Lecture10

Lecture10 - Lecture 10: Bounding Transcriber: Andy Parrish...

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Unformatted text preview: Lecture 10: Bounding Transcriber: Andy Parrish Goal: Show that (1- / 2) 2 t i 6 =0 (1- i / 2) 2 t n (1- 2 / 4) t , where t is any positive integer. For a positive integer t and a vertex x , let f t,x = x Z t- , where x ( v ) = 1 if x = v and 0 otherwise, Z = I + P 2 is the probability transition matrix for a lazy random walker, and = ~ 1 D vol G is the stationary distribution. Fact 1 : d 1 / 2 x f t,x ( y ) d- 1 / 2 y = d 1 / 2 y f t,y ( x ) d- 1 / 2 x = x ( I- L 2- I ) * y , where I = * = JD vol G is projection onto the 0 eigenvector for L . Proof: d 1 / 2 x f t,x ( y ) d- 1 / 2 y = x D 1 / 2 ( Z t- ) D- 1 / 2 * y = x D 1 / 2 I + P 2 t- JD vol G ! D- 1 / 2 * y = x I + D- 1 / 2 AD- 1 / 2 2 ! t- D 1 / 2 JD 1 / 2 vol G ! * y = x I- L 2 t- I ! * y = x I- L 2- I t * y because I is idempotent and L I = I L = 0 Since the matrix here is symmetric, multiplying by x on the left and * y on the right is the same as multiplying by y on the left and * x on the right. Fact 2 : i 6 =0 (1- i / 2) 2 t = x d- 1 x y d y f 2 t,y ( x ) Proof: { (1- i / 2) 2 t } n- 1 i =0 are the eigenvalues for ( I- L 2 ) 2 t . To exclude the eigenvalue for i = 0, 1 we should subtract projection onto . Thus we get....
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This note was uploaded on 09/30/2011 for the course MATH 262 taught by Professor Aterras during the Fall '08 term at UCSD.

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Lecture10 - Lecture 10: Bounding Transcriber: Andy Parrish...

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