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Thread: estimation of parameter

  1. #1
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    estimation of parameter

    consider the probability mass function
    $\displaystyle p_{X} (x; \theta) = \theta ^{x} (1 - \theta )^{1-x}, x = 0,1$
    and suppose we observe a sample of 5 outcomes: 0, 0, 1, 0, 1.
    (a) show how maximum likelihood estimation can be used to estimate the parameter $\displaystyle \theta$
    (b) show how the method of moments can be used to estimate the parameter $\displaystyle \theta$
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
    consider the probability mass function
    $\displaystyle p_{X} (x; \theta) = \theta ^{x} (1 - \theta )^{1-x}, x = 0,1$
    and suppose we observe a sample of 5 outcomes: 0, 0, 1, 0, 1.
    (a) show how maximum likelihood estimation can be used to estimate the parameter $\displaystyle \theta$
    (b) show how the method of moments can be used to estimate the parameter $\displaystyle \theta$
    (a) Well what is the likelihood of exactly 2 1'1 in 5 trials?

    (b) Set the expected number of 1's in 5 trials to the observed result.

    CB
    Last edited by CaptainBlack; Dec 15th 2010 at 11:18 AM.
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    im still confused on how to start this question?
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    Quote Originally Posted by CaptainBlack View Post
    (a) Well what is the likelihood of exactly 2 1'1 in 5 trials?
    Consider the 1's as successes then and $\displaystyle $$N$ the number of successes in 5 trials:

    $\displaystyle L(\theta|N=2)=k \theta^N (1-\theta)^{5-N}=k \theta^2 (1-\theta)^{3}$

    Then the maximum liklyhood estimator is the largest $\displaystyle $$\Theta$ which solves:

    $\displaystyle \dfrac{d}{d\theta}L(\theta|N=2)=0$

    CB
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  5. #5
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    i got $\displaystyle \theta = \frac{2}{5}$ is this correct?
    also, im still confused as what to do for (b), how do i calculate the expected number of 1's in 6 trials?
    thanks so much!
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  6. #6
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    Quote Originally Posted by wik_chick88 View Post
    i got $\displaystyle \theta = \frac{2}{5}$ is this correct?
    Yes
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  7. #7
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    Quote Originally Posted by wik_chick88 View Post
    iim still confused as what to do for (b), how do i calculate the expected number of 1's in 6 trials?
    thanks so much!
    We are setting $\displaystyle p_X(1;\theta)=\theta=2/5$

    (we are setting the mean of X to the observed proportion in our sample, which is matching first monents of the distribution of number of 1's in a sample with the observed)

    CB
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  8. #8
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    im confused! how do i know (or guess) whether the 6th outcome will be a 0 or a 1?
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  9. #9
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    Quote Originally Posted by wik_chick88 View Post
    im confused! how do i know (or guess) whether the 6th outcome will be a 0 or a 1?
    The 6 is a typo for 5

    CB
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