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# solutions-2 - Solution to Homework#2 36-754 7 February 2006...

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Unformatted text preview: Solution to Homework #2, 36-754 7 February 2006 Exercise 5.3 (The Logistic Map as a Measure- Preserving Transformation) The logistic map with a = 4 is a measure-preserving transfor- mation, and the measure it preserves has the density 1 /π p x (1- x ) (on the unit interval. 1. Verify that this density is invariant under the action of the logistic map. 2. Simulate the logistic map with uniformly distributed X . What happens to the density of X t as t → ∞ ? a There are a number of ways to do this; here is one which will feed into later material. It does not use the fact that the invariant density is the β (1 / 2 , 1 / 2) distribution, or that it has a closed-form integral, Z x dy π p y (1- y ) = 2 π sin- 1 √ x though both of these are fine facts to have handy. Let’s write the mapping as F ( x ) = 4 x (1- x ). Solving a simple quadratic equation gives us the fact that F- 1 ( x ) is the set 1 2 ( 1- √ 1- x ) , 1 2 ( 1 + √ 1- x ) . Notice, for later use, that the two solutions add up to 1. Notice also that F- 1 ([0 ,x ]) = 0 , 1 2 ( 1- √ 1- x ) ∪ 1 2 ( 1 + √ 1- x ) , 1 . Now we consider P ( X n +1 ≤ x ), the cumulative distribution function of X n +1 . P ( X n +1 ≤ x ) = P ( X n +1 ∈ [0 ,x ]) = P ( X n ∈ F- 1 ([0 ,x ]) ) = P X n ∈ , 1 2 ( 1- √ 1- x ) ∪ 1 2 ( 1 + √ 1- x ) , 1 = Z 1 2 ( 1- √ 1- x ) ρ n ( y ) dy + Z 1 1 2 ( 1+ √ 1- x ) ρ n ( y ) dy 1 where ρ n is the density of X n . So we have an integral equation for the evolution of the density, Z x ρ n +1 ( y ) dy = Z 1 2 ( 1- √ 1- x ) ρ n ( y ) dy + Z 1 1 2 ( 1+ √ 1- x ) ρ n ( y ) dy This sort of integral equation is complicated to solve directly. Instead, take the derivative of both sides with respect to x ; we can do this through the fundamental theorem of calculus. On the left hand side, this will just give ρ n +1 ( x ), the density we want. ρ n +1 ( x ) = d dx Z 1 2 ( 1- √ 1- x ) ρ n ( y ) dy + d dx Z 1 1 2 ( 1+ √ 1- x ) ρ n ( y ) dy = ρ n 1 2 ( 1- √ 1- x ) d dx 1 2 ( 1- √ 1- x )- ρ n 1 2 ( 1 + √ 1- x ) d dx 1 2 ( 1 + √ 1- x ) = 1 4 √ 1- x ρ n 1 2 ( 1- √ 1- x ) + ρ n 1 2 ( 1 + √ 1- x ) Notice that this defines a linear operator taking densities to densities. (YouNotice that this defines a linear operator taking densities to densities....
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## This note was uploaded on 12/20/2011 for the course STAT 36-754 taught by Professor Schalizi during the Spring '06 term at University of Michigan.

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solutions-2 - Solution to Homework#2 36-754 7 February 2006...

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