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$f(x;\theta) = \frac{3\theta x^{3\theta -1}}{(1+x^{3})^{\theta +1}}, x>0, \theta>0 $ I came up with FOC:

$ \hat{\theta} = \frac{1}{-3\log(x)+\log(1+x^3)} $

Is this correct? Thanks:-)

I took log's and then derivative = 0.

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  • $\begingroup$ I assume you want the mle of $\theta$ rather than $f(x;\theta)$ as your title suggests. Also, you probably have $n$ indepndent sample which resutls in the maximum likelihood estimator of $\hat{\theta}=\frac{n}{\sum _i^n \left(\log \left(x_i^3+1\right)-3 \log (x_i))\right)}$. $\endgroup$
    – JimB
    Mar 23, 2019 at 16:37
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    $\begingroup$ it’s better if you share intermediate steps to prevent everyone looking at your question from doing the same calculations :) $\endgroup$
    – gunes
    Mar 23, 2019 at 16:38
  • $\begingroup$ @JimB, why would you assume that? $\endgroup$ Mar 23, 2019 at 17:25
  • $\begingroup$ @StatsStudent Not sure what you mean. It seems odd at best to estimate the density from a single observation. Or do you mean the assumption that there are $n$ independent samples? $\endgroup$
    – JimB
    Mar 23, 2019 at 17:48
  • $\begingroup$ @JimB, while it's true that the need to obtain an MLE from $n=1$ rarely occurs in practice, MLE's can in fact be obtained from a single observation. Given that this is a homework problem, it's even more likely that the OP is to obtain the MLE from a single observation for pedagogical reasons. $\endgroup$ Mar 23, 2019 at 17:59

1 Answer 1

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Here is the derivation for general case of $N$ i.i.d. samples:

$$\begin{align*} \hat{\theta}=\underset{\theta}{\text{argmax}}\prod_{n=1}^Nf(x_n;\theta)=\underset{\theta}{\text{argmax}}\sum_{n=1}^N\text{log}f(x_n;\theta) \end{align*}$$

$$\begin{align*} \sum_{n=1}^N\text{log}f(x_n;\theta) &= \sum_{n=1}^N\text{log}(3\theta x_n^{3\theta -1})-\text{log}(1+x_n^3)^{1+\theta},\\ &=\sum_{n=1}^N \text{log}(3)+\text{log}(\theta)+(3\theta -1)\text{log}x_n-(1+\theta)\text{log}(1+x_n^3),\\ \frac{\partial \sum_{n=1}^N\text{log}f(x_n;\theta)}{\partial \theta} &= \frac{N}{\theta}+\sum_{n=1}^N 3\text{log}x_n-\text{log}(1+x_n^3)=0,\\ \hat{\theta} &= \frac{N}{\sum_{n=1}^N \text{log}(1+x_n^3) - 3\text{log}x_n} \end{align*}$$

The special case $N=1$ would be:

$$\begin{align*} \hat{\theta} &= \frac{1}{\text{log}(1+x^3) - 3\text{log}x} \end{align*}$$

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  • $\begingroup$ This is the solution for $n$ independent samples having the distribution $f(x;\theta)$, but the OP's post doesn't indicate there is more than a single $x$. As a result, you have an unnecessary summation and an $N$ term in your MLE. $\endgroup$ Mar 23, 2019 at 17:23

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