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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: 2,452 Thanks: 1337 
$\text{Suppose the random variable }X \text{ has a CDF }F_X(x)$ 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 $X \sim U[a,b] \text{ where }U \text{ is the uniform distribution }$ $F_X(x) = \begin{cases} 0 &x < a\\ \dfrac{xa}{ba} &a \leq x \leq b\\ 1 &b < x \end{cases}$ $\mu_X = \dfrac{a+b}{2}$ 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 $\dfrac{2\delta}{ba} = \alpha$ $\delta = \dfrac{ba}{2} \alpha$ 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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cdf, confidence, confidence intervals, intervals 
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