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This article$^1$ on p.16 specifies Haldane's prior as:

$$p(\theta) = \frac{1}{θ(1−θ)}$$.

However, other$^2$ source on p.6 specifies Haldane's prior as proportional to $\frac{1}{θ(1−θ)}$, i.e. $$p(\theta) \propto \frac{1}{θ(1−θ)}$$.

Could anyone clarify which expression is the accurate one.

1. Approximation of improper priors

2. Bayesian Analysis of Some Common Distributions

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    $\begingroup$ This is an improper prior because its integral diverges. Since it is improper, constants of proportionality have no relevance because no normalizing constant exists. $\endgroup$
    – whuber ♦
    Commented Jan 31, 2018 at 21:02

2 Answers 2

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Haldane prior is beta distribution with parameters $\alpha = \beta = 0$. So it is

$$ f(p) = \frac{p^{\alpha-1} (1-p)^{\beta-1}}{B(\alpha, \beta)} = \frac{p^{-1}(1-p)^{-1}}{B(0, 0)} $$

where $B(0, 0)$ is the normalizing constant that is infinite as described in Wikipedia:

The function $p^{-1}(1-p)^{-1}$ can be viewed as the limit of the numerator of the beta distribution as both shape parameters approach zero: $\alpha, \beta \to 0$. The Beta function (in the denominator of the beta distribution) approaches infinity, for both parameters approaching zero, $\alpha, \beta \to 0$. Therefore, $p^{-1}(1-p)^{-1}$ divided by the Beta function approaches a 2-point Bernoulli distribution with equal probability $1/2$ at each Dirac delta function end, at $0$ and $1$, and nothing in between, as $\alpha, \beta \to 0$.

So Haldane prior is not a proper distribution. It is an abstract idea of what would be the beta distribution be if it had $\alpha = \beta = 0$ parameters. As a distribution, it is rather not usable, yet it can be used as an "uninformative" prior for binomial distribution. It is often described in it's approximate form $f(p) \propto p^{-1}(1-p)^{-1}$, since the normalizing constant is meaningless.

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  • $\begingroup$ Thanks @Tim; can it be shown how the Beta function approaches $\infty$ as $\alpha, \beta→0$. $\endgroup$
    – AlexMe
    Commented Jan 31, 2018 at 22:01
  • $\begingroup$ @AlexStat it follows from $\Gamma(0) = \infty$ $\endgroup$
    – Tim
    Commented Jan 31, 2018 at 22:30
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    $\begingroup$ @Tim: in which sense $f(p) \propto p^{-1}(1-p)^{-1}$ is approximate? There is no imprecision in defining the prior without a constant since all choices of a constant lead to the same posterior (if defined) or none( if undefined). $\endgroup$
    – Xi'an
    Commented Mar 2, 2018 at 18:35
  • $\begingroup$ @Xi'an as a distribution it lacks the constant. $\endgroup$
    – Tim
    Commented Mar 2, 2018 at 18:41
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    $\begingroup$ Imho, it is a measure (or the density of a measure against the Lebesgue measure onm $(0,1)$) rather than a distribution. As the notion of distribution is not set outside "probability distributions". $\endgroup$
    – Xi'an
    Commented Mar 2, 2018 at 18:48
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The second expression is correct as this is an improper distribution, i.e. it doesn't integrate to $1$. Thus it doesn't have a density and you can only specify it up to proportionality.

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