Suppose there is a probability distribution for values of $x$ greater than $0$:

$$p(x) \propto \frac{m}{(x+1)^{m+1}}$$

And we select from a sample of $\{X_1, X_2, \ldots ,X_n\}$ with all $X_i$ having this distribution. What is the maximum likelihood estimator of $m$?

I tried to do this using the log-likelihood function method but it doesn't work because the log-likelihood function is not well behaved so I ended up concluding that the MLE of $m$ is $m=\max(X_n)$, similar to the continuous uniform distribution. Is this correct?

  • $\begingroup$ What is 'strange' about this distribution? $\endgroup$ Commented Jun 18, 2020 at 15:19

1 Answer 1


$$ \mathcal{L} = m^n\prod_{i=1}^n\frac{1}{(x_i+1)^{m+1}} $$ thus $$ \ln \mathcal{L} = n\ln m - (m+1)\sum_{i=1}^n\ln(x_i+1) $$ so we find $$ \partial_m \ln \mathcal{L} = \frac{n}{m} - \sum_{i=1}^n\ln(x_i+1) = 0 $$ This leads to $$ m = \frac{n}{\sum_{i=1}^n\ln(x_i+1) } = \left(\frac{1}{n}\sum_{i=1}^n\ln(x_i+1)\right)^{-1} $$

  • 1
    $\begingroup$ Teeny typo: $i = 0$ should be $i = 1$ throughout. $\endgroup$ Commented May 12, 2017 at 1:47
  • $\begingroup$ @david I thought this post was going to be cast to 'zero pile' for all eternity. Thank you for pointing it out I will fix it now :) $\endgroup$
    – Chinny84
    Commented May 12, 2017 at 2:35

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