$95\,\%$ confidence interval for geometric distribution

I am analyzing data with a geometric distribution. Using maximum likelihood estimation, I can estimate $p$ to be $\displaystyle \hat p_{MLE} = \frac{N}{\sum_{i=1}^N x_i}$, where $N$ is the number of datapoints and each $x$ is the number of trials necessary for the first success, in each separate experiment.

However, it is not clear to me how 'accurate' this value is, and thus I'd like to construct a $95\,\%$ confidence interval. But my searches have been rather unsuccessful, as I can't find any worked out versions of a suitable confidence interval for this particular distribution. I'm pretty sure there has to be something out there, and I would be very thankful if someone could guide me towards it!

In frequentist inference, the proper confidence interval depends on what you consider your sample space: If you were to repeat your experiment/observational scheme, what would remain constant, what would be allowed to change. Specifically, did you always intend to observe N geometric RV's or was that just when the experiment ended? Are you always observing the same number of trials? If neither is controlled, then its hard to specify what type of process you've observed and how an interval would behave under repeated sampling.

Here are a couple suggestions:

1. Turn your observed geometric variables into an equivalent bernoulli series and apply binomial inference procedures to it. This assumes you had no pre-specified number of successes you needed to achieve (i.e. the sample size was not specified in advance).
2. Apply negative binomial confidence procedures to it, with a known number of successes, i.e., r=N, and you want an interval for p given N. This assumes you pre-specifed the sample size.
3. Perform boostrap resampling on your observed data vector, each time re-calculating the estimate for p, and see the distribution of your estimator relative to the actual value of the estimator for the original sample. Look up Boostrap confidence intervals (percentile and BCC methods).
• Thank you for your elaborate answer. I'll dive into all of this today, and see if I can get anywhere from that. Commented Jan 23, 2014 at 9:10
• I was too late to edit my previous comment, but I'm not sure I understand the distinction between N geometric RV's, and the number of trials. Isn't the outcome of each trial a random variable? In any case, I do not know the number of trials I will end up beforehand, no. So I'll start looking at number 1 and 3! Commented Jan 23, 2014 at 9:15
• Though, now that I think of it, I might not know the number of trials before obtaining the data, but I do know them before I start calculating what P is. So I'll look at 2 instead. Commented Jan 23, 2014 at 9:32
• @user3183724 For the non-bootstrapping approaches, when I say "trials" I do not mean the geometric RVs, but the string of 1's and 0's that correspond to your observed RVs. Then, you either treat this as as one large negative binomial sample or one large binomial sample...depending on which parameter you assume to control for future replications (time periods observed or number of transitions). Also, with 60,000 data points (is that time periods or geometric RVs?), you have a great basis for bootstrapping. Of course, use as many replicaitons as possible to minimize simulation error :)
– user76844
Commented Jan 23, 2014 at 17:13
• @user3183724 As an example for non-bootstrapping methods: if you have the geometric RVs {2,2,1} then the sample you will apply neg or regular binominal methods to is {01011}...see what I mean...you are aggregating into the raw observed data string then applying a method to the entire string.
– user76844
Commented Jan 23, 2014 at 17:14