  Actuarial Outpost ACTEX p241 (12th edition) Ex 8-22
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#1
 SuriKate Member CAS Join Date: Jun 2013 Studying for Exam 5 Posts: 88 ACTEX p241 (12th edition) Ex 8-22

Suppose W is 3-point discrete uniform random variable {1, 2, 3} with P[W=1]=P[W=2]=P[W=3]=1/3, and suppose that the conditional distribution of Y given W =w is exponential with mean w. Find the unconditional mean and variance of Y.

In the solution, it notes the probability of E[Y|W] is 1/3 for each of W=1, W=2, W=3. Can someone explain why please? I thought P[Y|W] would be taking the integration of exponential function for each of W=1, 2, and 3: for example, P[Y|W=w] => P[Y|w=1] = -e^(-y). Please help to explain how ACTEX gets 1/3.

Thanks!
#2 Gandalf Site Supporter Site Supporter SOA  Join Date: Nov 2001 Location: Middle Earth Posts: 31,374 Quote:
 Originally Posted by SuriKate Suppose W is 3-point discrete uniform random variable {1, 2, 3} with P[W=1]=P[W=2]=P[W=3]=1/3, and suppose that the conditional distribution of Y given W =w is exponential with mean w. Find the unconditional mean and variance of Y. In the solution, it notes the probability of E[Y|W] is 1/3 for each of W=1, W=2, W=3. Can someone explain why please? I thought P[Y|W] would be taking the integration of exponential function for each of W=1, 2, and 3: for example, P[Y|W=w] => P[Y|w=1] = -e^(-y). Please help to explain how ACTEX gets 1/3. Thanks!
They are saying nothing about P(Y|W).
They presumably are using the formula E(Y)=E(E(Y|W)) and are evaluating that as E(Y|W=1)*P(W=1)+E(Y|W=2)*P(W=2)+E(Y|W=3)*P(W=3). Each of those probabilities is 1/3.
#3
 ZtableIsLife CAS Join Date: Feb 2017 Studying for Exam 2/FM College: M.A. Mathematics Favorite beer: Stout/ Porter Posts: 27 Video Solution

Exam P/1 Exam FM/2 #4
 Academic Actuary Member Join Date: Sep 2009 Posts: 9,191 Unconditional variance = Mean of variances + variance of the means

= (1 + 4 + 9)/3 + [(1-2)^2 +(2-2)^2 + (3-2)^2]/3

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