# Confidence interval

1. Sep 11, 2008

### twoflower

Let's say we know this:

$$\sqrt{n}\left(\widehat{\theta} - \theta\right) \sim \mathcal{N}\left(0, \frac{1}{F(\theta)}\right)$$

How do we get from this information to this expression of confidence interval for $\theta$?

$$\left( \widehat{\theta} \pm u_{1-\frac{\alpha}{2}}\frac{1}{\sqrt{nF\left(\widehat{\theta}\right)}}\right)$$

Where $u_{1-\frac{\alpha}{2}}$ is appropriate quantil of standard normal distribution.

Thank you.

2. Sep 11, 2008

If $$a$$ is the value from $$Z$$ (standard normal) with area $${\alpha}/2$$ to its right, you know the value of

$$\Pr\left(-u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u)$$

because of your stated approximate normality result. That means the event

$$-u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u$$

has a known probability of occurring. What can you do with this inequality? (Try some work and include it with your next question if you are unsure.)