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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 $CI=\left[\dfrac{1}{2} a (\alpha +1)+\dfrac{1}{2} (1-\alpha ) b,~\dfrac{1}{2} a (1-\alpha )+\dfrac{1}{2} (\alpha +1) b\right]$ Tags cdf, confidence, confidence intervals, intervals Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post calypso Probability and Statistics 3 May 28th, 2016 09:36 AM Artus Advanced Statistics 0 November 14th, 2012 02:39 PM twobeatsoff Advanced Statistics 2 September 12th, 2012 06:49 AM mochisushi01 Advanced Statistics 1 April 13th, 2010 12:54 AM symmetry Advanced Statistics 1 March 29th, 2007 08:59 PM

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