# Thread: method of moments and max likelihood from a Rayleigh distribution

1. ## method of moments and max likelihood from a Rayleigh distribution

Let x1, x2, ...xn be a random sample from a Rayleigh distribution whoose probability density is given by:

f(x;a) = 2*a*x*e^(-a*(x^2)) x>=0 a>0

Find an estimator for the parameter a by the method of max likelihood and the method of moments?

For the maximum likelihood, from trying to work it out I got n/(sum xi^2). Is this right?

i have no idea on how to find the method of moments!

Help appreciated. Thanks

2. ## Re: method of moments and max likelihood from a Rayleigh distribution

Hey JDAWES.

May I ask you to show us your working? This will make it easy to answer your question since it will show the details and the thinking involved at each step which is the important thing (not simply the answer).

3. ## Re: method of moments and max likelihood from a Rayleigh distribution

ok so....

f(x;a) = 2*a*x*e^(-a*(x^2)) x>=0 a>0

L(a)= 2*a^n*(sum of Xi)* e^(-a*(sum of Xi^2))

LnL(a) = n*ln(2*a) + ln(sum of Xi) - a*(sum of Xi^2)

dLnL(a)/d(a) = n/a - (sum of Xi^2)

n/a - (sum of Xi^2) = 0 so n/a = (sum of Xi^2) and so a= n/(sum of Xi^2)

4. ## Re: method of moments and max likelihood from a Rayleigh distribution

The (sum of xi) should be (product of xi) in 2nd line.

5. ## Re: method of moments and max likelihood from a Rayleigh distribution

So apart from that error, is the answer correct?

With the method of moments, you have to find the expectation value E(X) but i get an error term when integrating e^(-a*x^2). How would this cancel?

6. ## Re: method of moments and max likelihood from a Rayleigh distribution

Yes that looks very good for the estimator of parameter a for MLE.

Can you show us the specifics for the MOM Estimator?