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Unformatted text preview: EE 562a Homework Solutions 1 September 10, 2009 1 1. Solution: The cummulative distribution function can be evaluated using conditional probabilities. P x ( u,t ) ( z ) = P { u : x ( u,t ) ≤ z } = P { u = u } P { x ( u,t ) ≤ z  u = u } + P { u = u 1 } P { x ( u,t ) ≤ z  u = u 1 } = 1 2 P { 2sin( πt ) ≤ z } + 1 2 P { 2 t ≤ z } Since the calculation has been “conditioned down to single points in the sample space, the probabilities in this last expression are either 1 or 0 depending on whether the event is is true or false (a function of the choice of random variable index t ). Hence P x ( u,t ) ( z ) = 1 2 U r ( z 2sin( πt )) + 1 2 U r ( z 2 t ) . Here U r ( · ) denotes the rightcontinuous unitstep function, U r ( z ) = , z < 1 , z ≥ . In the above computation we first considered calculations for a particular u ∈ { u ,u 1 } by conditioning on different sample space points being given. Let’s try the computation a different way by first fixing t . For example, consider the case in which t = 0. Then { u : x ( u, 0) ≤ z } = ϕ, z < { u ,u 1 } , z ≥ where ϕ denotes the empty set. Therefore P x ( u,t ) ( z ) = P { u : x ( u,t ) ≤ z } = P { ϕ } , z < P { u ,u 1 } , z ≥ = , z < 1 , z ≥ = U r ( z ) The following special cases special plotted in Figure 1: P x ( u, 0) ( z ) = U r ( z ) P x ( u, 1 4 ) ( z ) = 1 2 [ U r ( z √ 2) + U r ( z 1 2 )] P x ( u, 1 2 ) ( z ) = 1 2 [ U r ( z 2) + U r ( z 1)] P x ( u, 1) ( z ) = 1 2 [ U r ( z ) + U r ( z 2)] 2. This problem challenges the student to use distribution functions in the analysis of simple transformations of random variables. (a) The student should be able to prove that when the strictly monotoneincreasing function g ( z ) performs a onetoone transformation of the real line into itself. Hence every value of w = g ( z ) can be uniquely associated with the value of z that created it through an inverse function z = g 1 ( w ). (Don’t confuse this with the reciprocal [ g ( w )] 1 .) Generally P y ( u ) ( z ) , P { u : y ( u ) ≤ z } = P { u : g ( x ( u )) ≤ z } 2 EE 562a Homework Solutions 1 September 10, 2009 1 1 P x ( u, 0) ( z ) P x ( u, . 25) ( z ) 2 2 1 / 4 √ 2 1 1 P x ( u, . 5) ( z ) P x ( u, 1) ( z ) 1 2 2 Figure 1: Special case plots Using the strictly monotoneincreasing property reduces this to P y ( u ) ( z ) = P { u : x ( u ) ≤ g 1 ( z ) } = P x ( u ) ( g 1 ( z )) . This result holds for all possible values that y ( u ) can take on. We note that g L , lim z →∞ g ( z ) ≤ y ( u ) ≤ lim z →∞ g ( z ) , g U , and hence the complete answer is P y ( u ) ( z ) = , z < g L P x ( u ) ( g 1 ( z )) , g L ≤ z ≤ g U 1 , g U < z. (b) In this case some of the inequalities are reversed and that causes minor changes. Using the strictly monotonedecreasing property on the first equation in (a) gives the result P y ( u ) ( z ) = P { u : x ( u ) ≥ g 1 ( z ) } = 1 P { u : x ( u ) < g 1 ( z ) } = 1 lim z ↑ z P x ( u ) ( g 1 ( z...
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This note was uploaded on 04/03/2010 for the course EE 562a taught by Professor Toddbrun during the Fall '07 term at USC.
 Fall '07
 ToddBrun

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