method of moments and max likelihood from a Rayleigh distribution

• Jan 12th 2013, 02:14 AM
JDAWES
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
• Jan 12th 2013, 06:31 PM
chiro
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).
• Jan 14th 2013, 07:33 AM
JDAWES
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)
• Jan 14th 2013, 05:31 PM
chiro
Re: method of moments and max likelihood from a Rayleigh distribution
The (sum of xi) should be (product of xi) in 2nd line.
• Jan 15th 2013, 02:05 AM
JDAWES
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?
• Jan 15th 2013, 09:05 PM
chiro
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?