Let $f_{\delta}(y) = \frac{y}{\delta^2}*e^(\frac{-y^2}{2*\delta^2})$
with $y\geq 0$
I set up the Likelihood function as $$L(\delta) = (\frac{y}{\delta^2})^n *e^(\frac{-\sum y^2}{2*\delta^2})$$
and the log likelihood as $$l(\delta) = n*ln(\frac{y}{\delta^2})+\frac{\sum y^2}{2*\delta^2}$$
Taking first derivative $$l'(\delta) = -\frac{2*n}{\delta}-\frac{\sum y^2}{\delta^3}$$
Second derivative $$l''(\delta) = \frac{2*n}{\delta^3}+\frac{3*\sum y^2}{\delta^4}$$
But this second derivative seems incorrect, but where do I go wrong?
I want to calculate the Fisher Information with the second derivative and calculate a confidence interval.
Here is my attempt:
Let $f_{\delta}(y) = \frac{y}{\delta^2}e^{\frac{-y^2}{2\delta^2}}$ with $y\geq 0.$
Then $$L(\delta) =\Pi_{i=1}^n f_{\delta}(y_i) = \frac{r}{\delta^{2n}} e^{\frac{-\sum_i y_i^2}{2\delta^2}}$$ where $r$ denotes the product of $y_i.$
Log likelihood is as $$l(\delta) = -2n\log(\delta)-\frac{\sum y_i^2}{2\delta^2}+C$$
Taking first derivative $$l'(\delta) = -\frac{2n}{\delta}+\frac{\sum_i y_i^2}{\delta^3}$$
Second derivative $$l''(\delta) = \frac{2n}{\delta^2}-\frac{3\sum_i y_i^2}{\delta^4}.$$
So yes you missed out a minus sign but also I would suggest you write $(y_1,...,y_n)$ rather than $(y,y,..,y).$
Edit:
This is how I would do the expectation for this $l''(\delta)$:
We now treat only $y_i$ as random variables and so $E[l''(\delta)]=\frac{2n}{\delta^2}-\frac{3}{\delta^4}E[\sum_i Y_i^2].$
Then you use linearity and this will end up finding the expected value of $Y^2,$ you can do this with the standard integral. (Or perhaps thinking about $Var(Y)=E[Y^2]-(E[Y])^2)$ can help?)