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Let $x_1,\ldots,x_n$ be a random sample with pdf $$f(x)= \begin{cases} (\alpha+1)x^\alpha, & 0 \le x \le 1, \\ 0,& \mbox{otherwise}. \end{cases}$$ Find the MLE of $\alpha.$

So, I've found the log-likelihood and taken the derivative, but I'm not quite sure if I've done it right. This is what I have so far. Would I then just set the equation equal to zero and solve for $\alpha$? \begin{align*} \log L &= \sum_{i=1}^n \log(\alpha + 1) + \sum_{i=1}^n \log(x_i^\alpha) \\ &= n\log(\alpha+1) + \sum_{i=1}^n \log(x_i^\alpha)\\ &= \frac{n}{\alpha+1} + \alpha\sum_{i=1}^n \log(x_i) \end{align*}

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    $\begingroup$ Going from your second to your third line you have made multiple errors. I think this question should probably carry the self-study tag (see its tag-wiki info) $\endgroup$ – Glen_b Apr 9 '14 at 0:46
  • $\begingroup$ No, you didn't take the derivative, you only... half-took it (or you made an elementary mistake in differentiating the second term). Take your second line, use the properties of the logarithm, and calculate, fully and correctly this time, its derivative w.r.t to $\alpha$. And please delete your third line. $\endgroup$ – Alecos Papadopoulos Apr 9 '14 at 1:10
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To expand on my comment slightly, this:

\begin{align*} \log L &= \sum_{i=1}^n \log(\alpha + 1) + \sum_{i=1}^n \log(x_i^\alpha) \\ &= n\log(\alpha+1) + \sum_{i=1}^n \log(x_i^\alpha) \end{align*}

is okay. You could also bring the power, $\alpha$ in the second term down if you like. But then you need to actually take the derivative. That is, you need a new left hand side:

$$\frac{\partial\log L}{\partial \alpha}= \: ...$$

(your original just said "log L" at the LHS of line 3).

And then you need to properly take the derivative of the right hand side.

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