We will now find a confidence interval for the rate parameter, \(\lambda\), of a homogeneous Poisson process, based on a random sample of inter-event times.
Example
The values below are times between failures of an item.
487 | 18 | 100 | 7 | 98 | 5 | 85 | 91 | 43 | 230 | 3 | 130 |
If failures arise at random over time with a constant rate, \(\lambda\), the values are a random sample from an \(\ExponDistn(\lambda)\) distribution, and the maximum likelihood estimator of \(\lambda\) is the inverse of the sample mean,
\[ \hat{\lambda} \;\;=\;\; \frac n {\sum{X_i}} \]The sum of \(n\) independent exponential variables has an Erlang distribution,
\[ \sum_{i=1}^n{X_i} \;\;\sim\;\; \ErlangDistn(n, \lambda) \;=\; \GammaDistn(n, \lambda) \]This distribution depends on \(\lambda\) but the following quantity is a pivot:
\[ \lambda \sum_{i=1}^n{X_i} \;\;\sim\;\; \GammaDistn(n, 1) \]Since \(n = 12\), we can use Excel to find the 2½% and 97½% points of the \(\GammaDistn(12, 1)\) distribution — 6.201 and 19.682. A 95% confidence interval is therefore the values of \(\lambda\) for which
\[ 6.201 \;\;\lt\;\; \lambda \sum_{i=1}^n{x_i} \;\;\lt\;\; 19.682 \] \[ 6.201 \;\;\lt\;\; 1297\lambda \;\;\lt\;\; 19.682 \]Rearranging, a 95% CI for the failure rate, \(\lambda\), is
\[ 0.0048 \;\;\lt\;\; \lambda \;\;\lt\;\; 0.0152 \]