# October 26

$CI(\theta,\alpha)$
Need to satisfy the following to be considered a confidence interval:
(imagine a normal distribution graph)
Middle portion of the graph $=1-\alpha$
Left portion $=\alpha_1$
Left boundary $=Z_{1-\alpha+\alpha_1}=Z_{\alpha_1}$
Right portion $=\alpha_2$
Right boundary $=Z_{1-\alpha_2}$
$\alpha=\alpha_1+\alpha_2$

### Case 1

$\alpha_1=0$
$\alpha_2=\alpha$
(insert graphic of normal distribution)
Left boundary $Z_{\alpha_1}=-\infty$
Right boundary $Z_{1-\alpha_2}=Z_{1-\alpha}$
Right portion $\alpha$

Called a one-sided confidence interval.

### Case 2
$\alpha_1=\alpha$
$\alpha_2=0$

Left boundary $Z_{\alpha_1}=Z_\alpha$
Left area $\alpha$
Right boundary $Z_{1-\alpha_2}=Z_1=+\infty$

One-sided confidence interval

### One-sided Confidence Intervals Examples

#### Estimate the "mean income" of households.
$x_1...x_n \sim f_x(x;\theta)$

sample $n=4$
$x_1=0$
$x_2=60k$
$x_3=0$
$x_4=120k$

$\bar{X}=45k$
$CI[-5k,95k]*m$
$m=$ some multiplier

Bounded on the left since you cannot have negative income, so its a one-sided confidence interval.

#### How to solve

$\theta$ population
$x_1...x_n \sim^{iid} f_x(x,\theta)$

Want: reasonable idea for what values $\theta$ takes
$CI(\theta,\alpha)=[l_\alpha,U_\alpha]$
$\mathbb{P}(l_\alpha\le \theta\le U_\alpha)=1-\alpha$

Steps:
1. Pick an estimator
2. find the sampling distribution
3. find a transformation $H(\hat{\theta},\theta)$ such that $y=H(\hat{\theta},\theta)\sim f_y(y)$ doesn't depend on $\theta$
4. $\mathbb{P}(y_{\alpha_2}\le y\le y_{1-\frac{\alpha}{2}})$
5. Find $H^{-1}(\hat{\theta},x)$ and use it to compute:
$l_\alpha=H^{-1}(\hat{\theta},y_{\alpha_1})$
$U_\alpha=H^{-1}(\hat{\theta},y_{1-\alpha_2})$
$\mathbb{P}(H^{-1}(\hat{\theta},y_{\alpha_1})\le \theta \le H^{-1}(\hat{\theta},y_{1-\alpha_2}))=1-\alpha$

##### Examples

Unless the question explicitly asks to calculate it, you can use the central limit thereom to estimate.

###### 1
$x_1...x_n\sim N(\theta,\sigma^2)$
$CI(\theta,\alpha)=[\bar{X}-\tilde{Z}_{\alpha_2}\frac{\sigma}{\sqrt{n}},\bar{X}+\tilde{Z}_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}]$

###### 2
$x_1...x_n\sim N(\mu,\theta^2)$
$CI(\theta^2;\alpha)=[\frac{(n-1)s^2}{\gamma^2_{(n-1),\frac{\alpha}{2}}}]$

###### 3
$x_1...x_n\sim Ber(\theta)$
$\hat{\theta}=\bar{X}$
$CI(\theta,\alpha) \approx []$
###### 4
###### 5
###### 6