Haldane's prior Beta(0,0) - Part 1 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
 A: 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.
A: 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.
