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    Hi guys,We have that X1...Xn are i.i.d Gaussian Random Variables with mean mu and variance 1. Our null hypothesis is mu = 0 and our alternative hypothesis is mu = -1 or 1.I need to construct the likelihood ratio test. I have no idea how to do this: I know how to construct the ratio but not sure how to construct the test.I then need to check whether this is the uniformly most powerful test, and if it isn't, construct a better test. Absolutely no clue for this part.Any ideas? Thanks for reading.
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    (Original post by pineapplechemist)
    Hi guys,We have that X1...Xn are i.i.d Gaussian Random Variables with mean mu and variance 1. Our null hypothesis is mu = 0 and our alternative hypothesis is mu = -1 or 1.I need to construct the likelihood ratio test. I have no idea how to do this: I know how to construct the ratio but not sure how to construct the test.I then need to check whether this is the uniformly most powerful test, and if it isn't, construct a better test. Absolutely no clue for this part.Any ideas? Thanks for reading.
    Let's have a look at your likelihood ratio.

    You construct a LRT out of an LR by rejecting the null hypothesis if the LR is too small. (That is, if the likelihood of the alternative hypothesis is sufficiently larger than the likelihood of the null). So the LRT will look something like LR < a where a is chosen to make the test have the correct Type I error rate. For the purposes of exercises like this, you don't need to find the value of a - you just write down "LR < a for suitable choice of a".

    For the later parts, do you know about the Neyman-Pearson lemma?
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    (Original post by Gregorius)
    Let's have a look at your likelihood ratio.

    You construct a LRT out of an LR by rejecting the null hypothesis if the LR is too small. (That is, if the likelihood of the alternative hypothesis is sufficiently larger than the likelihood of the null). So the LRT will look something like LR < a where a is chosen to make the test have the correct Type I error rate. For the purposes of exercises like this, you don't need to find the value of a - you just write down "LR < a for suitable choice of a".

    For the later parts, do you know about the Neyman-Pearson lemma?
    Thanks for the reply.

    I know the Neyman-Pearson lemma yes: does it show that the test is uniformly most powerful? An earlier part of the question is to do with having a monotone likelihood ratio with respect to T(X) = 1/n(sum from i=1 to n of Xi). In light of this, should we use the Karlin-Rubin theorem? One thing that confuses me though is that our alternative hypothesis is that the mean could be equal to -1 or 1, whether these theorems seem to concern more simple cases where the alternative hypothesis is just made up of one number, for example. How does this affect the situation?
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    (Original post by pineapplechemist)
    Thanks for the reply.

    I know the Neyman-Pearson lemma yes: does it show that the test is uniformly most powerful? An earlier part of the question is to do with having a monotone likelihood ratio with respect to T(X) = 1/n(sum from i=1 to n of Xi). In light of this, should we use the Karlin-Rubin theorem? One thing that confuses me though is that our alternative hypothesis is that the mean could be equal to -1 or 1, whether these theorems seem to concern more simple cases where the alternative hypothesis is just made up of one number, for example. How does this affect the situation?
    I think that the first parts of the question may be a bit of a red herring here. It is true that the monotone likelihood ratio stuff (and Karlin-Rubin, for that matter) are used to extend Neyman-Pearson from simple-null-versus-simple-alternative to certain sorts of simple-null-versus-compound-alternatives. But they tend to work in the situation of

    \displaystyle H_{0}: \theta = \theta_{0}
    \displaystyle H_{1}: \theta &gt; \theta_{0}

    For the normal distribution, the case

    \displaystyle H_{0}: \theta = \theta_{0}
    \displaystyle H_{1}: \theta \ne \theta_{0}

    is a well known example where there is no UMP.

    Here you should have found that the likelihood ratio is

    \displaystyle \exp[n(\frac{1}{2} - \bar{x}) ] if  \bar{x} \ge 0

    \displaystyle \exp[n(\frac{1}{2} + \bar{x}) ] if  \bar{x} &lt; 0

    You can simplify that to get your likelihood ratio test and you should end up with something pleasantly familiar which has some nice symmetry; how might symmetry this be exploited?
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    (Original post by Gregorius)
    I think that the first parts of the question may be a bit of a red herring here. It is true that the monotone likelihood ratio stuff (and Karlin-Rubin, for that matter) are used to extend Neyman-Pearson from simple-null-versus-simple-alternative to certain sorts of simple-null-versus-compound-alternatives. But they tend to work in the situation of

    \displaystyle H_{0}: \theta = \theta_{0}
    \displaystyle H_{1}: \theta &gt; \theta_{0}

    For the normal distribution, the case

    \displaystyle H_{0}: \theta = \theta_{0}
    \displaystyle H_{1}: \theta \ne \theta_{0}

    is a well known example where there is no UMP.

    Here you should have found that the likelihood ratio is

    \displaystyle \exp[n(\frac{1}{2} - \bar{x}) ] if  \bar{x} \ge 0

    \displaystyle \exp[n(\frac{1}{2} + \bar{x}) ] if  \bar{x} &lt; 0

    You can simplify that to get your likelihood ratio test and you should end up with something pleasantly familiar which has some nice symmetry; how might symmetry this be exploited?
    Hi, shouldnt the likelihood ratio be

    \displaystyle \exp[n(\frac{1}{2} - \bar{x}) ] if  \bar{x} &lt; 0

    \displaystyle \exp[n(\frac{1}{2} + \bar{x}) ] if  \bar{x} \ge 0
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    (Original post by Namch)
    Hi, shouldnt the likelihood ratio be

    \displaystyle \exp[n(\frac{1}{2} - \bar{x}) ] if  \bar{x} &lt; 0

    \displaystyle \exp[n(\frac{1}{2} + \bar{x}) ] if  \bar{x} \ge 0
    No, I don't think so. The likelihood ratio should be large at \bar{x} = 0 , favouring H_{0} and should be small for large positive or negative values of \bar{x}. The LR test will end up being something like "reject H_{0} if  |\bar{x}| &lt; k ", where k is chosen to get the test size right.
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    (Original post by Gregorius)
    No, I don't think so. The likelihood ratio should be large at \bar{x} = 0 , favouring H_{0} and should be small for large positive or negative values of \bar{x}. The LR test will end up being something like "reject H_{0} if  |\bar{x}| &lt; k ", where k is chosen to get the test size right.
    Oh yh that makes sense. I worked out the ratio by putting the null hypothesis as denominator and got something else (different to what i wrote up there).

    Anyway then you eventually get reject null hypothesis if |x bar| > k for some appropriate k depending on the significance of the test.

    I think this is the UMP ? I constructed a proof similar to Neymann Pearson lemma.
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    Actually i have no idea if this is UMP
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    why is it okay for tsr users to just come on here, and ask for help with their uni work? (which is, to all other extents, cheating)??
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    (Original post by john2054)
    why is it okay for tsr users to just come on here, and ask for help with their uni work? (which is, to all other extents, cheating)??
    feisty
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    (Original post by Gregorius)


    For the later parts, do you know about the Neyman-Pearson lemma?
    I tried using Neyman-Pearson to construct a proof but it broke down.

    I tried to find a test more powerful but it turned out to be less powerful.

    So this question is tough :O
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    (Original post by john2054)
    why is it okay for tsr users to just come on here, and ask for help with their uni work? (which is, to all other extents, cheating)??
    Guidance is offered to undergraduates (and postgraduates!) on the same basis that it is offered to all other students. The forum rules apparently state that no complete solutions are allowed - and most users adhere to that guideline.

    The basis on which you make your accusation of "cheating" is unclear.
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    (Original post by Zacken)
    Why is it okay for you to come on here and post useless and unhelpful crap?

    Answer: it's not and you'll be stopped if you continue.
    thanks zacken, but to call my posts 'crap' is abusive, and quite frankly offensive. I can understand that you may not always like what i say, but please refrain from insulting me in this way in the future, okay? pm me if you want further clarification on this topic thanks! john
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    (Original post by john2054)
    thanks zacken, but to call my posts 'crap' is abusive, and quite frankly offensive. I can understand that you may not always like what i say, but please refrain from insulting me in this way in the future, okay? pm me if you want further clarification on this topic thanks! john
    Why did you thank him
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    (Original post by Namch)
    Why did you thank him
    I was being courteous.
 
 
 
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