# Likelihood Ratio Test & Likelihood Function

• Apr 13th 2009, 06:50 PM
kingwinner
Likelihood Ratio Test & Likelihood Function
Suppose that X_1,X_2,...,X_n1, Y_1,Y_2,...,Y_n2, and W_1,W_2,...,W_n3 are independent ranom samples from normal distributions with respective unknown means μ1, μ2, and μ3 and variances (σ1)^2, (σ2)^2, (σ3)^2.
Find the likelihood ratio test for H_o: (σ1)^2 = (σ2)^2 = (σ3)^2 against the alternative of at least one inequality.
[source: Wackerly #10.108a]
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I am clueless...can someone please go through the approach/steps to solving this problem? In particular, I am not sure how to find the likelihood function. If we simply have the random variables X_1,X_2,...,X_n, then the likelihood function is just the joint density f_X1,X2,...,Xn(x1,x2,...,xn), but here we have three sets of different random variables (X,Y, and W), what would be the likelihood function, then?

Thanks for helping!
• Apr 13th 2009, 10:50 PM
matheagle
I almost assigned that problem two weeks ago.
I can work it out, but it's a lot of typing.

Basically, and I'm sure you'll keep asking me the same question repeatedly, the
MLEs of populations means under BOTH hypotheses are the sample means.
The estimates of the population variances are of the form

$\hat\sigma_i^2={\sum_{k=1}^n (X_{ik}-\bar X_i)^2\over n_i}$ for each of these sets.

UNDER the null hypothesis we have as the estimate of the COMMOM variance as the pooled estimator.
JUST like in the two sample t case, giving us...

$\hat\sigma^2 ={\hat\sigma_1^2 +\hat\sigma_2^2 +\hat\sigma_3^2 \over n_1+n_2+n_3}$

You reject the null hypothesis when the ensuing ratio is large...

$\lambda ={(\hat\sigma_1^2)^{n_1/2} (\hat\sigma_2^2)^{n_2/2} (\hat\sigma_3^2)^{n_3/2} \over (\hat\sigma^2)^{(n_1+n_2+n_3)/2}}$
• Apr 14th 2009, 01:37 AM
kingwinner
That's a tough problem from Wackerly...

1) In general, is it always true that the maximum likelihood estimator (MLE) of populations mean is the sample mean?

2) To apply the likelihood ratio test, an important step is to find the likelihood function, but what is the likelihood function in this case??? If we simply have the random variables X_1,X_2,...,X_n, then I know that the likelihood function is just the joint density f_X1,X2,...,Xn(x1,x2,...,xn), but here we have THREE sets of different random variables (X's,Y's, and W's) which makes me confused...

Thanks!
• Apr 14th 2009, 07:21 AM
JohnQ
1) No. The average is the MLE for the expected value in the normal distribution. It may or may not be the MLE for other distributions. It depends on the likelihood function and which values of the parameters maximize it.

2) The log likelihood is the sum of the logs of the normal pdf. Each normal pdf involves its own RV, its own mu, and its own variance. It just so happens that the mu for all the X's is the same ( $\mu_X$), that it is the same for all the Y's ( $\mu_Y$) and for all the W's (( $\mu_W$). Under the null, the variance happens to be the same for all the pdf's. Under the alternate, it is, just like the mean, grouped by X, Y, and W.

Hope this helps.
• Apr 14th 2009, 02:38 PM
kingwinner
But in this case, is the likelihood function going to a product or sum? Should we just multiply the 3 likelihood functions for X's, Y's, and W's or add them?
• Apr 16th 2009, 06:43 AM
JohnQ
A likelihood is a pdf. It is a joint pdf of all of the RV's that are under consideration. It's a single function.

If the RV's involved are independent, then the pdf can be rewritten as a product of other pdf's. It makes the problem very convenient.

However, it's often inconvenient dealing with a large product. So usually, instead of working with the likelihood (which is a product), people work with the logarithm of the likelihood which is a sum.
• Apr 23rd 2009, 07:53 AM
kingwinner
Quote:

Originally Posted by JohnQ
A likelihood is a pdf. It is a joint pdf of all of the RV's that are under consideration. It's a single function.

Is it always ALL (X's,Y's,W's) of the RV's under consideration? Is this the definition of a likelihood function? Somehow I have trouble understanding this because for all of the problems I have seen, the likelihood only deals with one sample (i.e. only X1,X2,...,Xn).

Quote:

If the RV's involved are independent, then the pdf can be rewritten as a product of other pdf's. It makes the problem very convenient.
I have a technical question. X's are iid, Y's are iid, Z's are iid, and the three samples are independent, does this mean all the RV's (X's,Y's,Z's) are mutually independent?
The likelihood in our problem is f(x1,x2,...,x_n1, y1,y2,...,y_n2, w1,w2,...,w_n3), how can we break this into a product?

Thanks a lot!