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Unformatted text preview: EE 562a Homework Solutions 3 October 1, 2009 1 1. Solution: (a) In this problem we will use the shorthand notation of m i = the i th component of m y k ij = the ij th element of K y First consider event A : Event A y(u,2) y(u,1) In order to bound the probability with Chebychevs inequality, we use a little trick: P ( A ) = P { y ( u, 1) < } = P { y ( u, 1) m 1 < m 1 } P { y ( u, 1) m 1  > m 1 } E { y ( u, 1) m 1  2 } m 2 1 = k 11 m 2 1 = P ( A ) 2 / 10 2 = 0 . 02 . The first inequality is from the bound P { x ( u ) > a } P { x ( u )  > a } . For the event B : Event B 4040 y(u,2) y(u,1) y(u,2)=y(u,1)40 To bound the probability, consider the random variable z ( u ) = y ( u, 1) y ( u, 2) . The mean and variance of z ( u ) are given by m z = E { z ( u ) } = E { y ( u, 1) y ( u, 2) } = m 1 m 2 = 0 2 EE 562a Homework Solutions 3 October 1, 2009 V ar { z ( u ) } = E { ( z ( u ) m z ) 2 } = E { ([ y ( u, 1) m 1 ] [ y ( u, 2) m 2 ]) 2 } = E { ( y ( u, 1) m 1 ) 2 }  2 E { ( y ( u, 1) m 1 )( y ( u, 2) m 2 ) } + E { ( y ( u, 2) m 2 ) 2 } = k 11 2 k 12 + k 22 = 6 . Then P ( B ) = P { z ( u ) > 40 } P { z ( u )  > 40 } E { z ( u )  2 } 40 2 = 6 / 1600 = 0 . 00375 Event C corresponds to the outside of a circle of radius 10, centered at m y since k y ( u ) m y k > 10 k y ( u ) m y k 2 > 100 , k y ( u ) m y k 2 = ( y ( u, 1) m 1 ) 2 + ( y ( u, 2) m 2 ) 2 Event C 10 10 y(u,2) y(u,1) Chebychevs inequality applies directly in this case P ( C ) = P {k y ( u ) m y k > 10 } E {k y ( u ) m y k 2 } 10 2 = Tr ( K y ) 10 2 = P ( C ) 4 / 100 = 0 . 04 Remark : The easy way to determine the region of the event is to draw the boundary first  that is turn the inequality into an equality and plot it in the plane. This gives a line, circle, etc., so to find out on which side of the boundary the event lies test a simple point (the origin is the easiest if it is not on the boundary). (b) We have already evaluated the mean and variance of the random variables in this prob lem. Since y ( u, 1) is a Gaussian random variable, and since z ( u ) is Gaussian because it is generated by a linear operation on y ( u ), the answers here are simply evaluations of Gaussian integrals. First consider event A : P ( A ) = P { y ( u, 1) < } = Z p y ( u, 1) ( x ) dx = Z exp ( ( x m 1 ) 2 / 2 k 11 ) 2 k 11 dx Putting this integral in standard form by substituting z = ( x m 1 ) / k 11 gives P ( A ) = Z m 1 / k 11 exp( z 2 / 2) 2 dz = Z 10 / 2 exp( z 2 / 2) 2 dz exp( 7 . 07 2 / 2) 7 . 07 2 7 . 9 10 13 EE 562a Homework Solutions 3 October 1, 2009 3 Coming up with a numerical answer may have been difficult for you here. Of course computers are always available. But computers rely on numerical algorithms and some very good ones are available in Abramowitz and Stegun (see hint). I used the symmetry of the probability density of an elementary Gaussian random variable and relations...
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This note was uploaded on 04/03/2010 for the course EE 562a taught by Professor Toddbrun during the Fall '07 term at USC.
 Fall '07
 ToddBrun

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