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Maximum Likelihood Estimation

  • Thread starter superwolf
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  • #1
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[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]
 
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Answers and Replies

  • #2
392
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Mostly right, with some typos.

[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]
\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 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]
[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.
 
  • #3
176
0
I'll try.
 
Last edited:
  • #4
176
0
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?
 
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  • #5
392
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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:

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