mtsamp_sn

mtsamp_sn - EE 278 November 4, 2009 Statistical Signal...

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Unformatted text preview: EE 278 November 4, 2009 Statistical Signal Processing Handout #13 Sample Midterm Examination Problems 1. Inequalities. Label each of the following statements with =, , , or None . Label a statement with = if equality always holds. Label a statement with or if the corresponding inequality holds in general and strict inequality holds sometimes. If no such equality or inequality holds in general, label the statement as None . Justify your answers. a. P( A ) vs. 1- (P( A c ,B ) + P( A c ,B c )). b. E( X 1 X 2 | X 3 ) vs. E( X 1 | X 3 ) E( X 2 | X 3 ) if X 1 and X 2 are independent. c. E[Var( X | Y,Z )] vs. E[Var( X | Y )]. d. E[Var( X | Y )] vs. E[Var( X | g ( Y ))]. (Hint: use the result of part (c).) e. E Z [E( X 2 | Z ) E( Y 2 | Z )] vs. [E Z (Cov( X,Y | Z ))] 2 . f. E parenleftBig log 2 parenleftBig 1 + X parenrightBigparenrightBig vs. 1 if X 0 and E( X ) 1. g. P { ( XY ) 2 > 16 } vs. 1 / 8 if E( X 4 ) = E( Y 4 ) = 2. Solution a. =. By the law of total probability P( A ) = 1- P( A c ) = 1- (P( A c ,B ) + P( A c ,B c )) . b. None . Independence does not necessarily imply conditional independence. c. . By the law of conditional variances (and conditioning both sides on Y ), it follows that E[Var( X | Y )] = E[Var( X | Y,Z )] + E[Var(E( X | Y,Z ))] . Thus E[Var( X | Y )] E[Var( X | Y,Z )]. This makes sense because with more observations ( Y,Z ), the MSE of the best estimate of the signal X should be less than or equal to that observing only Y . d. . From the previous result, it follows that E[Var( X | Y,g ( Y ))] E[Var( X | g ( Y ))] . But g ( Y ) is completely determined by Y , thus E[Var( X | Y,g ( Y ))] = E[Var( X | Y )]. This result makes sense because in general Y provides better information about the signal X than any function of it. e. . First note that E( X 2 | Z ) Var( X | Z ) and E( Y 2 | Z ) Var( Y | Z ). Thus E( X 2 | Z ) E( Y 2 | Z ) Var( X | Z )Var( Y | Z ) . Now, using Schwarz inequality, we obtain Var( X | Z )Var( Y | Z ) (Cov( X,Y | Z )) 2 . Taking expectations of both sides, we obtain E [Var( X | Z )Var( Y | Z )] E bracketleftbig (Cov( X,Y | Z )) 2 bracketrightbig . But E [(Cov( X,Y | Z )) 2 ] [E(Cov( X,Y | Z ))] 2 . f. . We use Jensens inequality twice and the fact that E ( X ) 1 E parenleftBig log 2 (1 + X ) parenrightBig log 2 parenleftBig 1 + E parenleftBig X parenrightBigparenrightBig log 2 parenleftBig 1 + radicalbig E( X ) parenrightBig log 2 (1 + 1) 1 ....
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mtsamp_sn - EE 278 November 4, 2009 Statistical Signal...

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