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IAI Actuarial Core Principles · Actuarial Statistics · Jointly distributed random variables

Let X ~ Poisson(λ = 4) be the number of claims, and given X = n, the number of large claims Y is Binomial(n, 0.25). What is Var(Y)?

Using the law of total variance, E[Var(Y|X)] = 0.75 and Var(E[Y|X]) = 0.25, giving Var(Y) = 1.00. Indeed thinning a Poisson(4) with probability 0.25 gives a Poisson(1) variable.

  1. A0.75
  2. B1.00Correct
  3. C1.25
  4. D4.00
  5. 0.80

Explanation

E[Y|X] = 0.25X and Var(Y|X) = 0.25*0.75*X = 0.1875X. E[Var(Y|X)] = 0.1875*4 = 0.75. Var(E[Y|X]) = 0.0625*4 = 0.25. Total = 1.00. Option 0.75 omits the variance of the conditional mean. (Y is in fact Poisson(1).)

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