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what is the likelihood, log-likelihood and MLE of;

$$θ(θ+1)x^{θ−1}(1−x)$$

any help greatly appreciated

Anthony West
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  • i have an answer that I will not post until you have added more to the post i.e. attempts, and where have you got stuck? because i assume that you had simpler problems to tackle before you had to attempt this? so show some work and then we can help you further. – Chinny84 Feb 19 '15 at 12:42
  • i think the likelihood is $$ θ^n (θ+1)^n \prod(x)^{θ-1} (1-\prod(x))$$ giving a log likelihood of $$n\logθ + n\log{(θ+1)} + (θ-1)\sum\log{x_i} - \sum\log{x_i}$$ but I'm unsure if I've dealt with the $(1-x)$ part correctly for the likelihood – Anthony West Feb 19 '15 at 14:59
  • You are almost there, and you are correct that you didn't handle the $(1-x)$ correctly, you treat it as its own entity. So remove the second product sign, and in the log you have to have a term $\log(1-x_i)$. – Chinny84 Feb 19 '15 at 15:02
  • ok thanks :) but would the log likelihood of the $(1-x)$ bit be just $\log(1-x)$ or $n\log(1-x)$ ?? in the same way that for θ in the log likelihood becomes $n\log(θ)$ – Anthony West Feb 19 '15 at 15:28

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Just to tidy up some of your work (too long for a comment) $$ \begin{align} \text{Likelihood} &=& \theta^n(\theta+1)^n\prod x_i^{\theta-1}(1-x_1)\\ \text{Log-Likelihood} &=& n\log \theta + n\log (\theta+1) + \\ &+&(\theta-1)\sum x_i + \sum\ln(1-x_1) \end{align} $$ To answer your question in the comments, you cannot sum up the observations the same way you can sum with the estimator, this is because, we are assuming there is one estimator to describe the observations.

Now can you determine the MLE (bear in mind it will be the same if you utilize either of the above)

Chinny84
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