chebychev , hoffding , chernoff

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Chebychev , Hoffding , Chernoff. Histo 1. X = 2*Bin(300,1/2) – 300 E[X] = 0. Histo 2. Y = 2*Bin(30,1/2) – 30 E[Y] = 0. Histo 3. Z = 4*Bin(10,1/4) – 10 E[Z] = 0. Histo 4. W = 0 E[W] = 0. A natural question:. - PowerPoint PPT Presentation

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Page 1: Chebychev , Hoffding ,  Chernoff
Page 2: Chebychev , Hoffding ,  Chernoff

0.

0.1

-50 -40 -30 -20 -10 0 10 20 30 40 50

X = 2*Bin(300,1/2) – 300E[X] = 0

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0.

0.1

-50 -40 -30 -20 -10 0 10 20 30 40 50

Y = 2*Bin(30,1/2) – 30E[Y] = 0

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0.

0.1

-50 -40 -30 -20 -10 -38 10 20 30 38 48

Z = 4*Bin(10,1/4) – 10E[Z] = 0

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0.

0.1

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1.

-50 -40 -30 -20 -10 0 10 20 30 40 50

W = 0E[W] = 0

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• Is there a good parameter that allow to distinguish between these distributions?

• Is there a way to measure the spread?

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• The variance of X, denoted by Var(X) is the mean squared deviation of X from its expected value = E(X):

Var(X) = E[(X-)2]. The standard deviation of X, denoted

by SD(X) is the square root of the variance of X.

SD(X) = Var(X).

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2 2Var(X) = E(X ) E(X) .Proof:

E[ (X-)2] = E[X2 – 2 X + 2]

E[ (X-)2] = E[X2] – 2 E[X] + 2

E[ (X-)2] = E[X2] – 22+ 2

E[ (X-)2] = E[X2] – E[X]2

Claim: Var(X) = E[(X-)2].

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1. Claim: Var(X) >= 0.

Pf: Var(X) = (x-)2 P(X=x) >= 0

2. Claim: Var(X) = 0 iff P[X=] = 1.

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Theorem: X is random variable on sample space S, and P(X=r) it’sprobability distribution. Then for any positive real number r:

(proof in book)

2

( )(| ( ) | )

V XP X E X r

r

In words: the probability of finding a value of X farther away from the meanthan r is smaller than the variance divided by r^2.

r

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What is the probability that with 100 Bernoulli trials we find more than89 or less than 11 successes when the prob. of success is ½.

X counts number of successes.EX=100 x ½ =50

V(X) = 100 x ½ x ½ = 25.

P(|X-50|>=40)<=25/40^2 = 1/64

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“k standard deviations”

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Let be random variables that for

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, ,

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What is the probability of getting 25 or fewer heads in 100 coin tosses?