conjugate of a binomial

probability that a vaccinated person is shielded from COVID-19?

Over my flight to Montpellier last week, I read an arXival on a Bayesian analysis of the vaccine efficiency. Whose full title is “What is the probability that a vaccinated person is shielded from Covid-19? A Bayesian MCMC based reanalysis of published data with emphasis on what should be reported as `efficacy'”, by Giulio D’Agostini and Alfredo Esposito. In short I was not particularly impressed.

“But the real point we wish to highlight, given the spread of distributions, is that we do not have enough data for drawing sound conclusion.”

The reason for this lack of enthusiasm on my side is that, while the authors’ criticism of an excessive precision in Pfizer, Moderna, or AstraZeneca press releases is appropriate, given the published confidence intervals are not claiming the same precision, a Bayesian reanalysis of the published outcome of their respective vaccine trial outcomes does not show much, simply because there is awfully little data, essentially two to four Binomial-like outcomes. Without further data, the modelling is one of a simple graph of Binomial observations, with two or three probability parameters, which results in a very standard Bayesian analysis that does depend on the modelling choices being made, from a highly unrealistic assumption of homogeneity throughout the population(s) tested for the vaccine(s), to a lack of hyperparameters that could have been shared between vaccinated populations. Parts of the arXival are unrelated and unnecessary, like the highly detailed MCMC algorithm for simulating the posterior (incl. JAGS code) to the reminiscence of Bayes’ and Laplace’s early rendering of inverse probability. (I find both interesting and revealing that arXiv, just like medRxiv, posts a warning on top of COVID related preprints.)

Bernoulli factory in the Riddler

“Mathematician John von Neumann is credited with figuring out how to take a p biased coin and “simulate” a fair coin. Simply flip the coin twice. If it comes up heads both times or tails both times, then flip it twice again. Eventually, you’ll get two different flips — either a heads and then a tails, or a tails and then a heads, with each of these two cases equally likely. Once you get two different flips, you can call the second of those flips the outcome of your “simulation.” For any value of p between zero and one, this procedure will always return heads half the time and tails half the time. This is pretty remarkable! But there’s a downside to von Neumann’s approach — you don’t know how long the simulation will last.” The Riddler

The associated riddle (first one of the post-T era!) is to figure out what are the values of p for which an algorithm can be derived for simulating a fair coin in at most three flips. In one single flip, p=½ sounds like the unique solution. For two flips, p²,(1-p)^2,2p(1-p)=½ work, but so do p+(1-p)p,(1-p)+p(1-p)=½, and the number of cases grows for three flips at most. However, since we can have 2³=8 different sequences, there are 2⁸ ways to aggregate these events and thus at most 2⁸ resulting probabilities (including 0 and 1). Running a quick R code and checking for proximity to ½ of any of these sums leads to

[1] 0.2062997 0.7937005 #p^3
[1] 0.2113249 0.7886753 #p^3+(1-p)^3
[1] 0.2281555 0.7718448 #p^3+p(1-p)^2
[1] 0.2372862 0.7627143 #p^3+(1-p)^3+p(1-p)^2
[1] 0.2653019 0.7346988 #p^3+2p(1-p)^2
[1] 0.2928933 0.7071078 #p^2
[1] 0.3154489 0.6845518 #p^3+2p^2(1-p)
[1] 0.352201  0.6477993 #p^3+p(1-p)^2+p^2(1-p)
[1] 0.4030316 0.5969686 #p^3+p(1-p)^2+3(1-p)p^2
[1] 0.5

which correspond to

1-p³=½, p³+(1-p)³=½,(1-p)³+(1-p)p²=½,p³+(1-p)³+p²(1-p),(1-p)³+2(1-p)p²=½,1-p²=½, p³+(1-p)³+p²(1-p)=½,(1-p)³+p(1-p)²+p²(1-p)=½,(1-p)³+p²(1-p)+3p(1-p)²=½,p³+p(1-p)²+3(p²(1-p)=½,p³+2p(1-p)²+3(1-p)p²=½,p=½,

(plus the symmetric ones), leading to 19 different values of p producing a “fair coin”. Missing any other combination?!

Another way to look at the problem is to find all roots of the equations

where

(None of these solutions is rational, by the way, except p=½.) I also tried this route with a slightly longer R code, calling polyroot, and finding the same 19 roots for three flips, [at least] 271 for four, and [at least] 8641 for five (The Riddler says 8635!). With an imprecision in the exact number of roots due to rather poor numerical rounding by polyroot. (Since the coefficients of the above are not directly providing those of the polynomial, I went through an alternate representation as a polynomial in (1-p)/p, with a straightforward derivation of the coefficients.)

unbiased estimators that do not exist

When looking at questions on X validated, I came across this seemingly obvious request for an unbiased estimator of P(X=k), when X~B(n,p). Except that X is not observed but only Y~B(s,p) with s<n. Since P(X=k) is a polynomial in p, I was expecting such an unbiased estimator to exist. But it does not, for the reasons that Y only takes s+1 values and that any function of Y, including the MLE of P(X=k), has an expectation involving monomials in p of power s at most. It is actually straightforward to establish properly that the unbiased estimator does not exist. But this remains an interesting additional example of the rarity of the existence of unbiased estimators, to be saved until a future mathematical statistics exam!

an accurate variance approximation

In answering a simple question on X validated about producing Monte Carlo estimates of the variance of estimators of exp(-θ) in a Poisson model, I wanted to illustrate the accuracy of these estimates against the theoretical values. While one case was easy, since the estimator was a Binomial B(n,exp(-θ)) variate [in yellow on the graph], the other one being the exponential of the negative of the Poisson sample average did not enjoy a closed-form variance and I instead used a first order (δ-method) approximation for this variance which ended up working surprisingly well [in brown] given that the experiment is based on an n=20 sample size.

Thanks to the comments of George Henry, I stand corrected: the variance of the exponential version is easily manageable with two lines of summation! As

which allows for a comparison with its second order Taylor approximation: