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I just wanna verify if I answered this question correctly.

The following numbers are taken from a population having normal distribution with mean and variance :

5.3299 4.2537 3.1502 3.7032 1.6070 6.392 3.1181

6.5941 3.5281 4.7433 0.1077 1.5977 5.4920 1.7220

4.1547 2.2799

a. Find the maximum likelihood estimate of the mean

b. Find the maximum likelihood estimate of the variance

c. Find the method of moments estimate of the mean

d. Find the method of moments estimate of the variance

My answers are:

a. 3.611

b. 3.204

c. 3.611

d. 3.204

Thanks Again for your help.

I greatly appreciate it.


I am unsure cause MLE and MME for the mean and variance of the Normal Distribution is the same.

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up vote 1 down vote accepted

For the normal distribution, the MLEs and the method-of-moments estimates are the same. You can prove that by doing some algebra. (It's quite a different situation from what happens with the uniform distribution, where the method-of-moments estimate of the upper endpoint can actually sometimes be less than some of the observed values, making it absurd to use such an estimate in practice.)

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hi michael, are my answers correct? – Kaji Apr 7 '12 at 16:21
@Kaji: You should explain how you got those numbers. – Xabier Domínguez Apr 7 '12 at 16:28
it's hard to explain cause i dont know how to type those mathematical symbols... ok, i guess I'll study the symbols first. hehe – Kaji Apr 7 '12 at 16:29
michael, xabier, are my answers right? so here's what i have done, i first derived the formula by using the given independent random variables. just like we did earlier, i ended up with u^ = summation of xi / n. by using this, we can also get that the variance = summation of (x-xbar)^2 / n... that's what i did for the MLE. as per the MME, i just used the expectation principle... E(x) = xbar = summation of xi from i=1 to i=n over n. and then, from there, i used the formula of variance which is = E(x^2) - E(x)^2 then arrived to the same with what i have on the MLE. – Kaji Apr 7 '12 at 17:03
Your answers are numerically correct. Your computation of the MLEs seems fine. But I don't understand your derivation of the MMEs very well. You should identify the population moments with the sample moments. If you have two parameters to estimate (as it is the case) you need two equations, hence you put (sample mean)=(E(X)) and (sample variance)=(var(X)) [or maybe $(1/n)\sum X_i^2=E(X^2)$, depending on your textbook.] – Xabier Domínguez Apr 7 '12 at 17:42

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