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#1




ACTEX p241 (12th edition) Ex 822
Suppose W is 3point 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[YW] is 1/3 for each of W=1, W=2, W=3. Can someone explain why please? I thought P[YW] would be taking the integration of exponential function for each of W=1, 2, and 3: for example, P[YW=w] => P[Yw=1] = e^(y). Please help to explain how ACTEX gets 1/3. Thanks! 
#2




Quote:
They presumably are using the formula E(Y)=E(E(YW)) and are evaluating that as E(YW=1)*P(W=1)+E(YW=2)*P(W=2)+E(YW=3)*P(W=3). Each of those probabilities is 1/3. 
#3




Video Solution
Check out my Youtube Channel, I made a video solution to your question!
https://www.youtube.com/channel/UC1F...ocxx3Sh37tQ1jQ 
#4




Unconditional variance = Mean of variances + variance of the means
= (1 + 4 + 9)/3 + [(12)^2 +(22)^2 + (32)^2]/3 
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