I would like to compute the fisher indicator for this function: $$f(y;u)=\frac{1}{\sqrt{2\pi y^3}} e^{\frac{-(y-u)^2}{2u^2y}}$$ With $y>0$. I have computed the log likelihood function and the score function defined by $U$: $$l(y;u)=-\frac{n}{2}\sum_{i=1}^n\log(2\pi y_i^3) -\frac{1}{2u^2}\sum_{i=1}^{n}\frac{(y_i-u)^2}{y_i}$$ $$U=\frac{\partial l}{\partial u}=\frac{\sum_{i=1}^{n} \frac{y_i-u}{y_i}}{u^2}+\frac{\sum_{i=1}^{n} \frac{(y_i-u)^2}{y_i}}{u^3}$$ For this function $f(y;u)$ I have computed the $MLE$ estimator for $u$ defined by $\hat{u}=\frac{\sum_{i=1}^{n} y_i}{n}$ using the first order condition over $U$. Now, I am struggling with the fact that I need to show that $\hat{u}$ is unbiased by using the properties of score function and showing that the asymptotic variance of $\hat{u}$ is $\frac{u^3}{n}$. I know that $var(\hat{u})=I(u)^{-1}$, where $I(u)$ denotes the fisher indicator defined by: $$I(u)=-E(\frac{\partial^2 l}{\partial u^2})$$ I have surfed around this last equation but I can not get the value $\frac{u^3}{n}$ and even showing that $\hat{u}$ is unbiased using the score function. Could you please help me giving a direction about how can I get that value as the variance of the estimator and using score function to prove that $\hat{u}$ is unbiased. Many thanks for your help.

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    $\begingroup$ I am having difficulties reproducing your calculation of the derivative. I obtain $$-U = \frac{n}{u^2} - \frac{1}{u^3}\sum y_i.$$ $\endgroup$ – whuber Oct 28 '18 at 22:32

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