ECE109 Disc5.pdf

# ECE109 Disc5.pdf - J Connelly UC San Diego ECE 109...

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UC San Diego J. Connelly ECE 109 Discussion 5 Notes Problem 5.1 Suppose we know that a discrete random variable X has mean m and variance σ 2 . We also know that X is uniform over a set of 2 integers. (a) Find the PMF of X . (b) Now let Y = X - m σ . Find the PMF, the mean, and the variance of Y . (c) Find the variance of cos( πY ) . (d) In general, if the PMF P ( X = k ) is an even function, then what is the expected value of X ? Solutions (a) We know that there exist integers a > b such that P ( X = k )= braceleftBigg 1 / 2 if k = a, b 0 otherwise. So m = E [ X ]= a + b 2 and σ 2 = V ar [ X ]= E [ X 2 ] - E [ X ] 2 = a 2 + b 2 2 - a 2 + b 2 +2 ab 4 = a 2 + b 2 - 2 ab 4 = ( a - b ) 2 4 . Together these imply a + b =2 m and a - b =2 σ and so solving for a and b yields a = m + σ and b = m - σ (b) We have P ( Y = k )= P parenleftbigg X - m σ = k parenrightbigg = P ( X = σk + m )= braceleftBigg 1 / 2 if σk + m = m ± σ 0 otherwise = braceleftBigg 1 / 2 if k = ± 1 0 otherwise. So E [ Y ]= 1 2 (1 - 1)=0 , and V ar [ Y ]= E [ Y 2 ]= 1 2 (1 2 +( - 1) 2 )=1 . Please report any typos/errors to [email protected]

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Alternatively, if Z = aX + b , for some constants a and b , then by the linearity of expected value, we have E [ Z ]= E [ aX + b ]= aE [ X ]+ b and V ar [ Z ]= E [ Z 2 ] - E [ Z ] 2 = E [( aX + b ) 2 ] - E [ aX + b ] 2 = ( a 2 E [ X 2 ]+2 abE [ X ]+ b 2 ) - ( a 2 E [ X ] 2 +2 abE [ X ]+ b 2 ) = a 2 ( E [ X 2 ] - E [ X ] 2 ) = a 2 V ar [ X ] Since Y = X σ - m σ , if we take a =1 and b = - m/σ , we get E [ Y ]= 1 σ E [ X ] - m σ = m σ - m σ =0 and V ar [ Y ]= 1 σ 2 V ar [ X ]=1 .
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• Spring '08
• KennethZeger
• Probability theory, Cos

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