Deviation from Pr[exactly 50.5 Heads] = ? = 0 the Mean Pr[exactly - PowerPoint PPT Presentation

Mathematics for Computer Science Don’t expect the Expectation! MIT 6.042J/18.062J Toss 101 fair coins. E[#Heads] = 50.5 Deviation from Pr[exactly 50.5 Heads] = ? = 0 the Mean Pr[exactly 50 Heads] < 1/13 Pr[50.5 ± 1 Heads] < 1/7 Albert R Meyer, May 10, 2013 devintro.1 Albert R Meyer, May 10, 2013 devintro.2 Don’t expect the Expectation! Don’t expect the Expectation! As #tosses grows, Toss 1001 fair coins. #Heads gets less likely E[#Heads] = 500.5 to be within a fixed Pr[#H = 500 ] 1/39 < distance of the mean Pr[#H = 500.5 ± 1 ] < 1/19 smaller Albert R Meyer, May 10, 2013 Albert R Meyer, May 10, 2013 devintro.3 devintro.4 1

Within a % of the mean? Giving Meaning to the Mean Toss 1001 fair coins. of 1001 Let µ ::= E[R]. What is Pr[#H = 500.5 ± 1%] Pr[R far from µ]? = Pr[#H = 500.5 ± 10] Pr[ |R − µ | > x ] ≈ 0.49 R’s average deviation ? not so bad E[ |R − µ| ] ? Albert R Meyer, May 10, 2013 devintro.5 Albert R Meyer, May 10, 2013 devintro.6 Two Dice with Same Mean Two Dice with Same Mean deviation from the mean Fair Die 1 Fair • E[D 1 ] = 3.5 1.5 on average 0 Pr[D = i] Loaded Die throwing only 1 & 6: 1 • E[D 2 ] = (1+6)/2 = 3.5 also! Loaded 2.5 0 i: 0 1 2 3 4 5 6 7 Albert R Meyer, May 10, 2013 Albert R Meyer, May 10, 2013 devintro.7 devintro.8 2

Dice have Different Deviations Giving Meaning to the Mean The mean alone is not a good Fair Die: predictor of R’s behavior. E[ |D 1 − µ| ] = 1.5 We generally need more about Loaded Die: its distribution, especially E[ |D 2 − µ| ] = 2.5 probable deviation from its mean. Albert R Meyer, May 10, 2013 devintro.9 Albert R Meyer, May 10, 2013 devintro.11 3

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