acsuvaic9

Answered

2022-07-21

Negative Binomial and Geometric Distributions

An actuary has determined that the number of claims per month can take any number 0, 1, 2, 3,... and follows a negative binomial distribution with mean 3 and variance 12. Calculate the probability that the number of claims is at least three but less than six.

So by using some properties of the negative binomial we can derive that $p=0.25$, and $r=1$. It was my understanding that a geometric distribution is just a negative binomial distribution with $r=1$. Given this I tried using the pmf for geometric distribution and got the wrong answer. Can someone explain to me what is going on here.

An actuary has determined that the number of claims per month can take any number 0, 1, 2, 3,... and follows a negative binomial distribution with mean 3 and variance 12. Calculate the probability that the number of claims is at least three but less than six.

So by using some properties of the negative binomial we can derive that $p=0.25$, and $r=1$. It was my understanding that a geometric distribution is just a negative binomial distribution with $r=1$. Given this I tried using the pmf for geometric distribution and got the wrong answer. Can someone explain to me what is going on here.

Answer & Explanation

tun1t2j

Expert

2022-07-22Added 13 answers

Explanation:

Since the number of claims $X\sim \text{Geometric}(\frac{1}{4})$, the desired porbability is $P(X=3)+P(X=4)+P(X=5)={\left(\frac{3}{4}\right)}^{3}\frac{1}{4}+{\left(\frac{3}{4}\right)}^{4}\frac{1}{4}+{\left(\frac{3}{4}\right)}^{5}\frac{1}{4}=0.2438965$

Since the number of claims $X\sim \text{Geometric}(\frac{1}{4})$, the desired porbability is $P(X=3)+P(X=4)+P(X=5)={\left(\frac{3}{4}\right)}^{3}\frac{1}{4}+{\left(\frac{3}{4}\right)}^{4}\frac{1}{4}+{\left(\frac{3}{4}\right)}^{5}\frac{1}{4}=0.2438965$

Haley Madden

Expert

2022-07-23Added 7 answers

Step 1

There are too many ways to parameterize the negative binomial, but using Klugman-Panjer-Wilmott as most US actuaries do, you can write it as: $(}\genfrac{}{}{0ex}{}{r+k-1}{k}{\textstyle )}\frac{{\beta}^{k}}{{(1+\beta )}^{r+k}$ where the mean is $r\beta $ and the variance is $r\beta (1+\beta )$.

Step 2

So if $r\beta =3$ and $r\beta (1+\beta )=12$ then $(1+\beta )=4$ so $\beta =3$ and $r=1$. Now solve for $P(k\in 3,4,5)$

There are too many ways to parameterize the negative binomial, but using Klugman-Panjer-Wilmott as most US actuaries do, you can write it as: $(}\genfrac{}{}{0ex}{}{r+k-1}{k}{\textstyle )}\frac{{\beta}^{k}}{{(1+\beta )}^{r+k}$ where the mean is $r\beta $ and the variance is $r\beta (1+\beta )$.

Step 2

So if $r\beta =3$ and $r\beta (1+\beta )=12$ then $(1+\beta )=4$ so $\beta =3$ and $r=1$. Now solve for $P(k\in 3,4,5)$

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