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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 14.384 Time Series Analysis Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . More Empirical Process Theory 1 14.384 Time Series Analysis, Fall 2008 Recitation by Paul Schrimpf Supplementary to lectures given by Anna Mikusheva October 24, 2008 Recitation 8 More Empirical Process Theory This section of notes, especially the subsection on stochastic integrals are based on http://cscs.umich. edu/ ~ crshalizi/weblog/472.html . Convergence of Random Walks 1 [ T ] We never did formally show the weak convergence of to a Brownian motion. Lets fill that hole T t =1 now. Remember the functional central limit theorem. Theorem 1. Functional Central Limit Theorem : If 1. there exists a finite-dimensional distribution convergence of T to (as in ( ?? )) 2. , > there exists a partition of into finitely many sets, 1 ,... k such that lim sup P (max sup T ( 1 ) T ( 2 ) > ) < T i 1 , 2 i | | then T Finite dimensional convergence follows from a standard central limit theorem. We just need to verify the second condition, stochastic equicontinuity. It is not very hard to do this directly. = [0 , 1] is compact, so we can choose our partition to be a collection of intervals of length . This gives P (max sup T ( 1 ) T ( 2 ) > ) P ( sup T ( 1 ) T ( 2 ) > ) i 1 , 2 i | | | 1 2 | < | | 1 [( + ) T ] P ( sup t ) [0 , 1] | T | t =[ T ] Chebyshevs inequality says that if E [ x ] = and V ar ( x ) = 2 , then P ( x > ) 2 2 . Here, [( + ) T ] [( + ) T ] | | 1 1 E [ T t =[ T ] t ] = 0 and V ar ( T t =[ T ] t ) = [ T T ] 2 and these things do not depend on , so [( + ) T ] P ( sup 1 t > ) [ T ] 2 | T | T 2 [0 , 1] t =[ T ] Hence, setting 2 we have the desired result. 2 Stochastic Integrals 2 Stochastic Integrals Our interest in stochastic integrals comes from the fact that we would like to say something about the convergence of 1 1 y t t = T ( t/T ) T ( T ( t/T ) T (( t 1) /T )) T T We know that T W , a Brownian motion. If W were differentiable, wed also know that lim T T ( W ( T/T ) W (([ T ] 1) /T )) = W ( ). Finally, a sum like the above would usually converge to an integral T 1 1 f ( t/T ) g ( t/T ) f ( x ) dg ( x ) T t =1 We would like to generalize the idea of an integral to apply to random functions such as . There are a few problems. First, W is not differentiable. Nonetheless, we can still look at T 1 lim ( t/T )( W ( t/T ) W (( t 1) /T )) T T t =1 We must verify that this sum converges to something. We will call its limit a stochastic integral, and write 1 it as ( t ) dW ( t ). We then want to define stochastic integrals...
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rec08 - MIT OpenCourseWare http://ocw.mit.edu 14.384 Time...

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