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Math Help - random sample MLE question

  1. #1
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    random sample MLE question

    Given that X1 = 2.3, X2 = 1.9 and X3 = 4.6 is a random sample from
    fX(x) = x^3e^(-x/a)
    6a^4
    for x greater than or equal to 0, calculate the MLE for a.

    also, can anyone tell me how to copy and paste formulas such as the one above into the question on this website? i hate how messy it looks like this...thanks
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
    Given that X1 = 2.3, X2 = 1.9 and X3 = 4.6 is a random sample from
    fX(x) = x^3e^(-x/a)
    6a^4
    for x greater than or equal to 0, calculate the MLE for a.

    also, can anyone tell me how to copy and paste formulas such as the one above into the question on this website? i hate how messy it looks like this...thanks
    The likelihood function is f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}.


    Now solve \frac{\partial f}{\partial a} = 0 for a in terms of x_1, \, x_2 and x_3.

    I get a = \frac{x_1 + x_2 + x_3}{12} but I'm known to make careless mistakes.

    Now substitute x_1 = 2.3, \, x_2 = 1.9 and x_3 = 4.6.


    If you quote any reply you'll see the latex code. For what it's worth, here's the code I used to generate f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}:


    [tex]f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}[/tex].

    No doubt a latex guru will chip in on how to make it look better.
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    Quote Originally Posted by mr fantastic View Post
    The likelihood function is f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}.


    Now solve \frac{\partial f}{\partial a} = 0 for a in terms of x_1, \, x_2 and x_3.

    I get a = \frac{x_1 + x_2 + x_3}{12} but I'm known to make careless mistakes.

    Now substitute x_1 = 2.3, \, x_2 = 1.9 and x_3 = 4.6.
    [snip]
    I meant to add that you might find it easier to work with the log-likelihood function L = \ln f(x_1, \, x_2, \, x_3).
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    Quote Originally Posted by mr fantastic View Post
    I meant to add that you might find it easier to work with the log-likelihood function L = \ln f(x_1, \, x_2, \, x_3).
    im a bit confused on how to get the partial derivative of f with respect to a in terms of x_1, \, x_2, \, x_3 are you able to show me your steps? and also, if i used L = \ln f(x_1, \, x_2, \, x_3) does that mean i work out partial derivative of L with respect to a instead?
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    Quote Originally Posted by wik_chick88 View Post
    im a bit confused on how to get the partial derivative of f with respect to a in terms of x_1, \, x_2, \, x_3 are you able to show me your steps? and also, if i used L = \ln f(x_1, \, x_2, \, x_3) does that mean i work out partial derivative of L with respect to a instead?
    Yes.

    Use the log-likelihood function rather than the likelihood function to make the calculation simple.

    Note that using the usal log rules you have:

    L = \text{constants} - 12 \ln a - \frac{x_1 + x_2 + x_3}{a}.

    Finding \frac{\partial L}{\partial a} and solving \frac{\partial L}{\partial a} = 0 should be straightforward.
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    Quote Originally Posted by mr fantastic View Post
    Yes.

    Use the log-likelihood function rather than the likelihood function to make the calculation simple.

    Note that using the usal log rules you have:

    [tex]L = \text{constants} - 12 \ln a - math].

    Finding \frac{\partial L}{\partial a} and solving \frac{\partial L}{\partial a} = 0 should be straightforward.
    ok. after i derive it and solve for \frac{\partial L}{\partial a} = 0 i get a = 1/12. but you originally said that you got a = \frac{x_1 + x_2 + x_3}{12}. how do i explain how i substituted x_1 + x_2 + x_3 back in?
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  7. #7
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    Quote Originally Posted by wik_chick88 View Post
    ok. after i derive it and solve for \frac{\partial L}{\partial a} = 0 i get a = 1/12. but you originally said that you got a = \frac{x_1 + x_2 + x_3}{12}. how do i explain how i substituted x_1 + x_2 + x_3 back in?
    \frac{\partial L}{\partial a} = - \frac{12}{a} + \frac{x_1 + x_2 + x_3}{a^2}.

    The derivative of -\frac{k}{a} is not \frac{1}{a^2}, it's \frac{k}{a^2}. I'm guessing that's the mistake you've made ....
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