# Non-zero Poisson Distrubution

## Homework Statement

Non-Zero Poisson Probability distributions
Scenario: The following data represents the number of typographical errors (typos) a lecturer makes (per page) in a document they are producing. A document's pages was inspected and the number of typos per page was recorded. Unfortunately the total number of pages inspected was not recorded and only a record of the number of pages with at least one or more typo was recorded.
The following table of the number of pages with one or more typo errors ##(X = x)##, was recorded:
$$\begin{array}{c|c c c c c c c c c} X=x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text{Observed frequency} & 15 & 22 & 19 & 16 & 14 & 4 & 2 & 0 & 1 \end{array}$$
from which we deduce that there are ##n = 93## recorded observations.

1. Show that if the underlying distribution of number of typo errors per page ##X##, is a ##\text{Poisson}(\lambda)## distribution then the probability distribution function for each observation is:
$$\begin{split} \text{Pr}(X=x) &= \frac{e^{-\lambda}\lambda^x}{x!\left(1-e^{-\lambda}\right)} \\ &= \frac{\lambda^x}{x!\left(e^{\lambda}-1\right)}, \quad \text{for x > 0}. \end{split}$$
2. Show that under the assumption of independence that the PDF for all observations ##\mathbf{X} = \left(x_1,x_2,\ldots,x_{93}\right)^T## is given by:
$$f_{\mathbf{X}}(\mathbf{x};\lambda) = \frac{\lambda^{n\bar{x}}}{\left(\prod_{i=1}^{n} x_i!\right)\left(e^{\lambda}-1\right)^n}.$$
3. Hence, show that the log-likelihood equation is given by (ignoring constant terms):
$$\ell(\lambda) = n\bar{x} \log{(\lambda) - n\log{\left(e^{\lambda}-1\right)}}.$$
4 Show that the maximum likelihood estimate for these data is given by solving the following equation in terms of ##\lambda##.
$$\bar{x} = \frac{\lambda e^{\lambda}}{e^{\lambda}-1}.$$
5. Show that the second derivative of the log-likelihood function can be expressed as:
$$\frac{d^2\ell(\lambda)}{d^2\lambda} = -\frac{n}{\lambda^2}\left(\bar{x}-\frac{\lambda^2e^{\lambda}}{\left(e^{\lambda}-1\right)^2}\right).$$
6. Substitute the expression for ##\bar{x}## in part (4) into this equation to see that the maximum likelihood estimate is in fact a maximal value.
Hint: Upon substituting this term into arrange the equation above to see that at ##\bar{x} = \frac{\lambda e^{\lambda}}{e^{\lambda}-1}## you can arrange this to
$$\frac{d^2\ell(\lambda)}{d^2\lambda} = -n\frac{e^{\lambda}}{\lambda\left(e^{\lambda}-1\right)^2}\left(e^{\lambda}-1-\lambda\right).$$
This expression needs to be negative so this reduces to simple calculus of a component term of this expression.
7. What is the smallest possible variance any unbiased estimate for the parameter ##\lambda##?
8. Use an asymptotic approximation to give a ##95\%## confidence interval for this parameter.
9. Estimate how many zeroes would have been produced if that had in fact been recorded.
10. Round the answer above, and append this into the table of counts (above) for number of zeroes that would have been observed. This makes the new observed" table have zeroes. Compute a ##95\%## confidence interval for these new" data.
Hint: This is a, relatively, easy computation and is an example of the EM algorithm applied to missing data.

## The Attempt at a Solution

1-6 I'm good with.
Do I use lagrange multipliers for 7.???? If so what is the constraint???
For 8. according to wikipedia for the Non-Zero Poisson distribution the mean and variance are
##E(X) = \frac{\lambda e^{\lambda}}{e^{\lambda}-1}## and ##\text{Var}(X) = \frac{\lambda e^{\lambda}}{e^{\lambda}-1}\left(1-\frac{\lambda}{e^{\lambda}-1}\right)## but using Fisher's information I get
$$\begin{split} I(\lambda) &= -E\left(\frac{dU}{d\lambda}\right) \\ &= -E\left(\frac{d^2\ell(\lambda)}{d\lambda^2}\right) \\ &= -E\left(-\frac{n\bar{x}}{\lambda^2} + \frac{ne^{\lambda}}{\left(e^{\lambda}-1\right)^2}\right) \\ &= \frac{n}{\lambda^2}E(\bar{X}) - \frac{ne^{\lambda}}{\left(e^{\lambda}-1\right)^2} \\ &= \frac{n}{\lambda^2}\frac{\lambda e^{\lambda}}{e^{\lambda}-1} - \frac{ne^{\lambda}}{\left(e^{\lambda}-1\right)^2} \\ &= \frac{n}{\lambda}\frac{e^{\lambda}}{e^{\lambda}-1} - \frac{ne^{\lambda}}{\left(e^{\lambda}-1\right)^2} \\ &= \frac{ne^{\lambda}\left(e^{\lambda}-1\right)-\lambda ne^{\lambda}}{\lambda\left(e^{\lambda}-1\right)^2}. \end{split}$$
so the variance is ##\text{Var}(X) = \frac{1}{I(\lambda)}## which is not the same as above so my confidence interval is wrong.
I have no idea on the last 2.