Let $X \sim Bin(n, \pi)$ with $\pi \in (0,1)$ and known $n$.

Derive Jeffreys' prior for $\pi$ in the following parametrizations:

a) Original parametrization $\pi$

b) Parametrization with $\phi = \text{logit}(\pi)$

c) Compute the posteriors for $\pi$ and $\phi$.

What I have done:

Finding Jeffreys' prior ($f(\pi) \propto \sqrt{J(\pi)}$) for $\pi$ and finding the prior is fairly straightforward I got

$$f(\pi) \propto \left(\frac{1}{\pi(1-\pi)} \right)^{1/2}$$ Which gives to following posterior

$$f(\pi|x) \propto L(\pi)\cdot f(\pi) \propto \pi^x(1-\pi)^{n-x}\cdot \left(\frac{1}{\pi(1-\pi)} \right)^{1/2} = \pi^{x-1/2}(1-\pi)^{n-x-1/2}$$

I'm fairly sure this is correct what I'm uncertain about is $\phi$. I know that the Jeffreys prior is invariant under one-to-one reparametrizations and so is the likelihood function. Does this mean that for $\phi = \text{logit}(\pi)$ i could simply take

$$f(\phi) \propto \left(\frac{1}{\pi(1-\phi)} \right)^{1/2} \qquad L(\phi) = \binom{n}{x}\phi^x(1-\phi)^{n-x}$$

i.e just replace $\pi$ with $\phi$ and not needing to work with the $\text{logit}(\cdot)$ function at all.

  • 2
    $\begingroup$ Jeffreys was the family name (surname) of Sir Harold Jeffreys. The forms (a) Jeffreys prior, (b) Jeffreys' prior and (c) Jeffreys's prior are all defensible. Sir Harold himself preferred Jeffreys's to Jeffreys' (personal communication, c. 1976) but Jeffrey's is just a typo. While I am on this topic, note that Richard C. Jeffrey and William H. Jefferys are different people. $\endgroup$
    – Nick Cox
    Nov 29, 2020 at 10:58

1 Answer 1


The "invariance" of Jeffreys' priors should have been called "equivariance" to avoid confusion. It does not mean that it does not change under reparameterisation but rather that the principle remains consistent under reparameterisation, namely,

computing the Jeffreys prior on $\theta$ and deriving the prior on $\varphi=\phi(\theta)$ by the Jacobian formula is equivalent to computing directly the Jeffreys prior on $\varphi$.

This follows from the transformation formula on Fisher's information matrix $$\mathfrak I(\varphi) = \dfrac{\partial \theta}{\partial \varphi^\text{T}}\mathfrak I(\theta(\varphi)) \dfrac{\partial \theta^\text{T}}{\partial \varphi}$$


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