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

N is Bernoulli with P(N=1)=0.3. Given N=1, X has density 2e^(-2x) for x>0; given N=0, X has density e^(-x) for x>0. What is P(N=1 | X>1)?

The probability is 0.3e^(-2) divided by 0.3e^(-2)+0.7e^(-1). Survival probabilities beyond 1 are e^(-2) for N=1 and e^(-1) for N=0, and Bayes' theorem weights them by the prior 0.3 and 0.7.

  1. A0.3
  2. B0.3e^(-2)/(0.3e^(-2)+0.7e^(-1))Correct
  3. C0.3e^(-1)/(0.3e^(-1)+0.7e^(-2))
  4. D0.3e^(-2)/(0.3e^(-2)+0.7)
  5. 0.6e^(-2)/(0.6e^(-2)+0.7e^(-1))

Explanation

Given N=1, P(X>1)=e^(-2); given N=0, P(X>1)=e^(-1). By Bayes, P(N=1|X>1)=0.3e^(-2)/(0.3e^(-2)+0.7e^(-1)). The option using the density 2e^(-2) at 1 conditions on X=1, not X>1.

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