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Unformatted text preview: EE 278 October 23, 2009 Statistical Signal Processing Handout #9 Homework #4 Solutions 1. Two envelopes. A fixed amount a is placed in one envelope and an amount 5 a is placed in the other. One of the envelopes is opened (each envelope is equally probable), and the amount X is observed to be in it. Let Y be the (unobserved) amount in the other envelope. a. Find E parenleftBig Y X parenrightBig . b. Find E parenleftBig X Y parenrightBig . c. Find E( Y ) E( X ) . Solution (10 points) By the design of the experiment, ( X, Y ) = braceleftBigg ( a, 5 a ) with probability 1 2 (5 a, a ) with probability 1 2 We use this joint pmf to find the required expectations. a. E parenleftBig Y X parenrightBig = 1 2 parenleftBig 5 a a parenrightBig + 1 2 parenleftBig a 5 a parenrightBig = 5 2 + 1 10 = 13 5 . b. Symmetrically, E parenleftBig X Y parenrightBig = 13 5 . c. E( X ) E( Y ) = 1 2 (5 a + a ) 1 2 ( a + 5 a ) = 1. This is obvious in retrospect. 2. Schwarz inequality. a. Prove the following inequality, which is known as the Schwarz inequality . ( E( XY ) ) 2 ≤ E( X 2 ) E( Y 2 ) . Hint: Use the fact that E ( ( X + aY ) 2 ) ≥ 0 for every real number a . b. Prove that equality holds if and only if either Y = cX or X = cY for some constant c . c. Use the Schwarz inequality to show that the correlation coefficient ρ X,Y satisfies  ρ X,Y  ≤ 1. d. Show that E ( ( X + Y ) 2 ) ≤ ( radicalbig E( X 2 ) + radicalbig E( Y 2 ) ) 2 . This is called the triangle inequality . Solution (20 points) a. Consider the following quadratic equation in the parameter a : 0 = E(( X + aY ) 2 ) = E( X 2 ) + 2 a E( XY ) + a 2 E( Y 2 ) . Since it is the expected value of a nonnegative random variable, E(( X + aY ) 2 ) ≥ 0. If E(( X + aY ) 2 ) > 0 there are two imaginary solutions, while if E(( X + aY ) 2 ) = 0 there is one real solution. Thus the discriminant must satisfy 4(E( XY )) 2 4 E( X 2 ) E( Y 2 ) ≤ , which we can rewrite as (E( XY )) 2 ≤ E( X 2 ) E( Y 2 ) . Here is another proof. Let a = ± radicalBigg E( X 2 ) E( Y 2 ) . Plugging these value into E(( X + aY ) 2 ) ≥ 0, we obtain E( XY ) ≤ radicalbig E( X 2 ) E( Y 2 ) and E( XY ) ≤ radicalbig E( X 2 ) E( Y 2 ) Combining the two inequalities for E( XY ) yields  E( XY )  ≤ radicalbig E( X 2 ) E( Y 2 ) ⇒ (E( XY )) 2 ≤ E( X 2 ) E( Y 2 ) , which is what we set out to prove. b. If X = cY then (E( cY · Y )) 2 = c 2 (E( Y 2 )) 2 = E(( cY ) 2 ) E( Y 2 ) = E( X 2 ) E( Y 2 ) . and similarly for Y = cX . Conversely, if (E( XY )) 2 = E( X 2 ) E( Y 2 ) then the discriminant of the quadratic equation in part (a) is 0. Therefore E(( X + aY ) 2 ) = 0, which means that X + aY = 0 with probability 1. Hence X = aY with probability 1. Clearly, a = ± radicalBigg E( X 2 ) E( Y 2 ) ....
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This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.
 Fall '09
 BalajiPrabhakar
 Signal Processing

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