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Math Help - Likelihood Ratio Test??

  1. #1
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    Likelihood Ratio Test??

    Let X1, X2, ..., X10 be a random sample from a normal distribution
    with mean = zero and variance = sigma^2. Find a best test of size alpha = 5%
    for testing H_o: sigma^2 = 1 against H_a: sigma^2 = 2. Is this a best test of
    size alpha = 5% for testing H_o: sigma^2 = 1 against H_a: sigma^2 = 4?

    Is this a best test of
    size alpha = 5% for testing
    H_o: sigma^2 = 1 against H_a: sigma^2 = # where # greater than 1?

    I'm lost on how to find a best test for this. I'm wondering if I need to use Likelihood Ratio Test in order to compute it? Please help!!
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by AwesomeHedgehog View Post
    Let X1, X2, ..., X10 be a random sample from a normal distribution
    with mean = zero and variance = sigma^2. Find a best test of size alpha = 5%
    for testing H_o: sigma^2 = 1 against H_a: sigma^2 = 2. Is this a best test of
    size alpha = 5% for testing H_o: sigma^2 = 1 against H_a: sigma^2 = 4?

    Is this a best test of
    size alpha = 5% for testing
    H_o: sigma^2 = 1 against H_a: sigma^2 = # where # greater than 1?

    I'm lost on how to find a best test for this. I'm wondering if I need to use Likelihood Ratio Test in order to compute it? Please help!!
    You want to use the maximum likelihood to determine the nature of your test. For symmetric distributions like we have here the nature of your test is either going to be one sided, when the means are not equal, or two sided, when they are. You can convince yourself of this by examining the likelihood ratio of your H0 and H1 distribution.

    A best test then for identical mean RV's looks at the how much of the area of the distribution is centered vs. at the tails. Higher standard deviation distributions are going to have more area in the tails.

    You would first form the single normal rv corresponding to the sum of your 10 X's scaled by 1/10, i.e. the mean of your X's.

    Then you have a HO accept region from [-c,c] and an H0 reject region of $(\infty, -c) \cup (c, \infty)$

    You need to specify c such that the alpha of the test (false alarm, Type I error, etc.) is 0.05.

    This is the region where you've chosen H1 when H0 was actually true. In this case this is simply the integral of the H0 distribution for |x|>c.

    Given c you can then find the power of the test as the integral over this same region for the H1 distribution.

    Given all this do you think the c's will be equal if the the standard deviations change?
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    Re: Likelihood Ratio Test??

    http://www.stat.ucla.edu/~hqxu/stat105/pdf/ch09.pdf

    On this link on pages 58 - 59 it shows the Likelihood Ratio Test for the normal distribution. I'm wondering if I just insert the values of H_0 and H_a into the equation?
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by romsek View Post
    You want to use the maximum likelihood to determine the nature of your test. For symmetric distributions like we have here the nature of your test is either going to be one sided, when the means are not equal, or two sided, when they are. You can convince yourself of this by examining the likelihood ratio of your H0 and H1 distribution.

    A best test then for identical mean RV's looks at the how much of the area of the distribution is centered vs. at the tails. Higher standard deviation distributions are going to have more area in the tails.

    You would first form the single normal rv corresponding to the sum of your 10 X's scaled by 1/10, i.e. the mean of your X's.

    Then you have a HO accept region from [-c,c] and an H0 reject region of $(\infty, -c) \cup (c, \infty)$

    You need to specify c such that the alpha of the test (false alarm, Type I error, etc.) is 0.05.

    This is the region where you've chosen H1 when H0 was actually true. In this case this is simply the integral of the H0 distribution for |x|>c.

    Given c you can then find the power of the test as the integral over this same region for the H1 distribution.

    Given all this do you think the c's will be equal if the the standard deviations change?
    Oh, so I want to find the value of c such that the P[Type I Error] = 0.05?

    So, I would want to do: P[ Σxi >= c ] = 0.05

    You said to do an integral for it, so my idea would be to do the integral from 0 to 10 of the normal distribution function

    But, that doesn't seem correct and the reason I'm struggling with how to figure out the integral is because my teacher just always uses the computer software to find the values, he never showed us how to do it by hand.
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    Re: Likelihood Ratio Test??

    you want section 9.4 of your pdf. It describes the test for equal mean, different variance, hypotheses which is what you have here.
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by romsek View Post
    you want section 9.4 of your pdf. It describes the test for equal mean, different variance, hypotheses which is what you have here.
    So are you saying that I want to solve for what the chi-square distribution is? Just plug in my values into the equation it gives for chi-square?
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by AwesomeHedgehog View Post
    So are you saying that I want to solve for what the chi-square distribution is? Just plug in my values into the equation it gives for chi-square?
    I want you to push your pencil a bit. If I have to you should too.

    Derive what what the pdfs are for the joint distributions of N iid normal(0,$\sigma_0$) and normal(0,$\sigma_a$) rvs are

    take the likelihood ratio of that, i.e. the ratio of those two pdfs

    reduce all the algebra to see why the sum of the squares of your observations is the test statistic for this problem.

    this is a good exercise. you should do it.
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    Re: Likelihood Ratio Test??

    If I'm doing the likelihood ratio test, wouldn't it be this:

    L(0, sigma_0) = pi summation from i = 1 to n of 1/(sqrt[2*pi* σ^2]) * e^[-0.5 *((xi - 0)^2/( σ^2))]

    = [1/(sqrt(2*pi* σ^2)]^n * e^(- ∑(xi)^2/(2*σ^2))

    All of the sigmas above would be sigma_0

    And for L(0, sigma_1), it would be the same thing, except the sigmas would be sigma_1

    Then what I also did was take the ln(L(0, sigma_0))

    Then took the derivative of ln(L(0, sigma_0))

    After the derivative was taken, I was left with:

    ∑(xi)/σ^2 = 0 ==> ∑(xi) = 0

    Which I know is completely wrong.
    Last edited by AwesomeHedgehog; April 29th 2014 at 09:13 AM.
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by AwesomeHedgehog View Post
    If I'm doing the likelihood ratio test, wouldn't it be this:

    L(0, sigma_0) = pi summation from i = 1 to n of 1/(sqrt[2*pi* σ^2]) * e^[-0.5 *((xi - 0)^2/( σ^2))]

    = [1/(sqrt(2*pi* σ^2)]^n * e^(- ∑(xi)^2/(2*σ^2))

    All of the sigmas above would be sigma_0

    And for L(0, sigma_1), it would be the same thing, except the sigmas would be sigma_1

    Then what I also did was take the ln(L(0, sigma_0))

    Then took the derivative of ln(L(0, sigma_0))

    After the derivative was taken, I was left with:

    ∑(xi)/σ^2 = 0 ==> ∑(xi) = 0

    Which I know is completely wrong.
    you're killin me

    $\LARGE \dfrac {L(H_0)}{L(H_a)}=\dfrac{\dfrac 1 {\sigma_0^n}e^{-\frac 1 2 \sum_{i=1}^n \left(\frac {x_i}{\sigma_0}\right)^2}}{\dfrac 1 {\sigma_1^n}e^{-\frac 1 2 \sum_{i=1}^n \left(\frac {x_i}{\sigma_1}\right)^2}}=$

    $\LARGE \dfrac {\sigma_1^n}{\sigma_0^n}e^{-\frac 1 2 \left(\frac 1 {\sigma_0^2}-\frac 1 {\sigma_1^2} \right) \sum_{i=0}^n x_i^2}$

    this shows that your test statistic is ultimately a function of the sum of the squares of your observed random variables. The rest of it doesn't really matter that much since you are going to set your threshold based on alpha anyway. You can threshold on $S^2= \sum_{i=0}^n x_i^2$ as well as you can some function of it. The fact that the power of the test for a given alpha is stronger with greater difference between $\sigma_0$ and $\sigma_1$ is shown when you compute the power of the test.

    The fact that the test statistic is the sum of squares of normal variates gets you to using the Chi-Square test mentioned in your text.
    Thanks from AwesomeHedgehog
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    Re: Likelihood Ratio Test??

    Sorry, I don't mean to frustrate you. Trust me, it's frustrating me too. I wish my teacher actually taught instead of just plugging the numbers into the computer software and letting the computer do all of the work. I think I'm slowly starting to understand what's going on.

    So, since

    Ho: σ^2 = 1

    Ha: σ^2 = 2

    Would I insert them into the equation L(Ho) / L(Ha) ≤ k ?
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by AwesomeHedgehog View Post
    Sorry, I don't mean to frustrate you. Trust me, it's frustrating me too. I wish my teacher actually taught instead of just plugging the numbers into the computer software and letting the computer do all of the work. I think I'm slowly starting to understand what's going on.

    So, since

    Ho: σ^2 = 1

    Ha: σ^2 = 2

    Would I insert them into the equation L(Ho) / L(Ha) ≤ k ?
    Not really. The likelihood ratio gives you the form of the test statistic. You will then test that statistic against a threshold.

    You set the threshold, in these problems anyway, according to the Type I error $\alpha$. You are told what $\alpha$ you want to achieve and set the threshold based on that.

    So at the end of the day you just have

    $\begin{cases}S^2 \leq \eta(\alpha) &\text{Decide }H_0 \\S^2> \eta(\alpha) &\text{Decide }H_a\end{cases}$

    where $S^2 = \sum X^2_i$ and $\eta(\alpha)$ is the threshold that gives Type I error $\alpha$
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by romsek View Post
    Not really. The likelihood ratio gives you the form of the test statistic. You will then test that statistic against a threshold.

    You set the threshold, in these problems anyway, according to the Type I error $\alpha$. You are told what $\alpha$ you want to achieve and set the threshold based on that.

    So at the end of the day you just have

    $\begin{cases}S^2 \leq \eta(\alpha) &\text{Decide }H_0 \\S^2> \eta(\alpha) &\text{Decide }H_a\end{cases}$

    where $S^2 = \sum X^2_i$ and $\eta(\alpha)$ is the threshold that gives Type I error $\alpha$
    So would that be the answer to the problem?? And then all that's left to do would be to discuss whether or not that equation is the best test for alpha?
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by AwesomeHedgehog View Post
    So would that be the answer to the problem?? And then all that's left to do would be to discuss whether or not that equation is the best test for alpha?
    you know it's the most powerful test for Type I error of 0.05. That's the point of the likelihood ratio, they provide the best tests.

    You still have to find the threshold $\eta(\alpha)$ that gives you $\alpha = 0.05$
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by romsek View Post
    you know it's the most powerful test for Type I error of 0.05. That's the point of the likelihood ratio, they provide the best tests.

    You still have to find the threshold $\eta(\alpha)$ that gives you $\alpha = 0.05$
    Okay, now you lost me. So, I have to find the power that gives 0.05?
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    Re: Likelihood Ratio Test??

    Quote Originally Posted by AwesomeHedgehog View Post
    Okay, now you lost me. So, I have to find the power that gives 0.05?
    not the power. The threshold you compare your test statistic to to decide whether to declare H0 or Ha
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