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# Lecture07 - ECE 5510 Random Processes Lecture Notes Fall 2009 Lecture 7 Today(1 Method of Moments cts r.v.s(Y&G 3.7(2 Jaco bian Method cts

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Unformatted text preview: ECE 5510: Random Processes Lecture Notes Fall 2009 Lecture 7 Today: (1) Method of Moments, cts r.v.s (Y&G 3.7), (2) Jaco- bian Method, cts. r.v.s (Kay handout) • Application assignment 1 due today (by midnight) on WebCT • HW 3 is due Thu 5pm. OH today after class until 1:15, Thu 1-3pm. 0.1 Method of Moments, continued Last time : For discrete r.v. X , and a function Y = g ( X ), P Y ( y ) = P [ { Y = y } ] = P [ { g ( X ) = y } ] = P bracketleftbig { X ∈ g- 1 ( y ) } bracketrightbig Def’n: Many-to-One A function Y = g ( X ) is many-to-one if, for some value y , there is more than one value of x such that y = g ( x ), or equivalently, { g- 1 ( y ) } has more than one element. Def’n: One-to-One A function Y = g ( X ) is one-to-one if, for every value y , there is exactly one value x such that y = g ( x ). Bottom line : know that the set { g- 1 ( y ) } can have either one, or many, elements. Example: One-to-One Transform Let X be a discrete uniform r.v. on S X = { 1 , 2 , 3 } , and Y = g ( X ) = 2 X . What is P Y ( y )? Answer: We can see that S Y = { 2 , 4 , 6 } . Then P Y ( y ) = P [ { Y = y } ] = P [ { 2 X = y } ] = P [ { X = y/ 2 } ] = P X ( y/ 2) = braceleftbigg 1 / 3 , y = 2 , 4 , 6 , o.w. 0.2 Continuous r.v.s Method of Moments For continuous r.v. X and a transformation g ( X ), find f Y ( y ) by: 1. Find the CDF by the method of moments. F Y ( y ) = P [ { Y ≤ y } ] = P [ { g ( X ) ≤ y } ] = P bracketleftbig { X ∈ g- 1 ( { Y : Y ≤ y } ) } bracketrightbig For example (starting from f X ) integrate the pdf of X over the set of X which ‘causes’ in...
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## This note was uploaded on 09/15/2011 for the course ECE 5510 taught by Professor Chen,r during the Fall '08 term at University of Utah.

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Lecture07 - ECE 5510 Random Processes Lecture Notes Fall 2009 Lecture 7 Today(1 Method of Moments cts r.v.s(Y&G 3.7(2 Jaco bian Method cts

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