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MIT6_042JF10_lec16

# MIT6_042JF10_lec16 - "mcs-ftl 2010/9/8 0:40 page 431#437 16...

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16 “mcs-ftl” — 2010/9/8 — 0:40 — page 431 — #437 Independence 16.1 Definitions Suppose that we ﬂip two fair coins simultaneously on opposite sides of a room. Intuitively, the way one coin lands does not affect the way the other coin lands. The mathematical concept that captures this intuition is called independence : Definition 16.1.1. Events A and B are independent if Pr ŒB� D 0 or if ± ² Pr A j B D Pr ŒA�: (16.1) In other words, A and B are independent if knowing that B happens does not al- ter the probability that A happens, as is the case with ﬂipping two coins on opposite sides of a room. 16.1.1 Potential Pitfall Students sometimes get the idea that disjoint events are independent. The opposite is true: if A \ B ; , then knowing that A happens means you know that B D does not happen. So disjoint events are never independent—unless one of them has probability zero. 16.1.2 Alternative Formulation Sometimes it is useful to express independence in an alternate form: Theorem 16.1.2. A and B are independent if and only if Pr ŒA \ B� D Pr ŒA� Pr ŒB�: (16.2) Proof. There are two cases to consider depending on whether or not Pr ŒB� D 0 . Case 1 . Pr ŒB� D 0/ : If Pr ŒB� D 0 , A and B are independent by Definition 16.1.1 . In addition, Equation 16.2 holds since both sides are 0. Hence, the theorem is true in this case. Case 2 . Pr ŒB� > 0/ : By Definition 15.1.1 , ± ² Pr ŒA \ B� D Pr A j B Pr ŒB�:

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432 “mcs-ftl” — 2010/9/8 — 0:40 — page 432 — #438 Chapter 16 Independence So Equation 16.2 holds if ± ² Pr A j B D Pr ŒA�; which, by Definition 16.1.1 , is true iff A and B are independent. Hence, the theorem is true in this case as well. ± 16.2 Independence Is an Assumption Generally, independence is something that you assume in modeling a phenomenon. For example, consider the experiment of ﬂipping two fair coins. Let A be the event that the first coin comes up heads, and let B be the event that the second coin is heads. If we assume that A and B are independent, then the probability that both coins come up heads is: 1 1 1 Pr ŒA \ B� D Pr ŒA� Pr ŒB� D 2 2 D 4 : In this example, the assumption of independence is reasonable. The result of one coin toss should have negligible impact on the outcome of the other coin toss. And if we were to repeat the experiment many times, we would be likely to have A \ B about 1/4 of the time. There are, of course, many examples of events where assuming independence is not justified, For example, let C be the event that tomorrow is cloudy and R be the event that tomorrow is rainy. Perhaps Pr ŒC� D 1=5 and Pr ŒR� D 1=10 in Boston. If these events were independent, then we could conclude that the probability of a rainy, cloudy day was quite small: 1 1 1 Pr ŒR \ C� D Pr ŒR� Pr ŒC� D � D : 5 10 50 Unfortunately, these events are definitely not independent; in particular, every rainy day is cloudy. Thus, the probability of a rainy, cloudy day is actually 1=10 .
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MIT6_042JF10_lec16 - "mcs-ftl 2010/9/8 0:40 page 431#437 16...

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