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I am working on a MLE question to find $\lambda$ where x has the distribution:

$$\lambda^2e^{-2\lambda x^2}$$

This is what I have done: $$L(\lambda)=\prod{\lambda^2e^{-2\lambda x^2}}$$ $$=\lambda^{2n}e^{\sum-2\lambda x^2}$$ $$\log L(\lambda)=2n \log \lambda+\sum(-2\lambda x^2)$$ $$\frac{d \log L(\lambda)}{d(\lambda)}=\frac{2n}{\lambda}+\sum(-2 x^2)=0$$ $$\frac{2n}{\lambda}=-\sum(-2 x^2)$$ $$\frac{2n}{\lambda}=2n-\sum{x^2}$$ $$\lambda=-\sum{x^2}$$

May I ask is the answer correct please??

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    $\begingroup$ How exactly did you proceed after $$ \frac{2n}{\lambda}=-\sum(-2 x^2)\ \mathrm ?$$ $\endgroup$ – Math1000 Sep 9 '16 at 13:05
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    $\begingroup$ um...I think I have made a mistake...thanks $\endgroup$ – BigData Sep 9 '16 at 13:41
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I do not have enough reputation to comment, so I will place my answer here. Math1000 is correct, the derivation should be $$ \frac{2n}{\lambda} = - \sum - 2x^2 $$ $$ \frac{n}{\lambda} = \sum x^2 $$ $$ \hat{\lambda} = \frac{n}{\sum x^2} $$

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  • $\begingroup$ OK i see...thanks $\endgroup$ – BigData Sep 9 '16 at 13:42
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    $\begingroup$ It is good practice to check (e.g. by second derivative test) that $\hat\lambda$ is indeed a local maximum of the log-likelihood function :) $\endgroup$ – Math1000 Sep 9 '16 at 13:44

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