## Practice Problem Set 4 – estimating parameters of discrete distributions

This practice problem set is to reinforce the topic discussed in this post, the topic of estimating parameters of discrete distributions.

Other posts on parameter estimation focus on continuous distributions – this one and this one. Two practice problem sets, Practice Problem Set 2 and Practice Problem Set 3, are to reinforce these two previous posts.

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Practice Problem 4-A

The following table gives information on claim frequency data of a group of insureds.

Frequency # of Insureds
0 39
1 25
2 20
3 8
4 4
5 3
6 1
7+ 0

A Poisson distribution with mean $\lambda$ is fitted to the claim frequency data.

• Determine the maximum likelihood estimate of the parameter $\lambda$.
• Determine the probability of having at least one claim.

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Practice Problem 4-B

The following table gives information on claim frequency data of a group of insureds.

Frequency # of Insureds
0 39
1 25
2 20
3 7
4+ 9

A Poisson distribution with mean $\lambda$ is fitted to the claim frequency data using maximum likelihood estimation.

• Determine the log-likelihood function.
• Determine the equation obtained by setting the derivative of the log-likelihood function equal to zero. Note that this is the equation for determining the maximum likelihood estimate of $\lambda$. However, solving this equation requires using numerical methods.

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 Practice Problem 4-C The probability function for the number of losses for a given insured in a year is given by the following: $\displaystyle P[X=x]=\binom{r+x-1}{x} \ p^r \ (1-p)^x \ \ \ \ \ \ x=0,1,2,\cdots$ where $r=3$ and the parameter $p$ is unknown. The following shows the numbers of losses for five insureds in one year: 0, 2, 3, 1, 3. Use the method of maximum likelihood estimation to estimate the parameter $p$. Determine the probability of observing zero claims according to the fitted distribution.

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Practice Problem 4-D

The observed claim frequency data of a group of policyholders is given in the following table.

Frequency # of Insureds
0 94
1 64
2 32
3 7
4 3

A binomial distribution with parameters $m=4$ and $p$ is fitted to the given claim frequency data.

• Estimate the parameter $p$ using maximum likelihood estimation.
• Determine the probability of observing zero claims or 1 claim according to the fitted distribution.

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 Practice Problem 4-E A large group of insureds is made up of two groups – low risk group and high risk group. The annual claim frequency for an insured in the low risk group has a Poisson distribution with mean $\lambda$. The annual claim frequency for an insured in the high risk group has a Poisson distribution with mean $2 \lambda$. Ten insured are observed for 5 years (5 insureds in each group). Their claim counts are as follows: Low Risk Group: 0, 2, 1, 0, 3 High Risk Group: 1, 0, 2, 3, 1 Estimate the parameter $\lambda$ using maximum likelihood estimation.

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 Practice Problem 4-F The number of claims in a year for an insured has a distribution whose probability function is given by the following: $\displaystyle P[X=x]=\biggl(\frac{1}{1+\theta} \biggr) \ \biggl(\frac{\theta}{1+\theta} \biggr)^x \ \ \ \ \ \ x=0,1,2,3,\cdots$ Out of a group of 100 insureds that have been observed for one year, 55 of them have no claims, 25 of them have exactly 1 claim and 20 of them have 2 or more claims. Estimate the parameter $\theta$ using maximum likelihood estimation. Determine the probability that there are two or more claims in a year for a randomly chosen insured.

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 Practice Problem 4-G Two groups of insureds are pooled for the purpose of maximum likelihood estimation. The number of claims per year for Group 1 follows a binomial distribution with parameters $n=12$ and $p$. The number of claims per year for Group 2 follows a binomial distribution with parameters $n=20$ and $p$. In observing these two groups for 3 years, there are 15 claims from Group 1 and 28 claims from Group 2. Estimate the parameter $p$ using maximum likelihood estimation.

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 Practice Problem 4-H The number of claims from 5 policyholders are: 2, 3, 1, 5, 5 A zero-truncated geometric distribution is fitted to the claim data using maximum likelihood estimation. Determine the estimated probability that the number of claims is at least 2.

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Practice Problem 4-I

The observed claim frequency data for a group of 105 insureds is given below.

Frequency # of Insureds
0 40
1 24
2 20
3 9
4 5
5 4
6 3
7+ 0

Which of the (a,b,0) distributions (Poisson, binomial, negative binomial) is the most appropriate fit to the claim frequency data? Answer this question from the following two angles.

• Comparing the sample mean and the sample variance.
• Compute the ratio $k \frac{n_k}{n_{k-1}}$ where $n_k$ is the number of insureds having $k$ claims. Plot these ratio against $k$. Observe the slop of the plot.

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 Practice Problem 4-J When fitting a binomial distribution with both parameters unknown, the maximum likelihood estimation using log-likelihood profile is demonstrated in Example 7 and Example 8 in this post. Show that this approach does not work for the claim frequency data in Problem 4-D.

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4-A
• $\displaystyle \hat{\lambda}=\frac{126}{100}=1.26$
• $\displaystyle 1-e^{-1.26}=0.7163$
4-B
• $\displaystyle \ln(L)=86 \ \ln(\lambda)-91 \ \lambda+9 \ \ln \biggl(1-e^{-\lambda}-\lambda e^{-\lambda}-\frac{\lambda^2}{2} e^{-\lambda}-\frac{\lambda^3}{6} e^{-\lambda} \biggr)$
• $\displaystyle \frac{86}{\lambda}-91+\frac{9}{1-e^{-\lambda}-\lambda e^{-\lambda}-\frac{\lambda^2}{2} e^{-\lambda}-\frac{\lambda^3}{6} e^{-\lambda}} \bigg(\frac{\lambda^3}{6} e^{-\lambda} \biggr)=0$
4-C
• $\displaystyle \hat{p}=\frac{5}{8}$
• $\displaystyle \biggl(\frac{5}{8} \biggr)^3=0.24414$
4-D
• $\displaystyle \hat{p}=\frac{161}{800}=0.20125$
• $\displaystyle P(X=0,1)=0.8172770109$
4-E
• $\displaystyle \hat{\lambda}=\frac{13}{15}$
4-F
• $\displaystyle \hat{\theta}=\frac{13}{16}$
• $\displaystyle \biggl(\frac{13}{29}\biggr)^2=0.20095$
4-G
• $\displaystyle \hat{p}=\frac{43}{96}$
4-H
• $\displaystyle \hat{p}=\frac{5}{16}$
• $\displaystyle \frac{11}{16}$

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