#StackBounty: #probability #mathematical-statistics #confidence-interval #inference Confidence Intervals: how to formally deal with $P(…

Bounty: 100

I was thinking about the formal definition of confidence intervals.

Given a Random Sample $textbf{X} = X_1,X_2,dots,X_n$ a level $alpha$ confidence interval for the population parameter $theta$ is defined as a pair of estimators (function of $textbf{X}$), namely $L(textbf{X})$ and $U(textbf{X})$ with $L(textbf{X}) leq U(textbf{X})$, with the property:
$$P(L(textbf{X}) leq theta leq U(textbf{X})) = 1-alpha$$

My interpretation of this this last equality is the joint probability of $L(textbf{X})$ and $U(textbf{X})$, i.e.

$$P(L(textbf{X}) leq theta, U(textbf{X})geqtheta) = 1-alpha$$

My question is how to work with this expression to find $L(textbf{X})$ and $U(textbf{X})$?

If I call $f_{L,U}(l,u)$ the joint pdf of $L(textbf{X})$ and $U(textbf{X})$ it should be something like:

$$int_{- infty}^{theta}int_{theta}^{+infty}f_{L,U}(l,u)dl du = 1 – alpha$$

and then I am stucked. I don’t know how to go on.

My first question is: since both $L(textbf{X})$ and $U(textbf{X})$ are function of the same $textbf{X}$ does something like the joint density even make sense?

I don’t know if my calculation is right but I found that something similar to a joint CDF for $L(textbf{X})$ and $U(textbf{X})$ could be (if $X in rm {I!R} $)

$$F_{L,U}(l,u)= F_{textbf{X}}bigg(minbig(L^{-1}(l),U^{-1}(u)big)bigg)$$
if that makes any sense at all. What is the correct way to think about this?
$$———————————–$$
I know that there are methods like pivotal quantities but the difference with respect to my case is that when we have Pivot the probabilistic statement is expressed it terms of only one random variable so I don’t have a joint density.

Suppose that I call $Q(textbf{X},theta)$ my pivot, I can find say $l$ and $u$ such that:
$$Pbig(lleq Q(textbf{X},theta) leq u big) = 1 – alpha$$
and then I can invert this relationship wrt $theta$. I suppose this equality could be rewritten as:

$$F_{Q}(u) – F_{Q}(l) = 1 – alpha$$

This last equation has infinite solutions since I have 2 unknowns $l$ and $u$. My understanding is that, in order to solve it, one has to choose value for one between $l$ and $u$. For instance I could choose $l$ such that $F_{Q}(l) = 2%$ and solve fo $u$ so that $F_{Q}(u) = 1 – alpha + 2%$

And here is my second question: so there the $level- alpha$ confidence intervals are infinite? And the difference between them is their length? Is my reasoning correct?

Any help would be much appreciated! Thank you.


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