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Math Help - Confidence Interval for Variances

  1. #1
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    Confidence Interval for Variances



    I understand that

    \frac{(n-1)S^2}{\sigma ^2} = \frac{(n-1)\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2}{\sigma ^2} ~ \chi ^2(n-1)

    I do not understand why

    \frac{n\hat{\sigma}^2}{\sigma ^2} = \frac{n\frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2}{\sigma ^2} ~ \chi ^2(n-2)
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  2. #2
    MHF Contributor matheagle's Avatar
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    I need to know more about this setting.
    It's unclear why you have n-2 degrees of freedom, most likely you're estimating TWO parameters.

    And what is \hat\sigma^2 here?
    \hat\sigma^2 can be any estimator of \sigma^2
    that most likely will answer the n-2 df question.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    I need to know more about this setting.
    It's unclear why you have n-2 degrees of freedom, most likely you're estimating TWO parameters.

    And what is \hat\sigma^2 here?
    \hat\sigma^2 can be any estimator of \sigma^2
    that most likely will answer the n-2 df question.
    Thanks so much, I figured that wouldn't be enough, but I wasn't sure.

    I'm working on regressions. So, x's are the known constants, and the y's are the random variables.

    \hat\sigma^2 = \sum_{i = 1}^{n}(y_i - \hat y)^2 ,

    and,

     \hat y = \hat\alpha + \hat\beta (x - \bar x)
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  4. #4
    MHF Contributor matheagle's Avatar
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    I figured that.
    You should have said you were estimating two parameters.
    That's why you have 2 degrees of freedom in the chi-square.
    \hat\sigma^2 is the MSE here AND not the usual sample variance.
    And I'm not so sure about the n in the formula.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    I figured that.
    You should have said you were estimating two parameters.
    That's why you have 2 degrees of freedom in the chi-square.
    \hat\sigma^2 is the MSE here AND not the usual sample variance.
    And I'm not so sure about the n in the formula.
    Thanks! So basically, whats the rule with chi-squared distributions and degrees of freedom. My book is getting very frustrating because they're starting to use certain concepts a little to liberally without explaining where they came from or why they are in the form the are; this is a good example.

    Thanks again.
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  6. #6
    Super Member Random Variable's Avatar
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     Y_{i} follows a normal distribution.

    so  \Big(\frac{Y_{i}-\bar{Y_{i}}}{\sigma}\Big)^{2} follows a chi-square distribution with 1 degree of freedom

    and  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\bar{Y_{i}})}{\sigma}\Big)^{2} follows a chi-square distribution with n degrees of freedom (since you are summing n independent chi-square variables)

    so you would expect that  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\hat{Y_{i}})}{\sigma}\Big)^{2} = \frac{n \hat{\sigma^{2}}}{\sigma^{2}} to follow a chi-square distribution with n-2 degrees of freedom because you lose two degrees of freedom estimating the two regression coefficients
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  7. #7
    MHF Contributor matheagle's Avatar
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    Just to clarify...
    when you write something like \hat\sigma^2 the idea is that it estimates \sigma^2

    so really you should have something like \hat\sigma^2={\sum(Y_i-\hat Y)^2\over n}

    or really the MSE here is \hat\sigma^2={\sum(Y_i-\hat Y)^2\over n-2}

    since MSE=SSE/(n-2) is unbiased for \sigma^2
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  8. #8
    MHF Contributor matheagle's Avatar
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    I don't agree with this...........

    Quote Originally Posted by Random Variable View Post
     Y_{i} follows a normal distribution.

    so  \Big(\frac{Y_{i}-\bar{Y_{i}}}{\sigma}\Big)^{2} follows a chi-square distribution with 1 degree of freedom

    and  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\bar{Y_{i}})}{\sigma}\Big)^{2} follows a chi-square distribution with n degrees of freedom (since you are summing n independent chi-square variables)

    so you would expect that  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\hat{Y_{i}})}{\sigma}\Big)^{2} = \frac{n \hat{\sigma^{2}}}{\sigma^{2}} to follow a chi-square distribution with n-2 degrees of freedom because you lose two degrees of freedom estimating the two regression coefficients
    I agree with ....

     \Big(\frac{Y_{i}-E{Y_{i}}}{\sigma}\Big)^{2} follows a chi-square distribution with 1 degree of freedom

    and
    and \frac{\sum_{i=1}^{n}(Y_{i}-\bar{Y_{i}})^2}{\sigma^2} follows a chi-square distribution with n-1 degrees of freedom

    while

    and  {\sum_{i=1}^{n}(Y_{i}-E(Y))^2\over \sigma^2} follows a chi-square distribution with n degrees of freedom (since you are summing n independent chi-square variables)
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  9. #9
    Super Member Random Variable's Avatar
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    In response to matheagle, let me change a few things:

     Y_{i} follows a normal distribution.

    so  \Big(\frac{Y_{i}-\alpha - \beta x_{i}}{\sigma}\Big)^{2} follows a chi-square distribution with 1 degree of freedom

    and  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\alpha-\beta x_{i}) }{\sigma}\Big)^{2} follows a chi-square distribution with n degrees of freedom (since you are summing n independent chi-square variables)

    so you would expect that  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\hat{\alpha}-\hat{\beta} x_{i})}{\sigma}\Big)^{2} = \frac{n \hat{\sigma^{2}}}{\sigma^{2}} to follow a chi-square distribution with n-2 degrees of freedom because you lose two degrees of freedom estimating the two regression coefficients
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  10. #10
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    Quote Originally Posted by Random Variable View Post
    In response to matheagle, let me change a few things:

     Y_{i} follows a normal distribution.

    so  \Big(\frac{Y_{i}-\alpha - \beta x_{i}}{\sigma}\Big)^{2} follows a chi-square distribution with 1 degree of freedom
    Got it.

    Quote Originally Posted by Random Variable View Post
    and  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\alpha-\beta x_{i}) }{\sigma}\Big)^{2} follows a chi-square distribution with n degrees of freedom (since you are summing n independent chi-square variables)
    Sure.

    Quote Originally Posted by Random Variable View Post
    so you would expect that  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\hat{\alpha}-\hat{\beta} x_{i})}{\sigma}\Big)^{2} = \frac{n \hat{\sigma^{2}}}{\sigma^{2}} to follow a chi-square distribution with n-2 degrees of freedom because you lose two degrees of freedom estimating the two regression coefficients
    What? How do you lose degrees of freedom? I think I'm lacking a fundamental understanding of just what a degree of freedom is. I understand how it relates to the Gamma distribution and it is actually the mean of the Chi-square distribution. But I wouldn't know something followed a Chi-square distribution if it was staring me in the face, let alone how many degrees of freedom it employs.
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  11. #11
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Random Variable View Post
    In response to matheagle, let me change a few things:

     Y_{i} follows a normal distribution.

    so  \Big(\frac{Y_{i}-\alpha - \beta x_{i}}{\sigma}\Big)^{2} follows a chi-square distribution with 1 degree of freedom

    and  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\alpha-\beta x_{i}) }{\sigma}\Big)^{2} follows a chi-square distribution with n degrees of freedom (since you are summing n independent chi-square variables)

    so you would expect that  \Big(\frac{\sum_{i=1}^{n}(Y_{i}-\hat{\alpha}-\hat{\beta} x_{i})}{\sigma}\Big)^{2} = \frac{n \hat{\sigma^{2}}}{\sigma^{2}} to follow a chi-square distribution with n-2 degrees of freedom because you lose two degrees of freedom estimating the two regression coefficients

    Don't you think that the square should be inside the sum not outside the sum?
    Then I agree, because \alpha is a parameter (an expected value) while \hat\alpha is a statistic.
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  12. #12
    Super Member Random Variable's Avatar
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    Degrees of freedom can be looked at as the number of independent pieces of information that go into estimating a parameter. By first estimating the 2 regression coefficients, you are constraining the residuals to lie within a certain vector space defined by the 2 normal equations. Therefore, you are working with two less independent pieces of information.
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  13. #13
    MHF Contributor matheagle's Avatar
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    and MSE is really SSE/(n-2) not SSE/n.
    That makes it unbiased for \sigma^2
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  14. #14
    Super Member Random Variable's Avatar
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    Quote Originally Posted by matheagle View Post
    Don't you think that the square should be inside the sum not outside the sum?
    Yes. Anything else I did wrong?
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