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Unformatted text preview: " ) = Pr( n )exp( x n ) n = N # = Pr( n )exp[ d ( N $ 2 n )] n = N # . Using the formula for Pr( n ), you can simply the generating function to the closed form solution ( ) = e dN T N , where the term T depends on p , d , and . Provide the value of T and the value of ( ) evaluated at = . ( ) = exp[ dN ] C ( N , n ) p n (1 # p ) N # n e # 2 dn n = N $ = exp[ dN ] C ( N , n )( pe # 2 d ) n (1 # p ) N # n n = N $ = exp[ dN ][ pe # 2 d + 1 # p ] N T = pe ! 2 d + 1 ! p ( = 0) = 1 (normalization) 4. (3 pts). The mean position after N hops can be obtained as x = d ln ! ( " ) / d evaluated at = . What is ln ( ) and what is x in simplified form? ln ( ) = dN + N ln[ pe # 2 d + 1 # p ] d ln ( ) / d = dN + N # 2 dpe # 2 d pe # 2 d + 1 # p x = dN # 2 dpN = Nd (1 # 2 p ) which makes sense because the mean position is Nd for p = 1 (all hops to the left) and Nd for p = 0 (all hops to the right)....
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This note was uploaded on 03/30/2010 for the course SBE 580.429 taught by Professor Joelbader during the Fall '09 term at Johns Hopkins.
 Fall '09
 JoelBader

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