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Old 06-16-2013, 04:15 PM
SuriKate SuriKate is offline
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Default 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.

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Old 06-16-2013, 04:34 PM
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Gandalf Gandalf is offline
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Quote:
Originally Posted by SuriKate View Post
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.
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Old 04-08-2017, 10:53 AM
ZtableIsLife ZtableIsLife is offline
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Thumbs up Video Solution

Check out my Youtube Channel, I made a video solution to your question!

https://www.youtube.com/channel/UC1F...ocxx3Sh37tQ1jQ

Exam P/1 Exam FM/2
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  #4  
Old 04-10-2017, 09:01 PM
Academic Actuary Academic Actuary is offline
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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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