Poisson Distribution Example_MLE

Example) Mathematical Statistics and Data Analysis, 3ED, Chapter 8. Q3. 

One of the earliest applications of the Poisson distribution was made by Student(1907) in studying errors made in counting yeast cells or blood corpuscles with a haemacytometer. In this study, yeast cells were killed and mixed with water and gelatin; the mixture was then spread on a glass and allowed to cool. Four different concentrations were used. Counts were made on 400 squares and the data are summarized in the following table; (we're deal with only one data set) 

(#cells, Concentration 2)

= (0, 103) (1, 143) (2, 98) (3, 42) (4, 8) (5, 4) (6 2) (7, 0) (8, 0) (9, 0) (10, 0) (11, 0) (12, 0)

a) log-likelihood for $\lambda$ 

b) maximum likelihood? 

c) Calculate maximum likelihood?

d) Find an approximate 95% confidence interval for $\lambda $

$P(Y=y)= \frac {e^{-\lambda} \lambda^y}{y!}$ 

$\triangleright$ Solution (a)

$l(\lambda) = \sum y_{i} \cdot \log \lambda = \sum \log(y_{i})-n\lambda$ 

$\triangleright$ Solution (b)

$l'(\lambda) = \frac {\sum y_{i}}{\lambda} -n=(set)0\rightarrow \hat {\lambda}=\frac {\sum y_i}{n} = \bar{y}$ 

$\triangleright$ Solution (c)

$\hat{\lambda}=\bar{y}= \frac {529}{400}=1.3225$ 

$\triangleright$ Solution (d)

We can find a variance by using Fisher information: 

$- \frac {d^2}{d\lambda^2}[\sum y_i \log \lambda-n\lambda]=\frac {\sum y_{i}}{\hat {\lambda}^2}= \frac {n^2}{\sum y_{i}} \rightarrow \frac {\sum y_{i}}{n^2}\approx \sigma ^2_{\hat{\lambda}}$ 

$\hat {\lambda} \pm Z_{0.975} \frac {\sqrt{\sum y_{i}}}{n} = 1.3225 \pm1.96 \cdot \frac {23}{400}=(1.210, 1.435)$