IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Compound distributions and their applications in risk modelling
A compound Poisson random variable S has Poisson parameter 4 and claim amounts X with E[X] = 3,000 and E[X^2] = 15,000,000 (in rupees). What is the variance of S?
For a compound Poisson distribution the variance equals lambda times the second moment of the claim size: 4 x 15,000,000 = Rs 60,000,000. Using the variance of X instead of E[X^2] would understate the result.
- ARs 36,000,000
- BRs 60,000,000Correct
- CRs 48,000,000
- DRs 12,000
- Rs 96,000,000
Explanation
For compound Poisson, Var(S) = lambda E[X^2] = 4 x 15,000,000 = 60,000,000. Using lambda Var(X) = 4 x 6,000,000 = 24,000,000 would be a mistake, since it ignores the mean-squared part.
Did you get it right without looking?
One question tells you little. A timed set on Compound distributions and their applications in risk modelling shows your real accuracy, how long you take and where you lose marks.
More Compound distributions and their applications in risk modelling questions
- If S is compound Poisson with parameter lambda and the claim size X has moment generating function M_X(t), which expression gives the MGF of…
- The number of claims N is binomial with n = 10 and p = 0.3. Claim sizes are independent with mean 100 and variance 400, independent of N. Wh…
- For a compound Poisson aggregate claims S with Poisson parameter λ and claim size X with moment generating function M_X(t), which expression…
- In the Panjer recursion for aggregate claims S = X1 + ... + XN, where the individual claim amounts are discrete on the positive integers (in…
- The number of claims N on a motor portfolio is Poisson with mean 20 per year. Claim sizes are independent with mean ₹5,000 and variance 4,00…
- For a compound Poisson aggregate loss S with Poisson parameter 3 and claim size X with E[X]=2000 and E[X^2]=6,000,000 (Rs squared), what is …