Lindley's paradox

Description of the paradox

Consider the result x of some experiment, with two possible explanations, hypotheses H 0 and H 1 , and some prior distribution π representing uncertainty as to which hypothesis is more accurate before taking into account x .

Lindley's paradox occurs when

1. The result x is "significant" by a frequentist test of H 0 , indicating sufficient evidence to reject H 0 , say, at the 5% level, and
2. The posterior probability of H 0 given x is high, indicating strong evidence that H 0 is in better agreement with x than H 1 .

These results can occur at the same time when H 0 is very specific, H 1 more diffuse, and the prior distribution does not strongly favor one or the other, as seen below.

Numerical example

We can illustrate Lindley's paradox with a numerical example. Imagine a certain city where 49,581 boys and 48,870 girls have been born over a certain time period. The observed proportion x of male births is thus 49,581/98,451 ≈ 0.5036. We assume the number of male births is a binomial variable with parameter θ . We are interested in testing whether θ is 0.5 or some other value. That is, our null hypothesis is H 0 : θ = 0.5 and the alternative is H 1 : θ ≠ 0.5 .

Frequentist approach

The frequentist approach to testing H 0 is to compute a p-value, the probability of observing a fraction of boys at least as large as x assuming H 0 is true. Because the number of births is very large, we can use a normal approximation for the fraction of male births X ∼ N ( μ , σ 2 ) , with μ = n p = n θ = 98 , 451 × 0.5 = 49 , 225.5 and σ 2 = n θ ( 1 − θ ) = 98 , 451 × 0.5 × 0.5 = 24 , 612.75 , to compute

We would have been equally surprised if we had seen 49,581 female births, i.e. x ≈ 0.4964 , so a frequentist would usually perform a two-sided test, for which the p-value would be p ≈ 2 × 0.0117 = 0.0235 . In both cases, the p-value is lower than the significance level, α, of 5%, so the frequentist approach rejects H 0 as it disagrees with the observed data.

Bayesian approach

Assuming no reason to favor one hypothesis over the other, the Bayesian approach would be to assign prior probabilities π ( H 0 ) = π ( H 1 ) = 0.5 and a uniform distribution to θ under H 1 , and then to compute the posterior probability of H 0 using Bayes' theorem,

After observing k = 49 , 581 boys out of n = 98 , 451 births, we can compute the posterior probability of each hypothesis using the probability mass function for a binomial variable,

where B ( a , b ) is the Beta function.

From these values, we find the posterior probability of P ( H 0 ∣ k ) ≈ 0.95 , which strongly favors H 0 over H 1 .

The two approaches—the Bayesian and the frequentist—appear to be in conflict, and this is the "paradox".

Reconciling the Bayesian and frequentist approaches

However, at least in Lindley's example, if we take a sequence of significance levels, α_n, such that α_n=n^−k with k>½, then the posterior probability of the null converges to 0, which is consistent with a rejection of the null. In this numerical example, taking k=½, results in a significance level of 0.00318, so the frequentist would not reject the null hypothesis, which is roughly in agreement with the Bayesian approach.

If one uses an uninformative prior and tests a hypothesis more similar to that in the frequentist approach, the paradox disappears.

For example, if we calculate the posterior distribution P ( θ ∣ x , n ) , using a uniform prior distribution on θ (i.e., π ( θ ∈ [ 0 , 1 ] ) = 1 ), we find

If we use this to check the probability that a newborn is more likely to be a boy than a girl, i.e., P ( θ > 0.5 ∣ k , n ) , we find

In other words, it is very likely that the proportion of male births is above 0.5.

Neither analysis gives an estimate of the effect size, directly, but both could be used to determine, for instance, if the fraction of boy births is likely to be above some particular threshold.

The lack of an actual paradox

The apparent disagreement between the two approaches is caused by a combination of factors. First, the frequentist approach above tests H 0 without reference to H 1 . The Bayesian approach evaluates H 0 as an alternative to H 1 , and finds the first to be in better agreement with the observations. This is because the latter hypothesis is much more diffuse, as θ can be anywhere in [ 0 , 1 ] , which results in it having a very low posterior probability. To understand why, it is helpful to consider the two hypotheses as generators of the observations:

Most of the possible values for θ under H 1 are very poorly supported by the observations. In essence, the apparent disagreement between the methods is not a disagreement at all, but rather two different statements about how the hypotheses relate to the data:

The ratio of the sex of newborns is improbably 50/50 male/female, according to the frequentist test. Yet 50/50 is a better approximation than most, but not all, other ratios. The hypothesis θ ≈ 0.504 would have fit the observation much better than almost all other ratios, including θ ≈ 0.500 .

For example, this choice of hypotheses and prior probabilities implies the statement: "if θ > 0.49 and θ < 0.51, then the prior probability of θ being exactly 0.5 is 0.50/0.51 ≈ 98%." Given such a strong preference for θ = 0.5 , it is easy to see why the Bayesian approach favors H 0 in the face of x ≈ 0.5036 , even though the observed value of x lies 2.28 σ away from 0.5. The deviation of over 2 sigma from H 0 is considered significant in the frequentist approach, but its significance is overruled by the prior in the Bayesian approach.

Looking at it another way, we can see that the prior distribution is essentially flat with a delta function at θ = 0.5 . Clearly this is dubious. In fact if you were to picture real numbers as being continuous, then it would be more logical to assume that it would impossible for any given number to be exactly the parameter value, i.e., we should assume P(theta = 0.5) = 0.

A more realistic distribution for θ in the alternative hypothesis produces a less surprising result for the posterior of H 0 . For example, if we replace H 1 with H 2 : θ = x , i.e., the maximum likelihood estimate for θ , the posterior probability of H 0 would be only 0.07 compared to 0.93 for H 2 (Of course, one cannot actually use the MLE as part of a prior distribution).

Recent discussion

The paradox continues to be a source of active discussion.