Hi. I really don't find my lecture notes useful, and I can find precisely zero good guides online for how to deal with this.
The question is:
Given X1,...,Xn are Exp(a) and Y1,...,Yn are Exp(b) random variables, find the likelihood ratio test of size alpha of Ho: a=b against H1: a=/=b, expressing it in the form of the statistic:
and the quantiles of a standard distribution.
My thoughts so far: Absolutely nothing. I have no idea whatsoever how to go about solving this question. And this is only Q2...
Thanks for any help.
i'm not sure what you want to know
Did you perform the likelihood ratio test?
Then you can use the fact that a sum of iid exponentials is a gamma
then the ratio of a gamma over sum of itself and another gamma should be a beta.
Thanks for the reply.
Originally Posted by matheagle
I don't understand how to deal with the fact that b is itself 'random' so to speak. I mean, we're testing the X's, and I can do that in a very simple case. But how do I use the Y's? I just don't understand what to do with them. Could you explain how you would answer the question? You don't need to give all the details, just push me in the right direction so I know what to do...
a and b are not random
They are two fixed parameters
you're testing whether the parameter is a or b in an exponential setting.
It like testing if the mean is 2 or 3.
And since the mean of an exponential is theta, that is exactly what you're doing.
Ok. So the likelihood ratio test should be:
f(x;b)/f(x,c) where c=/=b
=(b/c)^n * exp(-(b+c)Sum(x_i)) , which could be increasing or decreasing depending on the value of c.
How does this bear any resemblance to the T given? What's the point of including Y when it isn't involved in the ratio I've been told to work out?
Originally Posted by phillips101
Well, to be frank you haven't answered a single question I've asked in a way that I've found even remotely useful. I'm not going to 'hit thanks' when I have very little to be thankful for other than increasing frustration with this question, am I? Clearly you taking the time to post here is self-serving, so if you want a 'thanks' you should earn it. Not that I particularly care anymore, I doubt I'm going to bother asking for help here again.