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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 1:003:00. Homework 2 is posted online, it is a little tougher than HW 1. Todays reading: 1.51.9, 2.12.2. For Tuesday, read: 2.12.5, 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 RongRong Chen) There are three coins in a box. One is a twoheaded 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 twoheaded 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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 Fall '08
 Chen,R

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