lecture04 - ECE 5510: Random Processes Lecture Notes Fall...

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Unformatted text preview: ECE 5510: Random Processes Lecture Notes Fall 2008 Lecture 4 Today: (1) Conditional Probability II, (2) Combinations and Permutations, (3) Discrete random variables Announcements: Homework 1 due today at 5pm, solutions will be posted to- morrow just after 5pm. OH today are 1:00-3:00. Homework 2 is posted online, it is a little tougher than HW 1. Todays reading: 1.5-1.9, 2.1-2.2. For Tuesday, read: 2.1-2.5, 3.1, 3.2. 1 Conditional Probabilities II Independence of Sets: Sets A and B are indep. if P [ A B ] = P [ A ] P [ B ]. Conditional probability of event A given event B : P [ A | B ] = P [ B A ] P [ B ] . 1.1 Bayes Rule We know that P [ A | B ] = P [ A B ] P [ B ] so it is also clear that P [ A | B ] P [ B ] = P [ A B ]. But equivalently, P [ A B ] = P [ B | A ] P [ A ] So Defn: Bayes Rule P [ A | B ] = P [ B | A ] P [ A ] P [ B ] Not a whole lot different from definition of Conditional Probabil- ity. But, it does say explicitly how to get from P [ B | A ] to P [ A | B ]. 1.2 Trees A graphical method for organizing prior, posterior information. Example: (From Rong-Rong Chen) There are three coins in a box. One is a two-headed coin, another is a fair coin, ECE 5510 Fall 2008 2 Figure 1: Tree. and the third is a biased coin that comes up heads 75 per- cent of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the prob- ability that it was the two-headed coin? Answer: P [ C 2 | H ] = P [ H | C 2 ] P [ C 2 ] P [ H | C 2 ] P [ C 2 ] + P [ H | C f ] P [ C f ] + P [ H | C b ] P [ C b ] = 1 / 3 1 / 3 + 1 / 2(1 / 3) + 3 / 4(1 / 3) = 4 / 9 1.3 Conditional and Independence We know that for any two sets A and B , that P [ A B ] = P [ A | B ] P [ B ]....
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lecture04 - ECE 5510: Random Processes Lecture Notes Fall...

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