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# lecture5 - Lecture-5 Functions of a Random Variable Let X...

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1 Lecture-5 Functions of a Random Variable Let X be a r.v defined on the model and suppose g ( x ) is a function of the variable x . Define Is Y necessarily a r.v? If so what is its PDF pdf Clearly if Y is a r.v, then for every Borel set B , the set of for which must belong to F . Given that X is a r.v, this is assured if is also a Borel set, i.e., if g ( x ) is a Borel function. In that case if X is a r.v, so is Y , and for every Borel set B ), , , ( P F ). ( X g Y = (1) ), ( y F Y ? ) ( y f Y ξ B Y ) ( ) ( 1 B g - )). ( ( ) ( 1 B g X P B Y P - = (2)

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2 In particular Thus the distribution function as well of the density function of Y can be determined in terms of that of X . To obtain the distribution function of Y , we must determine the Borel set on the x -axis such that for every given y , and the probability of that set. At this point, we shall consider some of the following functions to illustrate the technical details. (29 . ] , ( ) ( )) ( ( ) ) ( ( ) ( 1 y g X P y X g P y Y P y F Y -∞ = = = - ξ (3) y x g - ) ( 1 ) ( X g Y = b aX + 2 X | | X X ) ( | | x U X X e X log X 1 X sin
3 Example 5.1: Solution: Suppose and On the other hand if then and hence b aX Y + = (4) . 0 a (29 ( 29 . ) ( ) ( ) ( ) ( - = - = + = = a b y F a b y X P y b aX P y Y P y F X Y ξ (5) . 1 ) ( - = a b y f a y f X Y (6) , 0 < a ( , 1 ) ( ) ( ) ( ) ( - - = - = + = = a b y F a b y X P y b aX P y Y P y F X Y (7) . 1 ) ( - - = a b y f a y f X Y (8)

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4 From (6) and (8), we obtain (for all a ) Example 5.2: If then the event and hence For from Fig. 5.1, the event is equivalent to . | | 1 ) ( - = a b y f a y f X Y (9) . 2 X Y = (29 . ) ( ) ( ) ( 2 y X P y Y P y F Y = = ξ (10) (11) , 0 < y {} , ) ( 2 φ = y X . 0 , 0 ) ( < = y y F Y (12) , 0 y } ) ( { } ) ( { 2 y X y Y = . ) ( 2 1 x X x 2 X Y = X y 2 x 1 x Fig. 5.1
5 Hence By direct differentiation, we get If represents an even function, then (14) reduces to In particular if so that (29 . 0 , ) ( ) ( 2 1 ) ( - + = y y f y f y y f X X Y (14) ) ( x f X ). ( 1 ) ( y U y f y y f X Y = (15) ), 1 , 0 ( N X , 2 1 ) ( 2 / 2 x X e x f - = π (16) . 0 ), ( ) ( ) ( ) ( ) ( ) ( 1 2 2 1 - - = - = = y y F y F x F x F x X x P y F X X X X Y ξ (13)

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6 and substituting this into (14) or (15), we obtain the p.d.f of to be On comparing this with (3.36), we notice that (17) represents a Chi-square r.v with n = 1, since Thus, if X is a Gaussian r.v with
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lecture5 - Lecture-5 Functions of a Random Variable Let X...

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