Question

Let’s in fact generalize to a c-confidence interval. Let the common rate parameter be a. (Note that the mean of an exponential distribution with rate parameter a is 1/a.)

First find the cdf of the sum of n such i.i.d. random variables. Use that to compute a c-confidence interval on the sum. Note that the max likelihood estimate (MLE) of the sum is n/a, ie, n times the mean of a single draw.

Background: This comes up in a program I’m writing to make time estimates via random samples. If I take samples according to a Poisson process (ie, the gaps between samples have an exponential distribution) and n of them happen during Activity X, what’s a good estimate for the duration of Activity X? I’m pretty sure the answer is the answer to this question.

Answer 1

As John D. Cook hinted, the sum of i.i.d. exponential random variables has a gamma distribution.
Here’s the cdf of the sum of n exponential random variables with rate parameter a (expressed in Mathematica):

F[x_] := 1 - GammaRegularized[n, a*x]; 

http://mathworld.wolfram.com/RegularizedGammaFunction.html

The inverse cdf is:

Fi[p_] := InverseGammaRegularized[n, 1 - p]/a; 

The c-confidence interval is then

ci[c_, a_, n_] := {Fi[a, n, (1-c)/2], Fi[a, n, c+(1-c)/2]} 

Here is some code to empirically verify that the above is correct:

(* Random draw from an exponential distribution given rate param. *) getGap[a_] := -1/a*Log[RandomReal[]]  betw[x_, {a_, b_}] := Boole[a <= x <= b]  c = .95; a = 1/.75; n = 40; ci0 = ci[c, a, n]; N@Mean@Table[betw[Sum[getGap[a], {n}], ci0], {100000}]  ----> 0.94995