Estimating \(\lambda\)
In Poisson processes, the rate of events, \(\lambda\), is usually an unknown parameter. If a random sample, \(\{x_1, x_2, \dots, x_n\}\) of times between events in the process is available — a random sample from an \(\ExponDistn(\lambda)\) distribution — the parameter can be estimated by either the method of moments or maximum likelihood. We will now derive the maximum likelihood estimator.
The logarithm of the exponential distribution's probability density function is
\[ \log f(x\;|\;\lambda) \;\;=\;\; \log(\lambda) - \lambda x \]and the log-likelihood is the sum of this over the data values,
\[ \ell(\lambda) \;\;=\;\; \sum_{i=1}^n \log f(x_i\;|\;\lambda) \;\;=\;\; n\log(\lambda) - \lambda \sum {x_i} \]The maximum likelihood estimate maximises this and is where \(\ell'(\lambda) = 0\).
\[ \ell'(\lambda) \;\;=\;\; \frac n{\lambda} - \sum {x_i} \;\;=\;\; 0 \]so
\[ \hat{\lambda} \;\;=\;\; \frac n{\sum {x_i}} \;\;=\;\; \frac 1 {\overline{x}}\]Aircraft air-conditioner failures
The table below shows the number of operating hours between successive failures of air-conditioning equipment in ten aircraft.
Aircraft number | |||||||||
---|---|---|---|---|---|---|---|---|---|
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 12 | 13 |
413 14 58 37 100 65 9 169 447 184 36 201 118 34 31 18 18 67 57 62 7 22 34 |
90 10 60 186 61 49 14 24 56 20 79 84 44 59 29 118 25 156 310 76 26 44 23 62 130 208 70 101 208 |
74 57 48 29 502 12 70 21 29 386 59 27 153 26 326 |
55 320 65 104 220 239 47 246 176 182 33 15 104 35 |
23 261 87 7 120 14 62 47 225 71 246 21 42 20 5 12 120 11 3 14 71 11 14 11 16 90 1 16 52 95 |
97 51 11 4 141 18 142 68 77 80 1 16 106 206 82 54 31 216 46 111 39 63 18 191 18 163 24 |
50 44 102 72 22 39 3 15 197 188 79 88 46 5 5 36 22 139 210 97 30 23 13 14 |
359 9 12 270 603 3 104 2 438 |
487 18 100 7 98 5 85 91 43 230 3 130 |
102 209 14 57 54 32 67 59 134 152 27 14 230 66 61 34 |
Assuming that each aircraft has the same failure rate for its air-conditioning equipment, and the occurrence of a failure in any hour is independent of whether or not the equipment has just been repaired, then failures will be a Poisson process with rate \(\lambda\) per hour for each aircraft and the times above will be a random sample of size \(n = 199\) from an \(\ExponDistn(\lambda)\) distribution.
The maximum likelihood estimate of \(\lambda\) is
\[ \hat{\lambda} \;\;=\;\; \frac 1 {\overline{x}} \;\;=\;\; \frac 1 {90.92} = 0.0110 \text{ failures per hour}\]