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consider the following PDF

$$ \begin{eqnarray} f(x;\theta) &=& \left\{\begin{array}{ll} 2\frac{\theta^2}{x^3} & \theta \leqslant x\\ 0 & x< \theta; 0 < \theta \end{array}\right.\\ \end{eqnarray} $$ Now the answer stats $X_{1:n}$ so the minimum of $X$, but this cannot be deduced from the answer via my normal method (log likelihood), how am I to approach this, and other questions like it?

WiseStrawberry
  • 817
  • " this cannot be deduced from the answer via my normal method " Yes it can. What you can't is just equal the derivative to zero and expect that this gives you the MLE. In general, the global maximum of a function is not necessarily on a critical point (you must check for points where the function is not differentiable, and -as is the case here- for the borders of the domain). – leonbloy May 23 '13 at 19:22
  • So in general, check for bounds within the domain. And don't expect all points to be differentiable? – WiseStrawberry May 23 '13 at 19:23

1 Answers1

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Hint: At what value of $\theta$ is the likelihood maximized keeping in mind that $\theta \le x_i$ for all $i$?

response
  • 5,071
  • I would say that $\theta$ should be as high as possible. because $\sum [log(2)+log(\theta)2-3log(x_i) ] $is the log likelihood function. this would be higher if theta was higher. And seeing as $\theta \leq x_i$ it should be the minimum of $x_i$, is this reasoning correct? – WiseStrawberry May 23 '13 at 19:15
  • Yes, it is correct. – response May 23 '13 at 19:18