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As said in the title I need to prove that an event is independent from all other events iff its 0 or 1. One side is pretty simple, if I assume the event is 0 or 1 probability the answer is immediate.

I'm having trouble formulating the other side, i.e if I assume there is an event which is independent from all other, I need to show its probability is 0 or 1. I understand that if I assume such an event exists its occurrence will never affect the probability of every other event, therefore if I assume by contradiction that its probability is not 0 or 1, an observer cannot be sure that such an event is independent from all others, because he will not be sure of when it will happen or not.

Frankly what I wrote doesn't sound exactly right to me, but I can't seem to formulate any mathematical proof, I'm not sure of how to represent the fact its independent from all the other events mathematically.

Hopefully you could give me a hint on where to start at least, thanks!

Xsy
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2 Answers2

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Let $A \subset \Omega$ be an event that is independent from all all other events, i.e. for every event $B \neq A$, $$ P(A \cap B) = P(A)P(B) \text{.} $$

In particular, for $B = A^C$, i.e. the complement of $A$,$$ 0 = P(\emptyset) = P(A \cap A^C) = P(A)P(A^C) = P(A)(1-P(A)) \text{.} $$ I leave the rest to you...

fgp
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  • ah. "All other events" referring to the event of the event not happening. I was confused why it had to be 0 or 1. makes perfect sense now. – Guy Mar 14 '14 at 09:54
  • Well that makes perfect sense, thank's a lot! Just one follow-up, does this mean that every event that is independent from its complement has a 0 or 1 probability, therefore every event that is independent from its complement is independent of all other events? Or am I missing something here? – Xsy Mar 14 '14 at 10:08
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    Yes. Note that events $A$ with probability $0$ or $1$ are even independent from themselves!. For events with probability $0$ or $1$, the formal notion of independence doesn't really match the intuitive notion that independence means "The occurence of one event doesn't provide information about the possible occurence of the other". Or, put differently, the formal notion of "information" behaves a bit counter-intuitively for events with probability $0$ or $1$ – fgp Mar 14 '14 at 11:01
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I'll retract this as an answer but leave it here for those similarly confused

Not true it seems to me.

Consider the probablity of getting heads or tails (H, T) on the toss of a coin and also getting a number (1 to 6) on the throw of a dice. p(H) = 1/2 and p(2) = 1/6. But the events are surely indenpendent ?

From the subsequent answer and comments I assume that I should be considering $\Omega $ as a probability space consisting of the combined events. So (H, 2) is not independent from all the other outcomes - correct ?

Tom Collinge
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