1
$\begingroup$

I try to get an intuition on, why pivotal quantities are used to construct confidence intervals.

First, I show how I understand the algorithm: For example let $x_1,...,x_n \in \mathbb{R}$ be realizations of the random variable $X\sim \mathcal{N}_{\sigma = 1, \mu}, \mu \in \mathbb{R}$. We want to construct a confidence interval that contains $\mu$ at least by a rate of $(1-\alpha)100\%,\ \alpha \in (0,1)$, if we repeat the sampling experiment (i.e. sampling $x_1,...,x_n$) many times.

This seems to be expressed as $Q([a,b]) \overset{!}{\geq} 1-\alpha$ with $a,b\in\mathbb{R}$ and $Q:= \mathcal{N}_{\sigma = 1, \mu}\circ T_\mu^{-1} = \mathcal{N}_{\sigma = 1,\mu = 0}$ and $T_\mu = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$. Then we need to find $a,b$. Since we now work with $\mathcal{N}_{0,1}$, we can use the inverse of $\Phi_{0,1}(x) = \int_\infty^x \mathcal{N}_{0,1}(\xi)d\xi$ to get $$a = \Phi_{0,1}^{-1}(\alpha/2) \\ b = \Phi_{0,1}^{-1}(1-\alpha/2).$$

Now the question: I dont grasp why that works, if we "normalize the problem" and solve it then w.r.t. $\mathcal{N}_{0,1}$. It seems that we dont loose any relevant information, even though we solve it in another space. Why is that? It must be obvious, since I never saw this being explained anywhere.

$\endgroup$

2 Answers 2

2
$\begingroup$

Suppose you have $n = 100$ observations taken at random from a normal population with unknown $\mu$ and $\sigma.$

You want to get a 95% confidence interval (CI) for $\mu:$ You know that $T = \frac{\bar X - \mu}{S/\sqrt{100}}$ has $P(-1.984 \le T \le 1.984) = 0.95).$

qt(.975, 99)
[1] 1.984217

In the continued equation below, the event in parentheses is exactly the same throughout. I have followed rules of arithmetic and inequalities to change the algebraic form of the event, but not the restrictions on the random variable $\bar X.$

$$0.95 = P(-1.984 \le T \le 1.984) = P\left(-1.984 \le \frac{\bar X - \mu}{S/\sqrt{100}}\le 1.984\right) = P\left(1.984 \ge \frac{\mu - \bar X}{S/\sqrt{100}}\ge -1.984\right) = P\left(-1.984 \le \frac{\mu - \bar X}{S/\sqrt{100}}\le 1.984\right) = P\left(-1.984\frac{S}{\sqrt{n}} \le \mu-\bar X \le 1.984\frac{S}{\sqrt{n}} \right) = P\left(\bar X-1.984\frac{S}{\sqrt{n}} \le \mu \le \bar X+1.984\frac{S}{\sqrt{n}} \right).$$

According to the definition of CI, the last form of the event still has probability $0.95$ and so the interval $\left(\bar X-1.984\frac{S}{\sqrt{n}},\, \bar X+1.984\frac{S}{\sqrt{n}}\right)$ is a 95% CI for $\mu.$ I'm not sure exactly which step in the continued equation is technically called a 'pivot', but the idea is that I started with two constants bounding an expression with a random variable and a parameter and ended with two expressions involving random variables bounding the parameter $\mu.$

$\endgroup$
1
  • $\begingroup$ It took me a while to realize, that we use the $t$ distribution not $\mathcal{N}(0,1)$ here. $\endgroup$
    – dba
    Jun 1, 2020 at 20:24
0
$\begingroup$

Hi: That's an interesting question especially since I've seen confidence intervals for a LONG TIME and never thought of looking at it the way you described it. The answer is that you don't lose any information because, aside from the integral limits, the scaled distribution of $X$ is the same as the distribution of $T_{\mu}$. It's as if you take the image of the original distribution of $X$ and scale the image so that it retains the same shape as the original but is a different ( standardized ) size with mean zero and standard deviation equal to one.

Fortunately, due to the central limit theorem, the same relation holds ( with some minor changes ) when $X$ is not normally distributed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.