# Non-zero Poisson Distribution: Estimating Parameters and Confidence Intervals

• squenshl
In summary, the conversation is about non-zero Poisson probability distributions and how to find the probability distribution function and log-likelihood equation for a set of observations. It also discusses finding the maximum likelihood estimate, the second derivative of the log-likelihood function, and how to use this information to estimate the parameter lambda. The conversation also mentions using asymptotic approximation and the EM algorithm to find confidence intervals for the parameter lambda. Finally, it asks for an estimate of how many zeroes would have been observed if they had been recorded and how to find a confidence interval for this new data.
squenshl

## 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.

I know that ##I(\lambda)## is now correct. Can I take the inverse of this to get the variance or because it is not unbiased then not?

## 1. What is a Non-zero Poisson Distribution?

A Non-zero Poisson Distribution is a probability distribution that models the number of events that occur in a fixed interval of time or space given that the events occur independently and at a constant rate. Unlike a regular Poisson Distribution, a Non-zero Poisson Distribution allows for the possibility of zero occurrences.

## 2. How is a Non-zero Poisson Distribution different from a regular Poisson Distribution?

In a regular Poisson Distribution, the probability of obtaining zero occurrences is always possible but very small. However, in a Non-zero Poisson Distribution, the probability of obtaining zero occurrences is higher due to the increased likelihood of events not occurring at a constant rate.

## 3. What is the use of a Non-zero Poisson Distribution in real life?

A Non-zero Poisson Distribution is often used to model rare events, such as natural disasters, machine failures, and rare diseases. It can also be used to analyze data in various fields, including finance, economics, and biology.

## 4. How is a Non-zero Poisson Distribution calculated?

The formula for calculating a Non-zero Poisson Distribution is: P(x; λ) = (e^-λ)(λ^x) / (x!(1-e^-λ)), where x is the number of occurrences, and λ is the expected rate of occurrences. This formula can also be used to calculate the probability of obtaining zero occurrences.

## 5. What are the limitations of a Non-zero Poisson Distribution?

A Non-zero Poisson Distribution assumes that the events occur independently and at a constant rate, which may not always be the case in real-life situations. Additionally, it does not account for external factors that may influence the rate of occurrences. Therefore, it is essential to carefully consider the assumptions and limitations of a Non-zero Poisson Distribution before applying it to a particular scenario.

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