Further Properties of the Compound Poisson Class

ADVANCED DISCRETE DISTRIBUTIONS

7.2 Further Properties of the Compound Poisson Class

This shows that adding, deleting, or modifying the probability at zero in the secondary distribution does not add a new distribution because it is equivalent to modifying the parameter𝜃of the primary distribution. Thus, for example, a Poisson primary distribution with a Poisson, zero-truncated Poisson, or zero-modified Poisson secondary distribution will still lead to a Neyman Type A (Poisson–Poisson) distribution.

EXAMPLE 7.6

Determine the probabilities for a Poisson–zero-modified ETNB distribution where the parameters are𝜆= 7.5,𝑝^𝑀₀ = 0.6,𝑟= −0.5, and𝛽= 1.

From Example 6.5, the secondary probabilities are 𝑓0 = 0.6, 𝑓1 = 0.341421, 𝑓2= 0.042678, and𝑓3= 0.010670. From (7.7),𝑔0= exp[−7.5(1 − 0.6)] = 0.049787.

For the Poisson primary distribution, 𝑎= 0and𝑏 = 7.5. The recursive formula in (7.5) becomes

𝑔_𝑘=

∑_𝑘

𝑗=1(7.5𝑗∕𝑘)𝑓_𝑗𝑔_𝑘−𝑗

1 − 0(0.6) =

∑𝑘 𝑗=1

7.5𝑗 𝑘 𝑓_𝑗𝑔_𝑘−𝑗. Then,

𝑔₁= 7.5(1)

1 0.341421(0.049787) = 0.127487, 𝑔₂= 7.5(1)

2 0.341421(0.127487) +7.5(2)

2 0.042678(0.049787) = 0.179161, 𝑔3= 7.5(1)

3 0.341421(0.179161) +7.5(2)

3 0.042678(0.127487) + 7.5(3)

3 0.010670(0.049787) = 0.184112.

Except for slight rounding differences, these probabilities are the same as those

obtained in Example 7.4. □

7.1.1 Exercises

7.1 Do all the members of the(𝑎, 𝑏,0)class satisfy the condition of Theorem 7.4? For those that do, identify the parameter (or function of its parameters) that plays the role of𝜃 in the theorem.

7.2 Show that the following three distributions are identical: (1) geometric–geometric, (2) Bernoulli–geometric, and (3) zero-modified geometric. That is, for any one of the distributions with arbitrary parameters, show that there is a member of the other two distribution types that has the same pf or pgf.

7.3 Show that the binomial–geometric and negative binomial–geometric (with negative binomial parameter𝑟a positive integer) distributions are identical.

from the fact that the Poisson distribution is often a good model to describe the number of claim-causing accidents, and the number of claims from an accident is often itself a random variable. There are numerous convenient mathematical properties enjoyed by the compound Poisson class. In particular, those involving recursive evaluation of the probabilities were also discussed in Section 7.1. In addition, there is a close connection between the compound Poisson distributions and the mixed Poisson frequency distributions that is discussed in Section 7.3.2. Here, we consider some other properties of these distributions. The compound Poisson pgf may be expressed as

𝑃(𝑧) = exp{𝜆[𝑄(𝑧) − 1]}, (7.9) where𝑄(𝑧)is the pgf of the secondary distribution.

EXAMPLE 7.7

Obtain the pgf for the Poisson–ETNB distribution and show that it looks like the pgf of a Poisson–negative binomial distribution.

The ETNB distribution has pgf

𝑄(𝑧) = [1 −𝛽(𝑧− 1)]^−𝑟− (1 +𝛽)^−𝑟 1 − (1 +𝛽)^−𝑟

for 𝛽 > 0, 𝑟 > −1, and 𝑟 ≠ 0. Then, the Poisson–ETNB distribution has, as the logarithm of its pgf,

ln𝑃(𝑧) =𝜆

{[1 −𝛽(𝑧− 1)]^−𝑟− (1 +𝛽)^−𝑟 1 − (1 +𝛽)^−𝑟 − 1

}

=𝜆

{[1 −𝛽(𝑧− 1)]^−𝑟− 1 1 − (1 +𝛽)^−𝑟

}

=𝜇{[1 −𝛽(𝑧− 1)]^−𝑟− 1},

where𝜇 = 𝜆∕[1 − (1 +𝛽)^−𝑟]. This defines a compound Poisson distribution with primary mean𝜇and secondary pgf[1 −𝛽(𝑧− 1)]^−𝑟, which is the pgf of a negative binomial random variable, as long as𝑟and, hence𝜇, are positive. This observation illustrates that the probability at zero in the secondary distribution has no impact on the compound Poisson form. Also, the preceding calculation demonstrates that the Poisson–ETNB pgf𝑃(𝑧), withln𝑃(𝑧) =𝜇{[1 −𝛽(𝑧− 1)]^−𝑟− 1}, has parameter space {𝛽 >0, 𝑟 >−1, 𝜇𝑟 >0}, a useful observation with respect to estimation and analysis

of the parameters. □

We can compare the skewness (third moment) of these distributions to develop an appre- ciation of the amount by which the skewness and, hence, the tails of these distributions can vary even when the mean and variance are fixed. From (7.9) (see Exercise 7.5) and Definition 3.2, the mean and second and third central moments of the compound Poisson distribution are

𝜇₁^′ =𝜇=𝜆𝑚^′₁,

𝜇2=𝜎²=𝜆𝑚^′₂, (7.10)

𝜇3=𝜆𝑚^′₃,

where𝑚^′_𝑗is the𝑗th raw moment of the secondary distribution. The coefficient of skewness is

𝛾1= 𝜇3

𝜎³ = 𝑚^′₃ 𝜆^1∕2(𝑚^′₂)^3∕2.

For the Poisson–binomial distribution, with a bit of algebra (see Exercise 7.6), we obtain 𝜇=𝜆𝑚𝑞,

𝜎²=𝜇[1 + (𝑚− 1)𝑞], (7.11)

𝜇₃= 3𝜎²− 2𝜇+ 𝑚− 2 𝑚− 1

(𝜎²−𝜇)²

𝜇 .

Carrying out similar exercises for the negative binomial, Polya–Aeppli, Neyman Type A, and Poisson–ETNB distributions yields

Negative binomial: 𝜇3= 3𝜎²− 2𝜇+ 2(𝜎²−𝜇)² 𝜇 Polya–Aeppli: 𝜇3= 3𝜎²− 2𝜇+3

2

(𝜎²−𝜇)² 𝜇 Neyman Type A:𝜇3= 3𝜎²− 2𝜇+(𝜎²−𝜇)²

𝜇 Poisson–ETNB:𝜇3= 3𝜎²− 2𝜇+𝑟+ 2

𝑟+ 1

(𝜎²−𝜇)² 𝜇

For the Poisson–ETNB distribution, the range of𝑟is−1< 𝑟 <∞,𝑟≠0. The other three distributions are special cases. Letting𝑟 →0, the secondary distribution is logarithmic, resulting in the negative binomial distribution. Setting𝑟 = 1defines the Polya–Aeppli distribution. Letting𝑟→∞, the secondary distribution is Poisson, resulting in the Neyman Type A distribution.

Note that for fixed mean and variance, the third moment only changes through the coefficient in the last term for each of the five distributions. For the Poisson distribution, 𝜇3 = 𝜆 = 3𝜎²− 2𝜇, and so the third term for each expression for 𝜇3 represents the change from the Poisson distribution. For the Poisson–binomial distribution, if 𝑚 = 1, the distribution is Poisson because it is equivalent to a Poisson–zero-truncated binomial as truncation at zero leaves only probability at 1. Another view is that from (7.11) we have

𝜇3= 3𝜎²− 2𝜇+𝑚− 2 𝑚− 1

(𝑚− 1)²𝑞⁴𝜆²𝑚² 𝜆𝑚𝑞

= 3𝜎²− 2𝜇+ (𝑚− 2)(𝑚− 1)𝑞³𝜆𝑚,

which reduces to the Poisson value for𝜇3when𝑚= 1. Hence, it is necessary that𝑚≥2 for non-Poisson distributions to be created. Then, the coefficient satisfies

0≤ 𝑚− 2 𝑚− 1 <1.

For the Poisson–ETNB, because𝑟 >−1, the coefficient satisfies 1< 𝑟+ 2

𝑟+ 1 <∞,

noting that when𝑟= 0this refers to the negative binomial distribution. For the Neyman Type A distribution, the coefficient is exactly1. Hence, these three distributions provide any desired degree of skewness greater than that of the Poisson distribution.

Table 7.1 The data of Hossack et al. [58].

Number of claims Observed frequency

0 565,664

1 68,714

2 5,177

3 365

4 24

5 6

6+ 0

EXAMPLE 7.8

The data in Table 7.1 are taken from Hossack et al. [58] and give the distribution of the number of claims on automobile insurance policies in Australia. Determine an appropriate frequency model based on the skewness results of this section.

The mean, variance, and third central moment are 0.1254614, 0.1299599, and 0.1401737, respectively. For these numbers,

𝜇3− 3𝜎²+ 2𝜇

(𝜎²−𝜇)²∕𝜇 = 7.543865.

From among the Poisson–binomial, negative binomial, Polya–Aeppli, Neyman Type A, and Poisson–ETNB distributions, only the latter is appropriate. For this distribution, an estimate of𝑟can be obtained from

7.543865 = 𝑟+ 2 𝑟+ 1,

resulting in 𝑟 = −0.8471851. In Example 15.12, a more formal estimation and selection procedure is applied, but the conclusion is the same. □ A very useful property of the compound Poisson class of probability distributions is the fact that it is closed under convolution. We have the following theorem.

Theorem 7.5 Suppose that𝑆_𝑖has a compound Poisson distribution with Poisson param- eter𝜆_𝑖 and secondary distribution{𝑞_𝑛(𝑖) ∶𝑛= 0,1,2,… }for𝑖= 1,2,3,…, 𝑘. Suppose also that𝑆1, 𝑆2,…, 𝑆_𝑘are independent random variables. Then,𝑆=𝑆1+𝑆2+⋯+𝑆_𝑘 also has a compound Poisson distribution with Poisson parameter𝜆=𝜆1+𝜆2+ ⋯ +𝜆_𝑘 and secondary distribution{𝑞_𝑛∶𝑛= 0,1,2,… }, where𝑞_𝑛= [𝜆₁𝑞_𝑛(1) +𝜆₂𝑞_𝑛(2) + ⋯+ 𝜆_𝑘𝑞_𝑛(𝑘)]∕𝜆.

Proof: Let𝑄_𝑖(𝑧) =∑∞

𝑛=0𝑞_𝑛(𝑖)𝑧^𝑛for𝑖= 1,2,…, 𝑘. Then,𝑆_𝑖has pgf𝑃_𝑆𝑖(𝑧) =E(𝑧^𝑆𝑖) =

exp{𝜆_𝑖[𝑄_𝑖(𝑧) − 1]}. Because the𝑆_𝑖s are independent,𝑆has pgf 𝑃_𝑆(𝑧) =

∏𝑘 𝑖=1

𝑃_𝑆𝑖(𝑧)

=

∏𝑘 𝑖=1

exp{𝜆_𝑖[𝑄_𝑖(𝑧) − 1]}

= exp [ _𝑘

∑

𝑖=1

𝜆_𝑖𝑄_𝑖(𝑧) −

∑𝑘 𝑖=1

𝜆_𝑖 ]

= exp{𝜆[𝑄(𝑧) − 1]}, where𝜆=∑_𝑘

𝑖=1𝜆_𝑖and𝑄(𝑧) =∑_𝑘

𝑖=1𝜆_𝑖𝑄_𝑖(𝑧)∕𝜆. The result follows by the uniqueness of

the generating function. □

One main advantage of this result is computational. If we are interested in the sum of independent compound Poisson random variables, then we do not need to compute the distribution of each compound Poisson random variable separately (i.e. recursively using Example 7.3), because Theorem 7.5 implies that a single application of the compound Poisson recursive formula in Example 7.3 will suffice. The following example illustrates this idea.

EXAMPLE 7.9

Suppose that 𝑘 = 2 and𝑆₁ has a compound Poisson distribution with 𝜆₁ = 2 and secondary distribution 𝑞1(1) = 0.2, 𝑞2(1) = 0.7, and 𝑞3(1) = 0.1. Also, 𝑆2 (inde- pendent of 𝑆1) has a compound Poisson distribution with 𝜆2 = 3 and secondary distribution𝑞2(2) = 0.25, 𝑞3(2) = 0.6, and𝑞4(2) = 0.15. Determine the distribution of 𝑆=𝑆1+𝑆2.

We have𝜆=𝜆₁+𝜆₂= 2 + 3 = 5. Then, 𝑞1= 0.4(0.2) + 0.6(0) = 0.08, 𝑞2= 0.4(0.7) + 0.6(0.25) = 0.43, 𝑞₃= 0.4(0.1) + 0.6(0.6) = 0.40, 𝑞4= 0.4(0) + 0.6(0.15) = 0.09.

Thus, 𝑆 has a compound Poisson distribution with Poisson parameter 𝜆 = 5 and secondary distribution𝑞1 = 0.08, 𝑞2 = 0.43, 𝑞3 = 0.40, and𝑞4 = 0.09. Numerical values of the distribution of𝑆may be obtained using the recursive formula

Pr(𝑆=𝑥) = 5 𝑥

∑𝑥 𝑛=1

𝑛𝑞_𝑛 Pr(𝑆=𝑥−𝑛), 𝑥= 1,2,…,

beginning withPr(𝑆= 0) =𝑒⁻⁵. □

In various situations, the convolution of negative binomial distributions is of interest. The following example indicates how this distribution may be evaluated.

EXAMPLE 7.10

(Convolutions of negative binomial distributions) Suppose that𝑁_𝑖has a negative bino- mial distribution with parameters𝑟_𝑖and𝛽_𝑖 for𝑖= 1,2,…, 𝑘and that𝑁1, 𝑁2,…, 𝑁_𝑘 are independent. Determine the distribution of𝑁 =𝑁1+𝑁2+⋯+𝑁_𝑘.

The pgf of𝑁_𝑖is𝑃_𝑁𝑖(𝑧) = [1 −𝛽_𝑖(𝑧− 1)]^−𝑟𝑖and that of𝑁is 𝑃_𝑁(𝑧) =∏_𝑘

𝑖=1𝑃_𝑁𝑖(𝑧) =∏_𝑘

𝑖=1[1 −𝛽_𝑖(𝑧− 1)]^−𝑟𝑖. If𝛽_𝑖=𝛽for𝑖= 1,2,…, 𝑘, then

𝑃_𝑁(𝑧) = [1 −𝛽(𝑧− 1)]^{−(𝑟1+𝑟2+⋯+𝑟𝑘)},

and𝑁 has a negative binomial distribution with parameters𝑟 = 𝑟₁+𝑟₂+⋯+𝑟_𝑘 and𝛽.

If not all the 𝛽_𝑖s are identical, however, we may proceed as follows. From Example 7.5,

𝑃_𝑁𝑖(𝑧) = [1 −𝛽_𝑖(𝑧− 1)]^−𝑟𝑖 =𝑒^{𝜆𝑖[𝑄𝑖(𝑧)−1]}, where𝜆_𝑖=𝑟_𝑖ln(1 +𝛽_𝑖), and

𝑄_𝑖(𝑧) = 1 −ln[1 −𝛽_𝑖(𝑧− 1)]

ln(1 +𝛽_𝑖) =

∑∞ 𝑛=1

𝑞_𝑛(𝑖)𝑧^𝑛 with

𝑞_𝑛(𝑖) = [𝛽_𝑖∕(1 +𝛽_𝑖)]^𝑛

𝑛ln(1 +𝛽_𝑖) , 𝑛= 1,2,… .

But Theorem 7.5 implies that𝑁 = 𝑁₁+𝑁₂+⋯+𝑁_𝑘 has a compound Poisson distribution with Poisson parameter

𝜆=

∑𝑘 𝑖=1

𝑟_𝑖ln(1 +𝛽_𝑖)

and secondary distribution 𝑞_𝑛=

∑𝑘 𝑖=1

𝜆_𝑖 𝜆𝑞_𝑛(𝑖)

=

∑_𝑘

𝑖=1𝑟_𝑖[𝛽_𝑖∕(1 +𝛽_𝑖)]^𝑛 𝑛∑_𝑘

𝑖=1𝑟_𝑖ln(1 +𝛽_𝑖) , 𝑛= 1,2,3,…. The distribution of𝑁may be computed recursively using the formula

Pr(𝑁=𝑛) = 𝜆 𝑛

∑𝑛 𝑘=1

𝑘𝑞_𝑘Pr(𝑁=𝑛−𝑘), 𝑛= 1,2,…,

beginning withPr(𝑁 = 0) = 𝑒^−𝜆 = ∏_𝑘

𝑖=1(1 +𝛽_𝑖)^−𝑟𝑖 and with𝜆 and𝑞_𝑛 as given

previously. □

It is not hard to see that Theorem 7.5 is a generalization of Theorem 6.1, which may be recovered with𝑞1(𝑖) = 1for𝑖= 1,2,…, 𝑘. Similarly, the decomposition result of Theorem 6.2 may also be extended to compound Poisson random variables, where the decomposition is on the basis of the region of support of the secondary distribution. For further details, see Panjer and Willmot [100, Sec. 6.4] or Karlin and Taylor [67, Sec. 16.9].

7.2.1 Exercises

7.4 For𝑖 = 1,…, 𝑛let𝑆_𝑖 have independent compound Poisson frequency distributions with Poisson parameter 𝜆_𝑖 and a secondary distribution with pgf 𝑃2(𝑧). Note that all 𝑛 of the variables have the same secondary distribution. Determine the distribution of 𝑆=𝑆₁+⋯+𝑆_𝑛.

7.5 Show that, for any pgf, 𝑃^(𝑘)(1) = E[𝑁(𝑁 − 1)⋯(𝑁−𝑘+ 1)] provided that the expectation exists. Here,𝑃^(𝑘)(𝑧)indicates the𝑘th derivative. Use this result to confirm the three moments as given in (7.10).

7.6 Verify the three moments as given in (7.11).