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Unformatted text preview: Lecture 7. More on Independence. Random Variables More on Independence. Comment . On the quiz last Thursday, more than half the class asserted incorrectly that“independent events are events that have no outcomes in common.” So, let us re view this idea. Two events A and B are said to be independent if the occurrence of one has no effect on the probability of the other. In terms of conditional probability, we may interpret the independence of A and B to mean P ( A ) = P ( A  B ): the probability of A is the same as the probability of A given B . But we might also interpret it to mean P ( B ) = P ( B  A ). Which is the right way? Fortunately, if P ( A ) and P ( B ) are both nonzero: P ( B ) = P ( B  A ) ⇐⇒ P ( B ) = P ( B ∩ A ) P ( A ) ⇐⇒ P ( A ) P ( B ) = P ( B ∩ A ) ⇐⇒ P ( A ) = P ( A ∩ B ) P ( B ) ⇐⇒ P ( A ) = P ( A  B ) Thus, either of these conditions implies the other. To make the symmetry of the indepen dence relation obvious, we use the condition P ( A ) P ( B ) = P ( B ∩ A ) as the definition. Let us repeat it: Definition. Events A and B are said to be independent if P ( A ) P ( B ) = P ( B ∩ A ). This definition has the advantage that it does not mention conditional probabilities, though of course it implies that the conditional probabilities are equal to the unconditioned ones, which was the idea that motivated the concept of independence. It has the advantage that we don’t need to worry if the probability of one of the events happens to be zero. The definition implies that an event of probability zero is independent of any other event. Caution. Two events are said to be mutually exclusive if they have no outcomes in common. If A and B are mutually exclusive, then A ∩ B is the empty event, and therefore P ( A ∩ B ) = 0. Accordingly, if A and B have nonzero probability and are mutually exclusive, then they are not independent. It is a common for learners to assume erroneously that the two concepts are the same. Example. Example 1.12 of our textbook reads: “An elevator with two passengers stops at the second, third and fourth floors. If it is equally likely that a passenger gets off at any of the three floors, what is the probability that the passengers get off at different floors.” Comment. Let’s say the passengers are Mary and Sarah. The problem stipulates that Mary has a 1 / 3 probability of getting off at the 2 nd floor, a 1 / 3 probability of getting off at the 3 rd floor and a 1 / 3 probability of getting off at the 4 th floor. The same is true 1 of Sarah. The sample space consists of all the ways the two of them may get off. The outcomes, therefore, correspond to the cells in the following table.outcomes, therefore, correspond to the cells in the following table....
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This note was uploaded on 11/29/2011 for the course MATH 3355 taught by Professor Britt during the Spring '08 term at LSU.
 Spring '08
 Britt
 Probability

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