Unknown parameter

In practical applications, we may be able to assume that the conditions for a Poisson process hold at least approximately, but the rate of events, \(\lambda\), is usually an unknown value that must be estimated from data.

We now give an example of one such data set — a rather unusual one!

Example

The following table shows the numbers of deaths by horse kicks in the Prussian army each year from 1875 to 1894, in each of 14 army corps.

Corps
G0 2 2 1 0 0 1 1 0 3 0 2 1 0 0 1 0 1 0 1
I00020302000111020310
II00020200110021100200
III00011120200010121000
IV01011110000100001100
V00002100100101111110
VI00102001201131110300
VII10100010110020021020
VIII10001001000010001101
IX00000211102110120100
X00110102020000213011
XI00002401301111213131
XIV11211304010321021100
XV0 1 0 0 0 0 0 1 0 1 1 0 0 0 2 2 0 0 0 0

It is reasonable to assume that events (i.e. horse-kick deaths) occur randomly over the 20 years at a constant rate — a homogeneous Poisson process within each corps. If we make a further assumption that the rate is the same for all 14 corps, the counts in the above table will be a random sample from a Poisson distribution.

However the rate of deaths per year in each corps, \(\lambda\), is unknown and must be estimated from the data.

Maximum likelihood

Given a random sample, \(\{x_1, x_2, \dots, x_n\}\), from a \(\PoissonDistn(\lambda)\) distribution, we will use maximum likelihood to estimate \(\lambda\). The logarithm of the Poisson probability function is

\[ \log(p(x | \lambda)) \;=\; x \log(\lambda) - \lambda - \log(x!) \]

so the log-likelihood function is

\[ \ell( \lambda) \;=\; \sum_{i=1}^n {x_i} \log(\lambda) - n\lambda + K \]

where \(K\) is a constant that does not depend on \(\lambda\). To find the maximum likelihood estimate, we set the derivative of the log-likelihood to zero,

\[ \ell'( \lambda) \;=\; \frac {\sum {x_i}} {\lambda} - n \;=\; 0 \]

so

\[ \hat{\lambda} \;=\; \frac {\sum {x_i}} n \;=\; \overline{x} \]