“General Binomial” distribution

Consider the following problems.

Problem A
Four biased coins are tossed where the probabilities of head are 0.53, 0.55, 0.57, 0.60. What is the probability of obtaining exactly j heads where j=1,2,3,4?

Problem B
Three independent insurance policies are randomly selected from a portfolio of insurance policies. Each policyholder has at most 1 claim in a policy period. The probabilities of incurring a claim in a policy period are 0.10, 0.15 and 0.20 for these three policyholders. What are the mean number of claims and the standard deviation of the number of claims among these three policyholders in the next policy period?

The problem at hand is an independent sum of Bernoulli distributions where the probability of success is not constant across the trials. Hence we cannot use the usual binomial distribution. For the lack of a better term, we call this “general binomial” distribution. There does not appear to be any quick route to finding the probability of exactly j successes, except to sum all possible combinations. This is a nice exercise for thinking about independent trials.

Problem A
Let X be the number of heads in the 4 trials. The following shows the calculation. The numbers in bold face are the probabilities of success (head).

\displaystyle \begin{aligned}P(X=0)&=0.47 \times 0.45 \times 0.43 \times 0.40 \\&=0.036378 \end{aligned}

\displaystyle \begin{aligned}P(X=1)&=\bold{0.53} \times 0.45 \times 0.43 \times 0.40 \\&\ \ \ +0.47 \times \bold{0.55} \times 0.43 \times 0.40 \\&\ \ \ + 0.47 \times 0.45 \times \bold{0.57} \times 0.40 \\&\ \ \ + 0.47 \times 0.45 \times 0.43 \times \bold{0.60} \\&=0.188273  \end{aligned}

\displaystyle \begin{aligned}P(X=2)&=\bold{0.53} \times \bold{0.55} \times 0.43 \times 0.40 \\&\ \ \ +\bold{0.53} \times 0.45 \times \bold{0.57} \times 0.40 \\&\ \ \ + \bold{0.53} \times 0.45 \times 0.43 \times \bold{0.60} \\&\ \ \ + 0.47 \times \bold{0.55} \times \bold{0.57} \times 0.40 \\&\ \ \ + 0.47 \times \bold{0.55} \times 0.43 \times \bold{0.60} \\&\ \ \ + 0.47 \times 0.45 \times \bold{0.57} \times \bold{0.60}\\&=0.364013  \end{aligned}

\displaystyle \begin{aligned}P(X=3)&=0.47 \times \bold{0.55} \times \bold{0.57} \times \bold{0.60} \\&\ \ \ +\bold{0.53} \times 0.45 \times \bold{0.57} \times \bold{0.60} \\&\ \ \ + \bold{0.53} \times \bold{0.55} \times 0.43 \times \bold{0.60} \\&\ \ \ + \bold{0.53} \times \bold{0.55} \times \bold{0.57} \times 0.40 \\&=0.3111643  \end{aligned}

\displaystyle \begin{aligned}P(X=4)&=\bold{0.53} \times \bold{0.55} \times \bold{0.57} \times \bold{0.60} \\&=0.099693 \end{aligned}

Problem B Answers
\displaystyle \begin{aligned}E(X)&=0 \times P(X=0) + 1 \times P(X=1)+ 2 \times P(X=2) + 3 \times P(X=3) \\&=0 \times 0.612 + 1 \times 0.329+ 2 \times 0.056 + 3 \times 0.003 \\&=0.45  \end{aligned}

\displaystyle \begin{aligned}E(X^2)&=0^2 \times P(X=0) + 1^2 \times P(X=1)+ 2^2 \times P(X=2) + 3^2 \times P(X=3) \\&=0 \times 0.612 + 1 \times 0.329+ 4 \times 0.056 + 9 \times 0.003 \\&=0.58  \end{aligned}

\displaystyle \begin{aligned}Var(X)&=E(X^2)-E(X)^2 \\&=0.58-0.45^2 \\&=0.3775  \end{aligned}

\displaystyle \sqrt{Var(X)}=\sqrt{0.3775}=0.61441

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