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??
I do not have enough reputation to comment, so I will place my answer here. Math1000 is correct, the derivation should be $$ \frac{2n}{\lambda} = - \sum - 2x^2 $$ $$ \frac{n}{\lambda} = \sum x^2 $$ $$ \hat{\lambda} = \frac{n}{\sum x^2} $$