Calculating Jeffreys prior - where's the mistake? Below you'll find a picture with the exercise I'm trying to do do.
My doubt is in c).
I've chosen the Jeffreys Prior as the prior for N.
Also, I've modelled $Y|N \sim \text{Uniform}\{1,...,N\}$, resulting in $P(Y=y|N=n)=\frac{1}{n}I_{\{1,...,n\}}(y)$. 
The problem occurs when calculating the second derivative of $log(p(y|n))$. If I assume ($y=203<n=N$), to apply the log to a non-zero value, I get $E\left(-\frac{d^2}{dN^2}log(P(y|n))|N\right)=-\frac{1}{N^2}$.
Where did I do a mistake?

Any help would be appreciated.
 A: It is likely that derivating/integrating wrt to discrete variables may be problematic. Nevertheless, considering the continuous analogous to the discrete model for derivated the associated Jeffreys is a situation discussed by Berger in https://www2.stat.duke.edu/~berger/papers/discrete.pdf (among others). Moreover, in this paper, it is stated that  the solution for your problem  is $p(N)\propto 1/N$ (first paragraph of Section 1.2.1). 
However I found the same result as yours. Nevertheless (assuming that everything go well with the Heaviside step function), the alternative definition of Fisher information  gives the expected result:
$$
(\frac{d log(f)}{dN})^2=\frac{1}{N^2}
$$
Then integrating:
$$
I(N)=\int_0^{\infty} \frac{1}{N^2} \frac{1}{N} 1(y<N) dy 
$$
gives $I(N)=\frac{1}{N^2}$ and finally $p(N) \propto \frac{1}{N} 1_{R^+}(N)$.
But I do not know why (but as the switch to continuous may be a bit problematic, I would not be so surprised that some of the conditions related to the Fisher information formulation are not met)
