### Bayes's Theorem in the *New York Times*

Siobhan Roberts' article "Thinking Like an Epidemiologist" ran on August 4, 2020 in the Science section of the *Times*. The sub-head reads "Don't worry, a little Bayesian analysis won't hurt you" and in fact the piece is about Bayes's Theorem and its applications, most explicitly to COVID testing.

Bayes's Theorem concernsRoberts: "Take diagnostic testing. In this scenario, the setup of Bayes's theorem might use events labeled 'T' for a positive test result and 'C' for the presence of Covid-19 antibodies:relative probability:$P(A)$ represents the probability of (condition, event, ...) $A$, whereas $P(A|B)$ represents the probability of $A$given that(condition, event, ...) $B$ (holds, has occurred, ...). Bayes's Theorem relates $P(A|B)$ to $P(B|A)$. This may seem counterintuitive but it follows directly from the common-sense equation $P(A ~\mbox{and}~ B)= $$~P(B)\times P(A|B) = $$~P(A)\times P(B|A)$: $$\mbox{Bayes's Theorem:}~~~ P(A|B)=\frac{P(B|A)\times P(A)}{P(B)}.$$

$$P(C|T)=\frac{P(T|C)\times P(C)}{P(T)}.$$Now suppose the prevalence of cases is 10 percent (that was so in New York City in the spring), and you have a positive result from a test with accuracy of 87.5 percent sensitivity and 97.5 percent specificity. Running numbers through the Bayesian gears, the probability that the result is correct, and that you do indeed have antibodies is 79.5%."

Here's how "running the numbers ..." works. The $87.5\%$ sensitivity means that $P(T|C)=.875$. The $97.5\%$ specificity means that the probability of aThe diagnostic test example can serve to illustrate how the article's opening statements are actually related to Bayes's Theorem. Roberts starts: "There is a statistician's rejoinder . . . that could hardly be a better motto for our times: 'Update your priors!' In stats lingo, 'priors' are your prior knowledge and beliefs, inevitably fuzzy and uncertain, before seeing evidence. Evidence prompts an updating; and then more evidence prompts further updating, so forth and so on. This iterative process hones greater certainty and generates a coherent accumulation of knowledge."false positive,i.e. $P(T|\mbox{not}~C)$, is $1-.975 = .025$. This allows us to calculate $P(T) $$= P(T|C)\times P(C)+P(T|\mbox{not}~C)\times P(\mbox{not}~C) =$$ .875\times .1 +.025\times .9=.11.$ Then $$P(C|T)=\frac{P(T|C)\times P(C)}{P(T)} = \frac{.875\times .1}{.11} = .795.$$

In the example, before you take the test you have "prior" probability $P(C)=.1$ of having COVID antibodies. Evidence comes from the test result $T$. Given that result, you mustFor the print version of this article theupdatethat probability to $P(C|T)=.795$.

*Times*'s typesetters left-justified Roberts's equations, which became, unfortunately, unintelligible. This has been rectified online.