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February 28th, 2019, 05:58 AM   #1
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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
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February 28th, 2019, 12:49 PM   #2
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$\text{Suppose the random variable }X \text{ has a CDF }F_X(x)$

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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