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

\[ \hat{\lambda} \;\;=\;\; \frac 1 {\overline{x}} \;\;=\;\; \frac 1 {90.92} = 0.0110 \text{ failures per hour}\]