# Thread: Rayleigh Distribution Confidence Interval

1. ## Rayleigh Distribution Confidence Interval

I am asked to calculate a confidence interval for a large sample based on a Rayleigh Distribtion: f_X(x;theta) = ((x^2)/(theta^2))exp((-x^2)/(2*theta^2)), x>=0, and theta>=0. I have found the maximum likelihood estimator for this distribution, but I am having difficulty finding the Fisher Information for the distribution. As soon as I have that figured out, I will have all I need to solve the problem.

Any ideas would be great. Thanks in advance.

2. Originally Posted by eigenvector11
I am asked to calculate a confidence interval for a large sample based on a Rayleigh Distribtion: f_X(x;theta) = ((x^2)/(theta^2))exp((-x^2)/(2*theta^2)), x>=0, and theta>=0. I have found the maximum likelihood estimator for this distribution, but I am having difficulty finding the Fisher Information for the distribution. As soon as I have that figured out, I will have all I need to solve the problem.

Any ideas would be great. Thanks in advance.
1. Do you have a typo in the definition of the density?

2. If yes, then: If $\displaystyle R\sim \text{Rayleigh}(\sigma)$ , then $\displaystyle R^2/\sigma^2 \sim \chi^2_2$.

3. You have a large sample, so use the large sample normal approximation for which you only need to know the mean and variance of the Rayleigh.

CB

3. No, there is not a typo in regards to the way the density function is defined. That is how it is defined in the question. The question asks me to derive the form of a large sample 95% confidence interval based upon the maximum likelihood estimator of theta hat. So from what I understand from the notes in class, this requires me to find the maximum likelihood estimator (which I've done), and then find the Fisher information for this distribution (which is where I need help). The Fisher information involves the Cramer-Rao lower bound (i.e. taking the natural log of the density, finding two derivatives, taking the negative expected value of the second derivative, taking the inverse of this, multiplying by n, and taking the square root, etc.).

4. Originally Posted by eigenvector11
No, there is not a typo in regards to the way the density function is defined. That is how it is defined in the question. The question asks me to derive the form of a large sample 95% confidence interval based upon the maximum likelihood estimator of theta hat. So from what I understand from the notes in class, this requires me to find the maximum likelihood estimator (which I've done), and then find the Fisher information for this distribution (which is where I need help). The Fisher information involves the Cramer-Rao lower bound (i.e. taking the natural log of the density, finding two derivatives, taking the negative expected value of the second derivative, taking the inverse of this, multiplying by n, and taking the square root, etc.).
I will put it more forcefully then. What you have does not agree with how the main online sources think the Rayleigh distribution is defined. Not a show stopper in itself since there is more than one definition out there. What is a show stopper is that it does not integrate up to 1.

Numerical evidence:

Code:
>dx=0.0001;
>x=dx/2:dx:20;
>theta=1;
>f=x^2/(theta^2)*exp((-x^2)/(2*theta^2));
>
>sum(f)*dx
1.25331
>
Symbolic evidence: see attachment

CB