In this paper you can see equation (11), the probability density, written as:

$$S_n(\xi) = \frac{1}{N}\frac{1}{2\pi\sigma_r(n)\sigma_g(n)} \sum\limits_{i=1}^{N}\, \exp\left[-\frac{1}{2}\left(\frac{x_i^2}{\sigma_r(n)}+\frac{y_i^2}{\sigma_g(n)} \right) \right]$$

never mind what each parameter represents, my question regards the next two steps, where it first defines the likelihood (eq. 12) as:

$L=\prod\limits_{n=1}^{M}\, S_n(\xi)$

and finally it applies a logarithm to that last equation to obtain (eq 13):

$$\ln \,L = -\sum\limits_{n=1}^{M} \ln\left\{N2\pi\sigma_r(n)\sigma_g(n)\,-\,\sum\limits_{i}\exp\left[ -\frac{1}{2}\left(\frac{x_i^2}{\sigma_r(n)}+\frac{y_i^2}{\sigma_g(n)}\right) \right] \right\}$$

Now, I'm either missing something really obvious here or that last equation is wrong and should be:

$$\ln L = -\sum\limits_{n=1}^{M} \ln\left\{\frac{N2\pi\sigma_r(n)\sigma_g(n)}{\sum\limits_{i}\exp\left[ -\frac{1}{2}\left(\frac{x_i^2}{\sigma_r(n)}+\frac{y_i^2}{\sigma_g(n)}\right) \right]} \right\}$$

or equivalently:

$$\ln L = -\sum\limits_{n=1}^{M} \left\{\ln(N2\pi\sigma_r(n)\sigma_g(n))\,-\,\ln\left(\sum\limits_{i}\exp\left[ -\frac{1}{2}\left(\frac{x_i^2}{\sigma_r(n)}+\frac{y_i^2}{\sigma_g(n)}\right) \right] \right) \right\}$$

Is equation (13) in that paper wrong or is my algebra terrible?


You are absolutely right. The log of a quotient is the difference of the logs. I think the author meant to express the second way you did but carelessly forgot to put the parentheses around the last term with the log in front of the parentheses, a fairly common simple mistake.

  • 2
    $\begingroup$ +1 Such slips, as we all know, are just the author's way of making sure people are reading the paper carefully. If you never find mistakes like this, you just aren't paying attention. :-) $\endgroup$
    – whuber
    May 4 '12 at 15:26
  • 3
    $\begingroup$ Notice that there's a very obvious typo (an extra mathematically ungrammatical '=') in the paper's version of Eq. 12, which increases my posterior probability that the mistake in Eq. 13 is also a typo. $\endgroup$
    – onestop
    May 4 '12 at 15:47

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