Negative Binomial Distribution

Study guides: Exam P-1, Exam MAS-I

The Negative Binomial Distribution X \sim \text{NegBin}(r, p) counts the number of independent Bernoulli trials needed to achieve exactly r successes. When r = 1 it reduces to the Geometric distribution.

P(X = k) = \binom{k-1}{r-1}p^r(1-p)^{k-r}

k = r, r+1, r+2, \ldots

\text{where } r = \text{number of successes required},\; p = \text{success probability}

E[X] = \frac{r}{p}, \qquad \text{Var}(X) = \frac{r(1-p)}{p^2}

• Check the parameterization first. As written, X counts trials and starts at r. The alternative counts failures before the r-th success, Y = X - r \in \{0,1,2,\ldots\}, with P(Y=k) = \binom{k+r-1}{k}p^r(1-p)^k and E[Y] = r(1-p)/p. Same variance, means differing by r — the same trap as the geometric.
• The exam tables' form. SOA's Tables for Exam C (the Loss Models appendix) write the failures count as \text{NegBin}(r, \beta) with p = 1/(1+\beta): P(Y=k) = \dfrac{r(r+1)\cdots(r+k-1)\,\beta^k}{k!\,(1+\beta)^{r+k}}, E[Y] = r\beta and \text{Var}(Y) = r\beta(1+\beta). Unlike the binomial's m, r is not required to be a whole number there.
• Overdispersion belongs to the failures form. \text{Var}(Y)/E[Y] = 1/p = 1+\beta > 1 always, where the binomial's ratio is below 1 and the Poisson's is exactly 1 — which is what makes Y the overdispersed alternative to the Poisson for counts. The trials count X does not share it: \text{Var}(X)/E[X] = (1-p)/p, below 1 whenever p > 1/2 (with r = 3, p = 0.75: E[X] = 4 but \text{Var}(X) = 4/3).

Claims Until Third Large Loss Each claim has a 25% probability of being a large loss. Find the probability the 3rd large loss occurs on the 7th claim.

Answer r=3, p=0.25, k=7: P(X=7) = \binom{6}{2}(0.25)^3(0.75)^4 = 15 \cdot 0.015625 \cdot 0.3164 \approx 0.0742