In Bayesian Data Analysis, PDF freely available, section 4.1 (page 84, bottom) there is a comment saying:

If we had instead constructed the normal approximation in terms of $p(\mu, \sigma^2)$, the second derivative matrix would be multiplied by the Jacobian of the transformation from $\log\sigma$ to $\sigma^2$ and the mode would change slightly, to $\tilde{\sigma}^2 = \frac{n}{n+2}\hat{\sigma}^2$.

My question is how do we compute the Jacobian from $\log\sigma$ to $\sigma^2$? I can differentiate one function with respect to the other but I can't reconcile the stated change in mode (which makes me think I'm mistaken).

Thanks in advance


1 Answer 1


For the parameter transform$$\eta\longmapsto\sigma^2=\exp\{2\eta\}$$the Jacobian is $$\frac{\text d\sigma^2}{\text d\eta}=2\exp\{2\eta\}=2\sigma^2$$and the posterior changes from $p(\mu,\eta)$ into$$p^\prime(\mu,\sigma^2)=p(\mu,\log(\sigma^2)/2)\times \frac{\text d\eta}{\text d\sigma^2} = p(\mu,\log(\sigma^2)/2) \frac{1}{2\sigma^2}$$which modifies the location of the mode: $$\arg\max_{\mu,\sigma^2}p(\mu,\log(\sigma^2)/2)\ne \arg\max_{\mu,\sigma^2}p(\mu,\log(\sigma^2)/2)\frac{1}{2\sigma^2}$$ Namely, $$\hat\sigma^2 = \arg\max_{\sigma^2} \left\{-n\log \sigma-\frac{n \hat\sigma^2}{2\sigma^2}\right\}$$ versus $${\tilde\sigma}^2 = \arg\max_{\sigma^2}\left\{-n\log \sigma-\frac{n \hat\sigma^2}{2\sigma^2}\underbrace{-\log \sigma^2}_{+\log \frac{\text d\eta}{\text d\sigma^2}}\right\}=\frac{n}{n+2}\hat\sigma^2$$

  • $\begingroup$ Thank you Xi'an - this is very helpful. Minor follow-up: when I said I can't reconcile the stated change in mode I meant the specific stated change in the book i.e. to $\frac{n}{n+2}\hat{\sigma}^2$. $\endgroup$ Sep 27, 2020 at 14:57
  • $\begingroup$ Wonderful, thank you. Final check: is there a typo in the Jacobian, it should be $\frac{d \eta}{d \sigma^2}$? $\endgroup$ Sep 28, 2020 at 13:57
  • 1
    $\begingroup$ Both are. The Jacobian of one transform $\eta\mapsto\sigma^2$ is the inverse of the reverse transform $\sigma^2\mapsto\eta$. (Typo corrected.) $\endgroup$
    – Xi'an
    Sep 28, 2020 at 16:32

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