My Math Forum how to get the confidence intervals in a CDF

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 February 28th, 2019, 05:58 AM #1 Newbie   Joined: Feb 2019 From: Poland Posts: 3 Thanks: 0 how to get the confidence intervals in a CDF Could anybody explain me how to obtain the confidence intervals for the points of a Cumulative Distribution Function (CDF)? I appreciate your help
 February 28th, 2019, 12:49 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,590 Thanks: 1434 $\text{Suppose the random variable }X \text{ has a CDF }F_X(x)$ then $P[x_l \leq x \leq x_h] = F_X(x_h) - F_X(x_l)$ Generally the confidence interval is centered about the mean. In this case we'd have $x_l = \mu_X - \delta,~x_h = \mu_X + \delta$ where $\delta$ is chose so $\alpha = F(\mu_X+\delta)-F(\mu_X-\delta)$ A simple example. Let $X \sim U[a,b] \text{ where }U \text{ is the uniform distribution }$ $F_X(x) = \begin{cases} 0 &x < a\\ \dfrac{x-a}{b-a} &a \leq x \leq b\\ 1 &b < x \end{cases}$ $\mu_X = \dfrac{a+b}{2}$ Suppose we're after confidence level $\alpha$ We need to solve for $\delta$ such that $F_X\left(\dfrac{a+b}{2} + \delta\right) - F_X\left(\dfrac{a+b}{2} -\delta\right) = \alpha$ after plugging in and simplifying $\dfrac{2\delta}{b-a} = \alpha$ $\delta = \dfrac{b-a}{2} \alpha$ and the $\alpha$ level confidence interval is thus $CI=\left[\dfrac{1}{2} a (\alpha +1)+\dfrac{1}{2} (1-\alpha ) b,~\dfrac{1}{2} a (1-\alpha )+\dfrac{1}{2} (\alpha +1) b\right]$

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