Lecture7_Spring11

Lecture7_Spring11 - Elec 210 Lecture 7 Expectation of a...

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Elec210 Lecture 7 1 Elec 210: Lecture 7 Expectation of a random variable Expected value of a function of a random variable Variance of a random variable Moments of a random variable The human brain weighs 1500 grams, on average .
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Interesting Fact… Einstein’s Brain Einstein’s brain weighs less than average , at 1230 grams!! Elec210 Lecture 7 2
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Definition of the Expected Value Expected value of a discrete random variable defined by The expected value is defined if the sum above converges absolutely: If this sum does not converge, then the expected value does not exist If we view p X ( x ) as the distribution of mass on the points x 1 , x 2 , x 3 , . .. then the expected value is the center of mass of the distribution. Elec210 Lecture 7 3 [] ( ) () X XX k X k xS k mE X x p x x p x == = kXk k xp x <∞ Synonyms: expectation, mean, first moment
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Example 3.11: Mean of Bernoulli RV The Bernoulli random variable assumes two values: 0 and 1 with probabilities Using the definition Note that we should NOT interpret the “expected value” as the value we expect to see on any single trial, since in the example above, the expected value is not even one of the possible values if 0 < p < 1. Elec210 Lecture 7 4 (0) (1 ) (1) XX pp p p =− = [] ( ) 0 ( 0 )1 ( 1 ) X X X xS mE X x p x p p p == = + ⋅=
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Example 3.12 Let X be the number of heads in three tosses of a fair coin. Find E[ X ]. Elec210 Lecture 7 5 Solution The sample space is S X = {0,1,2,3} The probability mass function is given by 33 11 88 8 8 (0) (1) (2) (3) XX pp == = = By the definition 8 8 [] ( ) 0 (0) 1 2 (2) 3 (3) (0 3 6 3) 012 3 1 . 5 8 X X xS XXX X EX xp x ppp p = =⋅ + +⋅ +++ =⋅+ ⋅+⋅+⋅= =
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Example 3.13: Uniform Discrete RV Let X be a discrete uniform random variable assuming values in the set S X = {0,1,2,. .., M -1}. Find E[ X ]. Elec210 Lecture 7 6 Solution Uniform pmf is given by: 1 ( ) for 0,1,. .., 1 X M pk k M == Substituting into the definition, 1 0 11 00 [] ( ) ( ) 1( 1 ) 1 22 X M XX xS k MM kk E Xx p x k p k MMM M ∈= −− =⋅ =  =    Example: If M = 3, S X ={0,1,2} and E[X] = 1
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Relative Frequency Interpretation E[X] corresponds to the “average of X ” in a large number of observations of X . Suppose that we perform an experiment n times. Let x ( j ) be the value of the RV in the j th experiment. Let N k ( n ) be the number of times that value x k is observed. We can compute the average either by summing the observations in order, or by counting the number of times each outcome occurs . Elec210 Lecture 7 7 11 2 2 12 2 () (1) (2) ( ) kn n n k kk k k k k xN n xx x n X nn Nn N n x n x f n x fn x x ++ + == =+ + + + + + + = 
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Relative Frequency Interpretation (cont.) From the previous page, the average value of X in n trials is As the number of trials, n , increases, This implies that Elec210 Lecture 7 8 () kk n k Xx f n = lim ( ) ( ) for all kX k n f np x k →∞ = ( ) [ ] k X k n n X xf n xp x EX =⎯ →= 
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Example 3.14 Suppose a player flips a fair coin 3 times Let X be the number of heads that appears Let Y be the reward, which is $1 for 2 heads, $8 for 3 heads, and nothing otherwise.
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This note was uploaded on 03/27/2011 for the course ELEC 202 taught by Professor ? during the Spring '11 term at HKUST.

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Lecture7_Spring11 - Elec 210 Lecture 7 Expectation of a...

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