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5.) An actuary discovers that policyholders are three times as likely to file 2 claims as to file 4 claims. If the number of claims filed follows a Poisson distribution, what is the standard deviation of the number of claims filed?

55.) An actuary discovers that policyholders are three times as likely to file 22 claims as to file 44 claims. If the number of claims filed follows a Poisson distribution, what is the standard deviation of the number of claims filed?

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Q. 55.) An actuary discovers that policyholders are three times as likely to file 22 claims as to file 44 claims. If the number of claims filed follows a Poisson distribution, what is the standard deviation of the number of claims filed?
  1. Denote probabilities and parameters: Let's denote the probability of filing 22 claims as P(2)P(2) and the probability of filing 44 claims as P(4)P(4). According to the given information, P(2)=3×P(4)P(2) = 3 \times P(4). In a Poisson distribution, the probabilities of different numbers of events are determined by the parameter λ\lambda (lambda), which is also the mean of the distribution. The standard deviation of a Poisson distribution is the square root of λ\lambda.
  2. Use Poisson PMF: To find the relationship between P(2)P(2) and P(4)P(4) in terms of λ\lambda, we use the Poisson probability mass function (PMF): P(k)=e−λ⋅λkk!P(k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}, where kk is the number of events (claims in this case), ee is the base of the natural logarithm, and k!k! is the factorial of kk.
  3. Set up and simplify equation: Using the Poisson PMF for P(2)P(2) and P(4)P(4), we have:\newlineP(2)=e(−λ)⋅λ22!P(2) = \frac{e^{(-\lambda)} \cdot \lambda^2}{2!}\newlineP(4)=e(−λ)⋅λ44!P(4) = \frac{e^{(-\lambda)} \cdot \lambda^4}{4!}\newlineGiven that P(2)=3⋅P(4)P(2) = 3 \cdot P(4), we can set up the equation:\newlinee(−λ)⋅λ22!=3⋅e(−λ)⋅λ44!\frac{e^{(-\lambda)} \cdot \lambda^2}{2!} = 3 \cdot \frac{e^{(-\lambda)} \cdot \lambda^4}{4!}
  4. Solve quadratic equation: Simplify the equation by canceling out the common terms and solving for λ\lambda:λ22=3×λ424\frac{\lambda^2}{2} = 3 \times \frac{\lambda^4}{24}λ22=λ48\frac{\lambda^2}{2} = \frac{\lambda^4}{8}Multiplying both sides by 88 to get rid of the denominator:4×λ2=λ44 \times \lambda^2 = \lambda^4
  5. Find lambda: We now have a quadratic equation in terms of λ2\lambda^2. Let's denote x=λ2x = \lambda^2 and solve for xx: \newline4x=x24x = x^2\newlinex2−4x=0x^2 - 4x = 0\newlinex(x−4)=0x(x - 4) = 0\newlineThis gives us two solutions for xx: x=0x = 0 or x=4x = 4. Since λ2\lambda^2 cannot be 00 (as it would imply no claims are ever filed, which is not reasonable in this context), we take λ2=4\lambda^2 = 4.
  6. Calculate standard deviation: Now that we have λ2=4\lambda^2 = 4, we can find λ\lambda by taking the square root:\newlineλ=4\lambda = \sqrt{4}\newlineλ=2\lambda = 2
  7. Calculate standard deviation: Now that we have λ2=4\lambda^2 = 4, we can find λ\lambda by taking the square root:\newlineλ=4\lambda = \sqrt{4}\newlineλ=2\lambda = 2 Since the standard deviation of a Poisson distribution is the square root of λ\lambda, and we have found λ\lambda to be 22, the standard deviation is:\newlineStandard deviation = λ\sqrt{\lambda}\newlineStandard deviation = 2\sqrt{2}

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