## Bernoulli factory in the Riddler

Posted in Books, Kids, R, Statistics with tags , , , , , , , , , , on December 1, 2020 by xi'an

“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 $2^{2^n}$ equations

$a_0p^n+a_1p^{n-1}(1-p)+\cdots+a_{n-1}p(1-p)^{n-1}+a_n(1-p)^n=1/2$

where

$0\le a_i\le{n \choose i}$

(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

Posted in Statistics with tags , , , , , , , on January 21, 2019 by xi'an

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

Posted in Books, Kids, pictures, R, Statistics with tags , , , , , , on February 7, 2017 by xi'an

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

$\text{var}(\exp\{-\bar{X}_n\})=\exp\left\{-n\theta[1-\exp\{-2/n\}]\right\}$

$-\exp\left\{-2n\theta[1-\exp\{-1/n\}]\right\}$

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

## simulating correlated Binomials [another Bernoulli factory]

Posted in Books, Kids, pictures, R, Running, Statistics, University life with tags , , , , , , , on April 21, 2015 by xi'an

This early morning, just before going out for my daily run around The Parc, I checked X validated for new questions and came upon that one. Namely, how to simulate X a Bin(8,2/3) variate and Y a Bin(18,2/3) such that corr(X,Y)=0.5. (No reason or motivation provided for this constraint.) And I thought the following (presumably well-known) resolution, namely to break the two binomials as sums of 8 and 18 Bernoulli variates, respectively, and to use some of those Bernoulli variates as being common to both sums. For this specific set of values (8,18,0.5), since 8×18=12², the solution is 0.5×12=6 common variates. (The probability of success does not matter.) While running, I first thought this was a very artificial problem because of this occurrence of 8×18 being a perfect square, 12², and cor(X,Y)x12 an integer. A wee bit later I realised that all positive values of cor(X,Y) could be achieved by randomisation, i.e., by deciding the identity of a Bernoulli variate in X with a Bernoulli variate in Y with a certain probability ϖ. For negative correlations, one can use the (U,1-U) trick, namely to write both Bernoulli variates as

$X_1=\mathbb{I}(U\le p)\quad Y_1=\mathbb{I}(U\ge 1-p)$

in order to minimise the probability they coincide.

I also checked this result with an R simulation

> z=rbinom(10^8,6,.66)
> y=z+rbinom(10^8,12,.66)
> x=z+rbinom(10^8,2,.66)
cor(x,y)
> cor(x,y)
[1] 0.5000539


Searching on Google gave me immediately a link to Stack Overflow with an earlier solution with the same idea. And a smarter R code.

## prayers and chi-square

Posted in Books, Kids, Statistics, University life with tags , , , , , , on November 25, 2014 by xi'an

One study I spotted in Richard Dawkins’ The God delusion this summer by the lake is a study of the (im)possible impact of prayer over patient’s recovery. As a coincidence, my daughter got this problem in her statistics class of last week (my translation):

1802 patients in 6 US hospitals have been divided into three groups. Members in group A was told that unspecified religious communities would pray for them nominally, while patients in groups B and C did not know if anyone prayed for them. Those in group B had communities praying for them while those in group C did not. After 14 days of prayer, the conditions of the patients were as follows:

• out of 604 patients in group A, the condition of 249 had significantly worsened;
• out of 601 patients in group B, the condition of 289 had significantly worsened;
• out of 597 patients in group C, the condition of 293 had significantly worsened.

Use a chi-square procedure to test the homogeneity between the three groups, a significant impact of prayers, and a placebo effect of prayer.

This may sound a wee bit weird for a school test, but she is in medical school after all so it is a good way to enforce rational thinking while learning about the chi-square test! (Answers: [even though the data is too sparse to clearly support a decision, esp. when using the chi-square test!] homogeneity and placebo effect are acceptable assumptions at level 5%, while the prayer effect is not [if barely].)