random variables - Suppose P(E | F = P(E i.e knowing F...

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Suppose P ( E | F ) = P ( E ), i.e., knowing F doesn’t help in predicting E . Then E and F are indepen- dent. What we have said is that in this case P ( E | F ) = P ( E F ) P ( F ) = P ( E ) , or P ( E F ) = P ( E ) P ( F ). We use the latter equation as a definition: We say E and F are independent if P ( E F ) = P ( E ) P ( F ) . Example. Suppose you flip two coins. The outcome of heads on the second is independent of the outcome of tails on the first. To be more precise, if A is tails for the first coin and B is heads for the second, and we assume we have fair coins (although this is not necessary), we have P ( A B ) = 1 4 = 1 2 · 1 2 = P ( A ) P ( B ). Example. Suppose you draw a card from an ordinary deck. Let E be you drew an ace, F be that you drew a spade. Here 1 52 = P ( E F ) = 1 13 · 1 4 = P ( E ) P ( F ). Proposition 3.1. If E and F are independent, then E and F c are independent. Proof. P ( E F c ) = P ( E ) - P ( E F ) = P ( E ) - P ( E ) P ( F ) = P ( E )[1 - P ( F )] = P ( E ) P ( F c ) . We say E , F , and G are independent if P ( E F G ) = P ( E ) P ( F ) P ( G ). Example. Suppose you roll two dice, E is that the sum is 7, F that the first is a 4, and G that the second is a 3. E and F
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