CS2800-Probability_part_c_v.2

# CS2800-Probability_part_c_v.2 - Discrete Math CS 2800 Prof...

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1 Discrete Math CS 2800 Prof. Bart Selman [email protected] Module Probability --- Part c) 1) Linearity of Expectation 2) St. Petersburg (Gambling) Paradox 3) Randomized Algorithms 4) Variance

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2 Linearity of Expectation
3 Linearity of Expectation Thm. 1 2 1 2 ( ... ) ( ) ( ) ... ( ) n n E X X X E X E X E X + + + = + + + Very useful result. Holds even when X_i’s are dependent! Proof: (case n=2) 1 2 1 2 ( ) ( )( ( ) ( )) s S E X X p s X s X s + = +

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2 1 2 1 ( )( ( ) ( ) ( ) ) s S E X s X X p s s X + + = Proof: (case n=2) (defn. of expectation; summing over elementary events in sample space ) 1 2 ( ) ( ) ( ) ( ) s S s S p s X s p s X s = + 1 2 ( ) ( ) E X E X + = ( 29 ( 29 x E X xP X x = = QED Aside, alternative defn. of expectation:
Example application Consider n coin flips of biased coin (prob. p of heads), what is the expected number of heads? Let X_i be a 0/1 r.v such that X_i is 1 iff i th coin flip comes up head. (X_i is called an “ indicator variable ”.) So, we want to know: 1 2 ( ... ) n E X X X + + + 1 2 ( ) ( ) ... ( ) n E X E X E X = + + + What is E(X_i) ? ( ) 1. 0.(1 ) i E X p p p = + - = 1 2 ( ) ( ) ... ( ) n E X E X E X np + + + = So, Linearity of Expectation! QED Holds even if coins are not flipped independently! Consider: all coins “glued” together. Either all “heads” or all “tails”. Still correct expectation! Why? (can be quite counter-intuitive)

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Example Consider n children of different heights placed in a line at random. Starting from the beginning of the line, select the first child. Continue walking, until you find a taller child or reach the end of the line. When you encounter taller child, also select him/her, and continue to look for next tallest child or reach end of line. Question: What is the expected value of the number of children selected from the line?? Hmm. Looks tricky… What would you guess? Lineup of 100 kids... [e.g. 15 more or less?] Lineup of 1,000 kids… [e.g. 25 more or less?]
Let X be the r.v. denoting the number of children selected from the line. 1 2 1 2 ( ) ( ... ) ( ) ( ) ... ( ) n n E X E X X X E X E X E X = + + + = + + + 1 2 ... n X X X X = + + + where i X = 1 1 if the tallest among the first i children. (i.e. will be selected from the line) 0 otherwise { By linearity of expectation, we have:

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What is E(X_i)? What is P(X_1 = 1)? Now, E(X_i) = 0 * P(X_i = 0) + 1 * P(X_i = 1) = 0 * (1 – 1/i) + 1 * (1/i) = 1/i. So, E(X) = 1 + 1/2 + 1/3 + 1/4 … + 1/n 1 ln( ) , 0.57722... ' . n where Euler s c s on n t γ γ + + = e.g. N = 100 E(X) ~ 5 N = 200 E(X) ~ 6 N = 1000 E(X) ~ 7 N = 1,000,000 E(X) ~ 14 N = 1,000,000,000 E(X) ~ 21 Note that the one tallest A: 1 What is P(X_n = 1)? A: 1/n What is P(X_i = 1)? A: 1/i small!! Consider doubling queue: What’s probablility tallest kid in first half? A: ½ (in which case 2 nd half doesn’t add anything!)
9 Indicator Random Variable Recall: Linearity of Expectation ) ( .... ) ( ) ( ) .... ( 2 1 2 1 n n X E X E X E X X X E + + + = + + + Y An indicator random variable is: 0/1 binary random variable. 1 corresponds to an event E, and 0 corresponds to the event did not occur Then E(Y) = 1*P(E) + 0*(1-P(E)) = P(E) Expected number of times event event occurs: E(X1+X2+…) = E(X1)+E(X2)+…. = P(E1)+P(E2)+….

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10 Define for i = 1, … n, a random variable: E [ X i i = 1 n ] = E[X i i = 1 n ] Suppose everyone (n) puts their cell phone in a pile in the middle of the room, and I return them randomly. What is the expected number of students who receive their own phone back? Guess??
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