I was reviewing the section of Andrew Gelman's "Bayesian Data Analysis" on uninformative priors, and came across this explanation for why Jeffreys' prior is invariant to parameterization.

Bayesian Data Analysis, 3e p. 53

My question is simply how Gelman reasoned from the first line of the equation to the second.


So using the chain rule, is this the correct reasoning? If we take $\theta(\phi)$ as a function of $\phi$, then

$$ \frac{d^2\log p(y \mid \phi)}{d\phi^2} = \frac{d}{d\phi} \left( \frac{d \log p(y\mid\theta(\phi))}{d \theta} \frac{d\theta}{d\phi} \right) = \frac{d^2 \log p(y\mid\theta(\phi))}{d \theta^2} \left|\frac{d\theta}{d\phi} \right|^2 $$

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Chain_rule $\endgroup$
    – whuber
    Commented Feb 11, 2017 at 20:18
  • 2
    $\begingroup$ I agree with William Huber. It simply amounts to the chain rule of calculus. $\endgroup$ Commented Feb 11, 2017 at 20:36
  • $\begingroup$ Thanks both, I got myself confused because it was to the right of the conditional. Added an updated explanation in an edit. $\endgroup$
    – ezhao15
    Commented Feb 11, 2017 at 20:39
  • $\begingroup$ This explanation is of course restricted to the unidimensional case. $\endgroup$
    – Xi'an
    Commented Feb 12, 2017 at 18:12

1 Answer 1


Regarding your edit, that's not right. You also need the product rule:

\begin{align*} \frac{d^2\log p(y | \phi)}{d\phi^2} &= \frac{d}{d\phi} \left( \frac{d \log p(y|\theta(\phi))}{d \theta} \frac{d\theta}{d\phi} \right) \tag{chain rule}\\ &= \left(\frac{d^2 \log p(y|\theta(\phi))}{d \theta d\phi}\right)\left( \frac{d\theta}{d\phi}\right) + \left(\frac{d \log p(y|\theta(\phi))}{d \theta}\right) \left( \frac{d^2\theta}{d\phi^2}\right) \tag{prod. rule} \\ &= \left(\frac{d^2 \log p(y|\theta(\phi))}{d \theta^2 }\right)\left( \frac{d\theta}{d\phi}\right)^2 + \left(\frac{d \log p(y|\theta(\phi))}{d \theta}\right) \left( \frac{d^2\theta}{d\phi^2}\right) \tag{chain rule} \end{align*} Then $\frac{d \log p(y|\theta(\phi))}{d \theta}$ (the "score function") is $0$ on average.


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