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Math Help - Another UMPT question.

  1. #1
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    Another UMPT question.

    I have this sample exam question and no model answer.

    Question.

    Let Y_1,...Y_n be a random sample from a distribution from a density function

    f_Y(y;\theta)=\frac{2y}{\theta}e^{\frac{-y^2}{\theta}}, y>0
    f_Y(y;\theta)=0, otherwise

    where \theta>0 is an unknown parameter.

    i. Show that the uniformly most powerful test (UMPT) for testing hypotheses

    H_0:\theta=1, H_1:\theta>1

    will reject Ho if and only if T>K for some constant K>0, where T=\Sigma_{i=1}^nY_i^2.

    ii. Find the distribution of the test statistic T under Ho. [You may use the fact that the \chi^2 distribution with k degrees of freedom has moment generating function (1-2t)^{-k/2}.

    Answer.

    i. The likelihood function is

    L_Y=(\frac{2}{\theta})^n\prod_{i=1}^n*y_i*e^{-\frac{1}{\theta}\Sigma_{i=1}^ny_i^2}

    The likelihood ratio from Neyman-Person lemma is

    \frac{L_Y(theta_1;y)}{L_Y(\theta_0;y)}=\frac{(\fra  c{2}{\theta})^n\prod_{i=1}^ny_ie^{-\frac{1}{\theta}\Sigma_{i=1}^ny_i^2}}{2^n\prod_{i=  1}^ny_ie^{\Sigma_{i=1}^ny_i^2}}=\theta^{-n}e^{{\Sigma_{i=1}^ny_i^2(-\frac{1}{\theta}-1)}}=\theta^{-n}e^{T(-\frac{1}{\theta}-1)}

    If this ratio is higher than some number k (determined by required confidence level), then we will reject Ho:

    -\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k

    take logs

    ln(-\theta^{-n})+T(-\frac{1}{\theta}-1)>ln(k)

    T\frac{-1-\theta}{\theta}>ln(k)-ln(-\theta^{-n})

    T>\frac{\theta}{1+\theta}(ln(-\theta^{-n}-ln(k))=K

    and the required T>K is the above inequality.

    does it make sense?


    ii. Distribution of T - let me think it over a bit longer. I am thinking to find MGF of y^2 (with theta=1 the density is simplifed) and to show that it is a form of MGF of the chi squared.
    Last edited by mr fantastic; March 14th 2011 at 04:04 AM. Reason: Title.
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  2. #2
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    part ii.

    ii. Distribution function of T

    Let Z=Y^2, T=\Sigma_{i=1}^nZ_i

    then M_Z(t)=E(e^{Zt})=E(e^{Y^2t})=\int_0^{\infty}e^{y^2  t}2ye^{-y^2}dy=\int_0^{\infty}e^{y^2(t-1)}2ydy=\int_0^{\infty}e^{z(t-1)}dz=\frac{1}{1-t}

    (I used the fact that \int_0^{\infty}e^{-u}du=1)

    Then Z_1,...Z_n are independent, identically distributed rvs, and therefore I can multiply their MGFs to get

    M_T(t)=(M_Z(t))^n=(\frac{1}{1-t})^n

    If I let k=2n, then this can be represented as

    (\frac{1}{1-t})^{k/2}=(1-t)^{-k/2}

    Now i need to find an error in above because this should be the 2xparameter in the expression... OR, could this still be a version of a chi-square distribution with some adjustment to [what? random variable? parameter?]
    Last edited by Volga; March 14th 2011 at 01:52 AM.
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  3. #3
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Volga View Post

    -\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k

    take logs

    ln(-\theta^{-n})+T(-\frac{1}{\theta}-1)>ln(k)
    This is not allowed since log of a negative number is not defined.

    Rather, write:


    -\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k

    e^{T(-\frac{1}{\theta}-1)} < k' (inequality changes since multiplying by a negative number)

    T(-\frac{1}{\theta}-1) < k''

    -T(\frac{1}{\theta}+1) < k'''

    T > k'''' (inequality changes since multiplying by a negative number)

    Calculate k'''' from here.

    NOTE: you need not write the simplified form of the constant in the RHS.
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  4. #4
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    Thanks! Good point!

    Quote Originally Posted by Sambit View Post
    NOTE: you need not write the simplified form of the constant in the RHS.
    Do you mean I don't need to work out the expression for K (as I did above), and it is sufficient to show that K is some constant?
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  5. #5
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Volga View Post
    Do you mean I don't need to work out the expression for K (as I did above), and it is sufficient to show that K is some constant?
    Yes that is sufficient. Finally you will determine k'''' from the required confidence level; the value of k is irrelevant there; only k'''' will suffice.
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  6. #6
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    Any help on part ii? I am still having t instead of 2t...
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  7. #7
    Senior Member Sambit's Avatar
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    I can't find any mistake in your solution. Your answer may not be wrong since the question does not say that the answer will involve 2t. They have only given a hint which you may or MAY NOT use. We did not use it.
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  8. #8
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    Oh cheers! I was wondering if I completely misunderstand the MGF concept )))
    then I would just say the distribution is defined by this MGF function (since it uniquely defines a distribution)
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