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Math Help - method of moments

  1. #1
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    method of moments


    f(y| \theta)= {e}^{[-(-(y-\theta))]}, y≥ \theta where theta is an unknown positive constant.

    Find a moment estimator for theta and adjust theta hat to find an unbiased estimator for theta.


    Solving for E(Y) I get y{e}^{y-x}-{e}^{y-x}+{e}^{-x} which is my sample average (Y-bar).

    Am I on the right track?
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  2. #2
    MHF Contributor matheagle's Avatar
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    I cannot understand your exponent.
    It looks like absolute values, but I doubt you want that.
    I bet you mean e^{-(y-\theta)} when y\ge \theta.
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  3. #3
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    That's right. Those are supposed to be braces
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  4. #4
    MHF Contributor matheagle's Avatar
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    BUT you have 2 negative signs, which means your exponent blows up.
    Hence you have infinite area.
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  5. #5
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    yes the integration used to obtain E(Y) goes from /theta to infinity. Oops I was going from 0 to y. This is tricky. I have two other problems related to this one. An explanation of this problem will hopefully help with the others.
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  6. #6
    MHF Contributor matheagle's Avatar
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    PLEASE type the correct density and the bounds on the variable.
    I don't know why you have 2 negative signs.
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  7. #7
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    ok my hw has a typo. Just got confirmation its e^(-(y-theta)) for the density function.
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  8. #8
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    now I have ye^(-y+x) + e^(-theta + x) as my E(Y)

    not exactly sure what the next step would be.
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  9. #9
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    setting the above equal to Ybar and solving for theta I get theta = -ln(Ybar-ye^(-y+x)) -x
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  10. #10
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Snarf? View Post
    ok my hw has a typo. Just got confirmation its e^(-(y-theta)) for the density function.
    I stated that several hours ago.
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  11. #11
    MHF Contributor matheagle's Avatar
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    This is getting weird.
    What's x now???
    ye^(-y+x) + e^(-theta + x) does not make sense.
    I BET you'e "integrating" x times f(y).
    THERE is only ONE variable.
    Either use x or y, it doesn't matter, these are dummy variables.
    JUST integrate yf(y) from theta to infinity.
    When you integrate y and plug in bounds, the y is no longer there.
    And this x is purely fictional.
    you need to review calculus.
    You should have ONLY a function of theta as the expectation of Y and you set that to the sample mean.
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  12. #12
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    Somewhere along the way during integration by parts I substituted x for theta and forgot to change it. I do that type of thing a lot. Anywho, now I have ye^(-y+theta) + 1 for my E(Y)
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  13. #13
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Snarf? View Post
    Somewhere along the way during integration by parts I substituted x for theta and forgot to change it. I do that type of thing a lot. Anywho, now I have ye^(-y+theta) + 1 for my E(Y)
    I assume you're using an integral calculator.
    No wonder I ban them in my classroom.
    ONCE again PLUG in your bounds.
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  14. #14
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    the bounds I used are from theta to infinity. Is this correct?

    integration by parts:

    ye^(-y+theta) - integral from theta to infinity of e^(-y+theta) dy

    = ye^(-y+theta) -(e^(-y+theta)| (from theta to infinity)

    = ye^-(y+theta)-(0-e^(-theta + theta)
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  15. #15
    MHF Contributor matheagle's Avatar
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    separate the exponent e^theta and e^(-y) and PULL that outside the integral
    then integrate ye^(-y) via parts and plug in theta and let y go to infinity for your bounds.
    WHEN you're done there can't be any more y's, only theta's.
    Then set that equal to the sample mean.
    AND
    e^(-theta + theta) =e^0=1
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