Fisher information of the Rayleigh distribution

Question

Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(\theta) = -E(\frac{\partial^2 \log L}{\partial \theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(\theta) = \frac{n}{\theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$\log L = \prod \log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$\log L = \sum_{i}(\log x_{i} -\frac{2}{\theta} - \frac{x_{i}^2}{2\theta^2}) $$ Then I take the partial derivative with respect to $\theta$ two times and obtain $$\frac{\partial^2 \log L}{\partial \theta^2} = \frac{2n\theta^2-3\sum_{i}x_{i}^2}{\theta^4} $$ So $I(\theta) = -E(\frac{2n\theta^2-3\sum_{i}x_{i}^2}{\theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $\theta$?

Answer 1

Continuing the above discussion, I find the following when computing $-E(\frac{2n\theta^2-3\sum_{i}x_{i}^2}{\theta^4} )$. $$-E(\frac{2n\theta^2-3\sum_{i}x_{i}^2}{\theta^4}) = -E(\frac{2n\theta^2+3\sum_{i}-x_{i}^2}{\theta^4})$$ now since I found that $$E(T(x)) = E(-x_{j}^2) = -2\theta^2 $$ for arbitrary $x_{j}$. $$E(\frac{2n\theta^2+3\sum_{i}-x_{i}^2}{\theta^4}) = \frac{2n}{\theta^2} + \frac{3nE(T(x))}{\theta^4} = \frac{-4n}{\theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(\frac{2n\theta^2+3\sum_{i}-x_{i}^2}{\theta^4}) = \frac{4n}{\theta^2}$$

Answer 2

The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.