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Maximum Likelihood Estimation

  1. Apr 18, 2009 #1
    [tex]
    L(x_1,x_2,...,x_n;\theta)=\Pi _{i=1}^n (\frac{\theta}{2})^x (1-\frac{\theta}{2})^{1-x} = (\frac{\theta}{2})^{\Sigma^n_{i=1}x_i}(1-frac{\theta}{2})^{n-\Sigma^n_{i=1}x_i}
    [/tex]

    Correct so far if [tex]f(x) = (\frac{\theta}{2})^x (1-\frac{\theta}{2})^{1-x} [/tex]?

    [tex]
    lnL(x_1,x_2,...,x_n;\theta) = \Sigma^n_{i=1} x_i ln(\frac{\theta}{2}) + (n-\Sigma_{i=1}x_i) ln(\frac{1}{2 - \theta})
    [/tex]

    [tex]
    \frac{d lnL}{d \theta}(x_1,x_2,...,x_n;\theta) = Sigma^n_{i=1}x_i \cdot \frac{1}{\theta - }(n-\Sigma^n_{i=1}x_i) \frac{1}{2-\theta}
    [/tex]
     
    Last edited: Apr 18, 2009
  2. jcsd
  3. Apr 18, 2009 #2
    Mostly right, with some typos.

    [tex]
    \ln L(x_1,x_2,...,x_n;\theta) = \Sigma_{i} x_i \ln(\frac{\theta}{2}) + (n-\Sigma_{i}x_i) \ln(\frac{2 - \theta}{2})
    [/tex]

    [tex]
    \frac{d\ln L}{d \theta}(x_1,x_2,...,x_n;\theta) =\frac{\Sigma_{i}x_i}{\theta}\,
    -\,\frac{n-\Sigma_{i}x_i}{2-\theta}
    [/tex]

    Now combine into a single fraction, set equal to zero, and solve for theta.
     
  4. Apr 18, 2009 #3
    I'll try.
     
    Last edited: Apr 18, 2009
  5. Apr 18, 2009 #4
    It worked! Great stuff, thanks!


    [tex]
    \hat{\theta}=\frac{2}{n}\Sigma^n_{i=1}X_i
    [/tex]

    Now, [tex]E(\hat{\theta}) = \theta[/tex] and [tex]Var(\hat{\theta})=\frac{2 \theta}{n}(1-\frac{\theta}{2})[/tex].

    Find a 95% confidence interval for [tex]\theta[/tex] by using that [tex]\frac{\hat{\theta} - \theta}{\sqrt{\frac{2 \hat{\theta}}{n}(1-\frac{\hat{\theta}}{2})}}[/tex]

    Can I use this formula?

    244qs9g.jpg

    With n=100 and [tex]\Sigma^n_{i=1}X_i = 32[/tex] I get

    [tex]\hat{\theta}=0.64[/tex], but using the formula above gives doesn't give me the correct interval, which is [0.46, 0.82]

    Edit: I get the right interval when I don't include the [tex]\sqrt{n}[/tex] in the formula above. Is the formula wrong?
     
    Last edited: Apr 18, 2009
  6. Apr 18, 2009 #5
    You used the wrong formula. You have to derive your own confidence interval formula.

    I assume you meant to say that [tex]\frac{\hat{\theta} - \theta}{\sqrt{\frac{2 \hat{\theta}}{n}(1-\frac{\hat{\theta}}{2})}}[/tex] is standard normal.

    Then [tex]P(-1.96<\frac{\hat{\theta} - \theta}{\sqrt{\frac{2 \hat{\theta}}{n}(1-\frac{\hat{\theta}}{2})}} <1.96 )=0.95[/tex]

    Now rearrange to get

    [tex]P(\text{thing}<-\theta<\text{other thing})=0.95[/tex]

    and then finally

    [tex]P(-\text{thing}>\theta>-\text{other thing})=0.95[/tex]
     
    Last edited: Apr 18, 2009
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