Say we have a simple random sample of size $n$ distributed according to a function of the family: $$ f(x|\theta)=\frac{\theta}{(1+x)^{1+\theta}} $$ where $\theta >0$.

We want to compute the maximum likelihood estimate of $\theta$.

I have tried the following. We want to calculate the $\theta$ which maximizes the likelihood function of a given sample $\vec{x}$, but that coincides with the maximum of the log of the likelihood function.

To calculate it, we try to find the zeros of its derivative: $$ \ln{L(\theta|\vec{x})} = n\ln{\theta}-(1-\theta)\sum_{i=1}^{n}\ln{(1+x_i)} $$ $$ \frac{\partial }{\partial \theta}\ln{L(\theta|\vec{x})} = \frac{n}{\theta} + \sum_{i=1}^{n}\ln{(1+x_i)} $$ The solution when we equate this expression to zero is $\theta=-\frac{n}{\sum_{i=1}^{n}\ln{(1+x_i)}}$, which is negative and thus cannot be the MLE.

What am I doing wrong?

  • 1
    $\begingroup$ Shouldn't the log-likelihood be $$n \log \theta - (1 + \theta) \sum_{i=1}^n \log (1+x_i)?$$ $\endgroup$ – heropup Jun 9 '16 at 15:28
  • $\begingroup$ @heropup oops my mistake. TY $\endgroup$ – Jsevillamol Jun 9 '16 at 15:30

Your question denominator has $1+\theta$, while in your calculations, you are using $1-\theta$. Once you fix this minus sign in your calculations, you will have the same result as you have, without the minus sign.


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