# Numerically finding confidence interval bounds

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I have a symmetric (about some value $\mu$) probability density function $f$ with support in $(-\infty, \infty)$. Let's suppose, for simplicity's sake, that it's the normal distribution $$f(x; \mu, \sigma) = \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left[-\left(\frac{x-\mu}{\sigma}\right)^2/2\right]\text{.}$$ Ignoring the fact that there's a dnorm function in R, let's say I wrote this out as a function:

f <- function(x, mu, sigma){
1/(sigma*sqrt(2*pi))*exp(-((x-mu)/sigma)^2/2)
}


I would like to solve the equation $$\int_{a}^{b}f(x; \mu, \sigma)\text{ d}x = \pi \in [0, 1]\text{ fixed.}$$ where $\mu = \dfrac{a+b}{2}$, for $a$ and $b$.

Without standardizing the normal distribution and without using any of the "normal"-supplied functions in R (e.g., dnorm, pnorm, qnorm, rnorm and related functions), how do I solve for $a$ and $b$ in R?

Here's what I do know:

1. R has an integrating function. But I'm not sure how to use this to help me solve for the bounds.
2. The normal distribution PDF, along with the PDF of distribution I am working with, do not have closed forms for the CDF.

I would guess that simulation is involved, but I wouldn't know where to start.

(Note: I know this seems silly, but I would like to know how this works for the normal distribution so that I can extend it to a general PDF.)

Yes use integrate to evaluate the integral and then e.g. uniroot to solve $\int_{\mu-a}^{\mu+a} f(x)dx - \pi = 0$ as follows (n.b. I changed the parameterisation slightly):

f <- function(x, mu, sigma) {
1/(sigma*sqrt(2*pi)) * exp(-((x-mu)/sigma)^2/2)
}

g<-function(a,mu,sigma,p) {
integrate(function(x){f(x,mu,sigma)}, mu-a, mu+a)$value - p } uniroot(g, interval=c(-10,10), mu=0,sigma=1, p =0.95)$root


Output:

[1] 1.959969


Note that the (-10,10) is the search interval. Check ?uniroot for more info.

N.b. no need for simulation. Also, even if a closed form solution is not available, many distributions have good numerical approximations to their inverses.

I am not familiar with R. But a (simple) solution is based on Monte Carlo: You can draw n samples from the distribution.

With the following algorithm you can find a and b:

• Divide the samples in two equally sized arrays:
• the samples greater as the median (gm) and
• the samples lower than the median (lm).
• to get a : sort (lm) descent. a is approximate the value of the $n * \pi/2$-item ($\pi$ is the value of the total integral)
• analog for b: sorting order asc
• I'm sorry, how does this help me find the interval $[a, b]$? – Clarinetist Dec 4 '15 at 16:56
• I updated my answer, because I misunderstood your question the first time. – chris elgoog Dec 4 '15 at 21:42