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IAI Actuarial Core Principles · Actuarial Mathematics for Modelling · Means and variances of assurance and annuity payments

A whole life assurance pays Rs 100,000 at the end of the year of death of a life aged x. The present value random variable has E[Z]/100,000 = A_x = 0.40 and second moment factor 2A_x = 0.22 (calculated at double the force of interest). What is the standard deviation of the present value of the benefit?

The standard deviation is Rs 24,494.90. The variance of the present value equals the sum assured squared times (2A_x minus A_x squared), which is 10^10 times 0.06, or 600 million; taking the square root gives the standard deviation.

  1. ARs 6,000.00
  2. BRs 24,494.90Correct
  3. CRs 42,426.41
  4. DRs 46,904.16
  5. Rs 600,000,000.00

Explanation

Var(Z) = S^2(2A_x - A_x^2) = 10^10 x (0.22 - 0.16) = 6 x 10^8. The standard deviation is the square root, which is Rs 24,494.90. Using 0.22 - 0.40 would give the wrong figure of Rs 42,426.41 (a mismatch of mean and squared mean), and Rs 600,000,000 is the variance, not the standard deviation.

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