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Unformatted text preview: CS70: Satish Rao: Administration. I Midterm regrades requests taken until today after lecture. (Watch piazza!) CS70: Satish Rao: Lecture 23. 1. Linearity of Expectation 2. Applications 3. Important Distributions and Expectations. Expectation. Expectation. Definition: The expectation of a random variable X is E [ X ] = ∑ a a × Pr [ X = a ] . Dice, dice! X- number of pips on 1 die in random roll. Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . X- number of pips on 2 die in random roll. Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . X- number of pips on 2 die in random roll. E [ X ] = 2 × 1 36 + Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . X- number of pips on 2 die in random roll. E [ X ] = 2 × 1 36 + ··· 7 × 1 6 + ··· Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . X- number of pips on 2 die in random roll. E [ X ] = 2 × 1 36 + ··· 7 × 1 6 + ··· + 12 × 1 36 Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . X- number of pips on 2 die in random roll. E [ X ] = 2 × 1 36 + ··· 7 × 1 6 + ··· + 12 × 1 36 = 7 . Dice, dice! X- number of pips on 1 die in random roll. E [ X ] = 1 × 1 6 + ··· + 6 × 1 6 = 7 2 . X- number of pips on 2 die in random roll. E [ X ] = 2 × 1 36 + ··· 7 × 1 6 + ··· + 12 × 1 36 = 7 . (Notice: twice the expected value of one die roll.) Fixed points. Pass out homeworks at random? Fixed points. Pass out homeworks at random? The expected number of students that get their homework back? Fixed points. Pass out homeworks at random? The expected number of students that get their homework back? “The expected number of fixed points in a random permutation.” Expected value of a random variable: E [ X ] = ∑ a a × Pr [ X = a ] . Fixed points. Pass out homeworks at random? The expected number of students that get their homework back? “The expected number of fixed points in a random permutation.” Expected value of a random variable: E [ X ] = ∑ a a × Pr [ X = a ] . For 3 students (permutations of 3 elements): Fixed points. Pass out homeworks at random? The expected number of students that get their homework back? “The expected number of fixed points in a random permutation.” Expected value of a random variable: E [ X ] = ∑ a a × Pr [ X = a ] . For 3 students (permutations of 3 elements): Pr [ X = 3 ] = 1 / 6 , Pr [ X = 1 ] = 1 / 2 , Pr [ X = ] = 1 / 3 . Fixed points....
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