What is the Confidence Interval Formula for Maximum Likelihood Estimation?

[superwolf]

<br /> 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}<br />

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

<br /> 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})<br />

<br /> \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}<br />

 

[Billy Bob]

Mostly right, with some typos.

<br /> 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})<br />

<br /> \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})<br />

<br /> \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}<br />

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

Now combine into a single fraction, set equal to zero, and solve for theta.

 

[superwolf]

I'll try.

 

[superwolf]

It worked! Great stuff, thanks!


<br /> \hat{\theta}=\frac{2}{n}\Sigma^n_{i=1}X_i<br />

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

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

Can I use this formula?

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

\hat{\theta}=0.64, 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 \sqrt{n} in the formula above. Is the formula wrong?

 

[Billy Bob]

You used the wrong formula. You have to derive your own confidence interval formula.

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

Then P(-1.96&lt;\frac{\hat{\theta} - \theta}{\sqrt{\frac{2 \hat{\theta}}{n}(1-\frac{\hat{\theta}}{2})}} &lt;1.96 )=0.95

Now rearrange to get

P(\text{thing}&lt;-\theta&lt;\text{other thing})=0.95

and then finally

P(-\text{thing}&gt;\theta&gt;-\text{other thing})=0.95