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anonymous
 5 years ago
Show that for binary random variables X, Y, the eventlevel independence implies random variable independence. (eg. Suppose getting heads on your first coin toss is independent of getting heads on your next coin toss. Show that any combination of heads or tails on the first toss is independent of any combination of heads or tails on the second toss.)
anonymous
 5 years ago
Show that for binary random variables X, Y, the eventlevel independence implies random variable independence. (eg. Suppose getting heads on your first coin toss is independent of getting heads on your next coin toss. Show that any combination of heads or tails on the first toss is independent of any combination of heads or tails on the second toss.)

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Owlfred
 5 years ago
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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the proof must be related to the fact that the p(tails) = 1 p(heads). So you could prove all four cases separately? (I'll use "T" to stand for independent because there's no upsidedown T symbol.) Let x^o be the event where we get heads on the first toss & y^o the event of heads on the second toss. x^1 will be tails on the first toss; y^1, tails on the second toss. We are given that (x^o T y^o). Looking at the first case, (x^1 T y^o) We can state that x^1 = 1x^o, so we could state the desired case as (1x^o T y^o), right? Now is the part I'm not sure about. What happens to constants (products) in independence relations? It seems like they should just factor out. If they do, we are left with (x^o T y^o)  our first equation. In other words, (x^1 T y^o) is true. Similar proofs should hold of the other two cases: (x^o T y^1) & (x^1 T y^1) . So since we can prove independence of all four cases, we should be able to conclude (X T Y). Does this sound right? Is there anything more rigorous I should be doing?
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