ would be a 95% credible interval for $\theta$.
The main issue with this method is that there is more than one way to get a 95% credible interval given $p(\theta | x)$ - technically, any interval $(L, U)$ with $P(L \le \theta \le U) = 0.95$ is a valid 95% credible interval. Statisticians have several ways to determine which one to use, but I'm going to show you one that can be easily done with most computer software.
Baseball Example
In the previous post, I explained how to use the beta-binomial model to get a posterior distribution for a batting average using a beta-binomial model. Let's take observer A, who for the batter with $15$ hits in $n = 50$ at-bats had a prior distribution of belief given by a beta distribution with parameters $\alpha = 1$ and $\beta = 1$
And a posterior distribution of belief given by a beta distribution with parameters $\alpha' = 16$ and $\beta' = 36$
To get a credible interval, we can take quantiles from the beta distribution. A quantile is a value $Q_{p}$ for a distribution so that $P(X \le Q_{p}) = p$. To get a 95% credible interval, we can take $Q_{0.025}$ as the lower boundary of the interval and $Q_{0.975}$ as the upper boundary of the interval (since 97.5% - 2.5% = 95%) so that the interval contains the middle 95% of the probability.
Since these do not have nice formulas to calculate by hand, it's easiest to use computer software to get the - any good statistical software should be able to give quantiles for common distributions. In the program $R$, the command to do this is
> qbeta(c(.025,.975),16,36)
[1] 0.1911040 0.4382887
So for observer A, a 95% credible interval for $\theta$ is given by $(0.191, 0.438)$. With quantiles, the posterior belief for observer A looks like
The area under the curve between the two vertical lines is 0.95 - and so the values of the vertical lines give the 95% credible interval.
What about observer B? Observer B used a beta distribution as their prior with $\alpha = 53$ and $\beta = 147$, for a posterior distribution that is beta with $\alpha' = 68$ and $\beta' = 182$. Quantiles from observer B's posterior distribution are
> qbeta(c(.025,.975),68,182)
[1] 0.2187438 0.3287111
So observer B's 95% credible interval for $\theta$ is $(0.219, 0.329)$. Note that observer B's interval is much more realistic - as baseball fans, we know that a $\theta = 0.400$ batting average is very, very unlikely - so good prior information can lead to improved inference.
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