# What is the Maximum-Likelihood Estimator of this strange distribution?

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?

## 1 Answer

$$\mathcal{L} = m^n\prod_{i=1}^n\frac{1}{(x_i+1)^{m+1}}$$ 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 $$m = \frac{n}{\sum_{i=1}^n\ln(x_i+1) } = \left(\frac{1}{n}\sum_{i=1}^n\ln(x_i+1)\right)^{-1}$$

• Teeny typo: $i = 0$ should be $i = 1$ throughout. – David G. Stork May 12 '17 at 1:47
• @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 :) – Chinny84 May 12 '17 at 2:35