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How to find the maximum likelihood estimator of $$f(y;\alpha)=\frac{5\alpha^5}{y^6}$$ for $$0<\alpha\leq y<\infty.$$ Otherwise, $$f(y;\alpha)=0.$$ I have tried, but my result is $\hat{\alpha}_{MLE} = +\infty$. Am I right?

Did
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MaxGaussian
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    No, you are not right, and the trouble is that you wrote the PDF incorrectly. The correct PDF to be maximized over $\alpha$ is $$f(y;\alpha)=5\alpha^5y^{-6}\mathbf 1_{0<\alpha\leqslant y}.$$ Can you find the value $\hat\alpha$ which maximizes this? – Did Feb 22 '15 at 16:17
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    You should solve $max \mathcal{L}$ subject to $0<\alpha\leq min{y_1,...y_n}$ – Math-fun Feb 22 '15 at 16:21
  • @MaxGaussian Indeed. – Did Feb 22 '15 at 16:35
  • Thanks, bro. I came up with $$\hat{\alpha}_{MLE}=\min{Y_i}$$ – MaxGaussian Feb 22 '15 at 18:08

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