Welcome to St. Petersburg! Breaking Vegas • Game set-up • Consider even money bet (e.g., bet “Red” in roulette) We have a fair coin (come up “heads” with p = 0.5) p = 18/38 you win $Y, otherwise (1 – p) you lose $Y Let n = number of coin flips before first “tails” Consider this algorithm for one series of bets: 1. Y = $1 You win $2 n 2. Bet Y • How much would you pay to play? 3. If Win, stop 4. if Loss, Y = 2 * Y, goto 2 • Solution Let Z = winnings upon stopping Let X = your winnings 2 18 20 18 20 18 E[Z] 1 2 3 4 i 1 1 ( 2 1 ) ( 4 2 1 ) ... 1 1 1 1 1 38 38 38 38 38 0 1 2 3 i E[X] = 2 2 2 2 ... 2 2 2 2 2 2 i i 20 18 i 18 20 18 1 i 0 i j 1 2 2 1 1 38 38 38 38 38 20 i 0 j 1 i 0 1 0 2 38 i I’ll let you play for $1 million... but just once! Takers? Expected winnings ≥ 0. Use algorithm infinitely often! Vegas Breaks You Variance • Why doesn’t everyone do this? • Consider the following 3 distributions (PMFs) Real games have maximum bet amounts You have finite money o Not be able to keep doubling bet beyond certain point Casinos can kick you out • But, if you had: No betting limits, and • All have the same expected value, E[X] = 3 Infinite money, and • But “spread” in distributions is different Could play as often as you want... • Variance = a formal quantification of “spread” • Then, go for it! And tell me which planet you are living on Variance Computing Variance m 2 • If X is a random variable with mean m then the Var ( X ) E [( X ) ] m variance of X , denoted Var( X ), is: 2 ( x ) p ( x ) Var( X ) = E [( X – m ) 2 ] x m m 2 2 ( x 2 x ) p ( x ) x m m • Note: Var( X ) ≥ 0 2 2 x p ( x ) 2 xp ( x ) p ( x ) x x x m m 2 2 Ladies and gentlemen, please E [ X ] 2 E [ X ] welcome the 2 nd moment! • Also known as the 2nd Central Moment, or m m 2 2 2 E [ X ] 2 square of the Standard Deviation m 2 ] 2 E [ X 2 2 E [ X ] ( E [ X ]) 1
Variance of 6 Sided Die Properties of Variance • Var(aX + b) = a 2 Var(X) • Let X = value on roll of 6 sided die Proof: • Recall that E[X] = 7/2 = E[(aX + b) 2 ] – (E[aX + b]) 2 Var(aX + b) • Compute E[X 2 ] = E[a 2 X 2 + 2abX + b 2 ] – (aE[X] + b) 2 1 1 1 1 1 1 91 = a 2 E[X 2 ] + 2abE[X] + b 2 – (a 2 (E[X]) 2 + 2abE[X] + b 2 ) 2 2 2 2 2 2 2 E [ X ] 1 2 3 4 5 6 6 6 6 6 6 6 6 = a 2 E[X 2 ] – a 2 (E[X]) 2 = a 2 (E[X 2 ] – (E[X]) 2 ) 2 2 Var ( X ) E [ X ] ( E [ X ]) = a 2 Var(X) • Standard Deviation of X, denoted SD(X), is: 2 91 7 35 X SD ( ) Var ( X ) 6 2 12 Var(X) is in units of X 2 SD(X) is in same units as X Jacob Bernoulli Bernoulli Random Variable • Experiment results in “Success” or “Failure” • Jacob Bernoulli (1654-1705), also known as “James”, was a Swiss mathematician X is random indicator variable (1 = success, 0 = failure) P(X = 0) = p(0) = 1 – p P(X = 1) = p(1) = p X is a Bernoulli Random Variable: X ~ Ber(p) E[X] = p Var(X) = p(1 – p) • Examples • One of many mathematicians in Bernoulli family coin flip • The Bernoulli Random Variable is named for him random binary digit • He is my academic great 11 -grandfather whether a disk drive crashed • Resemblance to Charlie Sheen weak at best Binomial Random Variable Three Coin Flips • Three fair (“heads” with p = 0.5) coins are flipped • Consider n independent trials of Ber(p) rand. var. X is number of successes in n trials X is number of heads X is a Binomial Random Variable: X ~ Bin(n, p) X ~ Bin(3, 0.5) n 3 1 0 3 i n i P ( X 0 ) p ( 1 p ) P ( X i ) p ( i ) p ( 1 p ) i 0 , 1 ,..., n 0 8 i 3 3 ( ) 1 1 2 By Binomial Theorem, we know that P X i P ( X 1 ) p ( 1 p ) 1 8 i 0 3 • Examples 3 2 1 P ( X 2 ) p ( 1 p ) 2 8 # of heads in n coin flips # of 1’s in randomly generated length n bit string 3 1 3 0 P ( X 3 ) p ( 1 p ) 3 8 # of disk drives crashed in 1000 computer cluster o Assuming disks crash independently 2
Error Correcting Codes Error Correcting Codes (cont) • Error correcting codes • Using error correcting codes: X ~ Bin(7, 0.1) Have original 4 bit string to send over network 7 0 7 P ( X 0 ) ( 0 . 1 ) ( 0 . 9 ) 0 . 4783 Add 3 “parity” bits, and send 7 bits total 0 Each bit independently corrupted (flipped) in transition 7 1 6 P ( X 1 ) ( 0 . 1 ) ( 0 . 9 ) 0 . 3720 with probability 0.1 1 X = number of bits corrupted: X ~ Bin(7, 0.1) P(X = 0) + P(X = 1) = 0.8503 But, parity bits allow us to correct at most 1 bit error • What if we didn’t use error correcting codes? • P(a correctable message is received)? X ~ Bin(4, 0.1) P(X = 0) + P(X = 1) P(correct message received) = P(X = 0) 4 0 4 P ( X 0 ) ( 0 . 1 ) ( 0 . 9 ) 0 . 6561 0 • Using error correction improves reliability ~30%! Genetic Inheritance Properties of Bin(n, p) • Person has 2 genes for trait (eye color) • We have X ~ Bin(n, p) Child receives 1 gene (equally likely) from each parent n n n n k k i n i k i n i E [ X ] i p ( 1 p ) i p ( 1 p ) i i Child has brown eyes if either (or both) genes brown i 0 i 1 n n 1 i n ! n ( n 1 )! Child only has blue eyes if both genes blue Noting that: i n i i ! ( n i )! ( i 1 )! (( n 1 ) ( i 1 ))! i 1 Brown is “dominant” (d) , Blue is recessive (r) n n 1 n 1 n 1 where Parents each have 1 brown and 1 blue gene k k 1 i 1 n i k 1 j n ( j 1 ) E [ X ] np i p ( 1 p ) np ( j 1 ) p ( 1 p ) , i 1 j i j 1 i 1 j 0 • 4 children, what is P(3 children with brown eyes)? k 1 npE [( Y 1 ) ], where Y ~ Bin ( n 1 , p ) Child has blue eyes: p = (½) (½) = ¼ (2 blue genes) Set k = 1 E[X] = np P(child has brown eyes) = 1 – (¼) = 0.75 Set k = 2 E[X 2 ] = npE[Y + 1] = np[(n – 1)p + 1] X = # of children with brown eyes. X ~ Bin(4, 0.75) Var(X) = np[(n – 1)p + 1] – (np) 2 = np(1 – p) 4 3 1 P ( X 3 ) ( 0 . 75 ) ( 0 . 25 ) 0 . 4219 3 • Note: Ber(p) = Bin(1, p) PMF for X ~ Bin(10, 0.5) PMF for X ~ Bin(10, 0.3) P(X=k) P(X=k) k k 3
Power of Your Vote • Is it better to vote in small or large state? Small: more likely your vote changes outcome Large: larger outcome (electoral votes) if state swings a (= 2n) voters equally likely to vote for either candidate You are deciding (a + 1) st vote n n 2 n 1 1 ( 2 n )! P ( 2 n voters tie ) 2 n n 2 2 n ! n ! 2 Use Stirling’s Approximation: n 1 / 2 n n ! n e 2 2 n 1 / 2 2 n ( 2 n ) e 2 1 P ( 2 n voters tie ) 2 n 1 2 n 2 n n e 2 2 n 1 c 2 a ( ac ) Power = P(tie) * Elec. Votes = ( a / 2 ) Larger state = more power 4
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