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This work is from a textbook where the answer is given as the MLE being $X_{1:n}$ where $1:n$ indicates the minimum value. The pdf of the distribution is given as: $$ f(x;\theta)=\begin{cases}2\theta^2x^{-3}\hspace{10pt}\theta\leq{x}\\0\hspace{33pt}x<\theta;0<\theta\end{cases} $$ Typically, I would calculate the log-likelihood which is $$ \ell(\theta)=n\ln(2)+2n\ln(\theta)-3\sum^n_{i=1}\ln(x_i), $$ and then take the derivative $$ \ell'(\theta)=\frac{2n}{\theta} $$ and then work out the maximum.

Obviously, in this case, it evaluates to zero and I'm not sure what I'm missing to understand how to get to the solution given. In this instance, how should I be calculating the MLE?

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  • $\begingroup$ Hint: this is a case in which you don't use derivatives to maximize the log-likelihood, but rather just inspect the log-likelihood and maximize directly. $\endgroup$
    – Cliff AB
    Aug 7, 2017 at 0:06
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    $\begingroup$ Oh, right, so because the limits on theta are that it is between 0 and x, the log-likelihood is maximized at the highest value it can take which is the lowest observed value of x? $\endgroup$
    – Syzorr
    Aug 7, 2017 at 0:13

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On inspection of the equation, it becomes obvious that the maximum of $\ell(\theta)$ is found for the largest possible value of $\theta$, therefore $$\hat{\theta}=X_{1:n}$$ as $0<\theta\leq{x}$

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