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Calculus and Beyond Homework Help
Conditional expectation
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[QUOTE="Kate2010, post: 2548449, member: 223479"] [h2]Homework Statement [/h2] An email is sent on the network in which the recipients (0,1,2,3,4,5} are in communication. 1 can send to 4 and 2 2 to 1,3,5 3 to 0,2,5 4 to 1, 5 5 to 0,2,4 0 to 3 and 5 If a message is sent to 2,3,4,5 it is forwarded randomly to a neighbour (even if this means a repeat). 0 and 1 never forward messages. Let e[SUB]k[/SUB] be the expected number of time that a message starting at k is passed on. Find e[SUB]4[/SUB]. [h2]Homework Equations[/h2] E(X) = [tex]\sum[/tex] E(X|A)P(A) [h2]The Attempt at a Solution[/h2] I think I need to partition this but I'm unsure on the partition. Let X be the number of times a message is sent on E(X) = E(X|1st move is to 0)P(1st move is to ) + E(X|1st move is to 1)P(1st move is to 1) + E(X|1st move is to 2)P(1st move is to 2) + E(X|1st move is to 3)P(1st move is to 3) + E(X|1st move is to 4)P(1st move is to 4) + E(X|1st move is to 5)P(1st move is to 5) I'm not sure how I'd work these out using a general k to start at. If I assume I start at 4, as I'm trying to find e[SUB]4[/SUB] then I get e[SUB]4[/SUB] = 0 + 1x(1/2) + 0 + 0 + e[SUB]5[/SUB] (1/2) 2e[SUB]4[/SUB] = 1 + e[SUB]5[/SUB] e[SUB]5[/SUB] = 1x(1/2) + 0 + e[SUB]2[/SUB] (1/3) + 0 + e[SUB]4[/SUB] (1/3) I don't really think I'm going about this the right way, I would have thought I need to find a formula for starting at a general k but I don't know how. [/QUOTE]
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