I have a likelihood function $$ L(\theta;\mathbf x) = \frac{\prod x_i^{\nu-1} \exp\left( -\sum x_i/\theta \right) }{\theta^{\nu n} [\Gamma(\nu)] } \qquad x>0 $$

It gets log-transformed into the following formula $$ \ln L(\theta;\mathbf x) = \text{constant} - \frac{n\overline x} \theta - \nu\theta\ln\theta $$

Two questions:

  1. I get the same result when I perform the transformation myself, except in addition to the above result I get an extra term $n\bar{x}(\nu-1)$ — why is it not supposed to be there?
  1. Also I get ${}-\text{const}$ rather than ${}+\text{const}$, but I suppose because it is an arbitrary constant value, then either $+$ or $-$ works?
  • 1
    $\begingroup$ BTW, the letters $v$ and $\nu$ are not the same thing. $\endgroup$ Commented Aug 15, 2020 at 16:44
  • $\begingroup$ I'm tentatively guessing that the numerator here was to be parsed as $$ \left( \prod_{i=1}^n x^{\nu-1} \right) \left( \exp\left( - \sum_{i=1}^n x_i/\theta \right) \right). $$ $\endgroup$ Commented Aug 19, 2020 at 13:24
  • $\begingroup$ It appears that your "extra term" should be $\displaystyle (\nu-1)\sum_i \ln x_i$ (which is a "constant" since it doesn't depend on $\theta$). $\qquad$ $\endgroup$ Commented Aug 19, 2020 at 13:27
  • $\begingroup$ Also, your last term should be $\nu n\ln\theta. \qquad$ $\endgroup$ Commented Aug 19, 2020 at 13:30
  • $\begingroup$ ok, My tentative guess is no longer tentative: I see where this came from. $\qquad$ $\endgroup$ Commented Aug 19, 2020 at 13:31

1 Answer 1


In this context, "constant" means not depending on $\theta.$ All terms that do not depend on $\theta$ are constant. In particular, often the next thing one does after taking the logarithm is differentiating with respect to $\theta,$ and then every term not depending on $\theta$ vanishes.


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