Module 2. Random Processes. Version 2 ECE IIT, Kharagpur

Transcription

1 Module Random Processes

2 Lesson 6 Functons of Random Varables

3 After readng ths lesson, ou wll learn about cdf of functon of a random varable. Formula for determnng the pdf of a random varable. Let, X be a random varable and ga ( ) s a functon of a real varable a. Then, the expresson = g( x) leads to a new random varable Y wth the followng connotaton: Let s ndcate an outcome of a random experment, as ntroduced earler n Lesson #5. For a gven s, x(s) s a real number and gxs [ ( )] s another real number specfed n terms of x(s) and g(a). Ths new number s the value s () = gxs [()], whch s assgned to the random varable Y. In bref, Y = g (X) ndcates ths functonal relatonshp between the random varables X and Y. The cdf F (b) of the new random varable Y, so formed, s the probablt of the event { b}, consstng of all outcomes s such that s () = gxs [()] b. Ths means, F ( b) = P{ b} = P{ g( s) b.6. For a specfc b, there ma be multple values of a for whch ga ( ) b. Let us assume that all these values of a for whch ga ( ) b, form a set on the a-axs and let us denote ths set as I. Ths set s known as the pont set. So, gxs [ ( )] bf x(s) s a number n the set I,.e. F ( b) = P{ x I}.6. Now, g(a) must have the followng propertes so that g(x) s a random varable : a) The doman of g(a) must nclude the range of the random varable X. b) For ever b such that ga ( ) b, the set I must consst of the unon and ntersecton of a countable number of ntervals snce then onl { b} s an event. c) The events {g(x) = ± } must have zero probablt. Cumulatve Dstrbuton Functon [cdf] of g(x) We wsh to express the cdf F (b) of the new random varable Y where = g(x) n term of the cdf F x (a) of the random varable X and the functon g(a). To do ths, we determne the set I on the a-axs so that ga ( ) band also the probablt that the random varable X s n ths set. Let us assume that F x (a) s contnuous and consder a few examples to llustrate the pont. Example #.6.

4 Let, = g(x) = c.x + d, where c and d are constants [Ths s an equaton of a straght lne]. To fnd F (b), we have to fnd the values of a such that, c.a + d b. b d For c > 0: ca + d b means a c b d b d So, F ( b) = P x = Fx c c b d Whle, for c < 0, ca + d b means a and so c b d b d F( b) = P x = Fx c c Example #.6. Let, = g(x) = x It s eas to see that, for b < 0, F (b) = 0 However, for b 0 a b for b a b and hence, F ( b) P b x b = Fx b Fx b = { } ( ) ( ) Example #.6.3 Let us consder the followng functon g(a): a+ c, a< c ga ( ) = 0, c a c a c, a> c It s a good dea to sketch g(a) versus a to gan a closer look at the functon. Note that, F (b) s dscontnuous at b= g( a) = 0 b the amount Fx( c) Fx( c) Further, for b 0, P{ b} = P{ x b + c} = Fx ( b + c) & for b < 0, P{ b} = P{ x b c} = Fx ( b c) Example #.6.4 Whle we wll dscuss more about lnear and non-lnear quantzers n the next Module, let us consder the smple transfer characterstcs of a lnear quantzer here: Let, ga ( ) = ns.,( n as ) < a nswhere s s a constant, ndcatng a fxed step sze and n s an nteger, representng the n-th quantzaton level. Then for = g( x), the random varable Y takes values bn = ns wth P{ = ns} = P{( n ) s < x ns} = Fx( ns) Fx( ns s)

5 Example #.6.5 a+ c, a 0 Let, ga ( ) =, where c s a constant. Plot g(a) versus a and see that a c, a< 0 g(a) s dscontnuous at a = 0, wth g(0 - ) = -c and g(0 + ) = +c. Ths mples that, F Y (b) = F X (0), for b c. Further, for b c, g( a) b for a b c; hence, F( b) = Fx( b c) c b c, g( a) b for a c; hence, F( b) = F x(0) b c, g( a) b for a b + c; hence, F ( b) = F ( b + c) x An mportant step whle dealng wth functons of random varables s to fnd the pont set I and thereb the cdf F Y (Y) when the functons g(x) and F X (X) are known. In terms of probablt, t s equvalent to fndng the values of the random varable X such that, FY( ) = P{ Y } = P{ X I}. We now brefl dscuss about a concse and convenent relatonshp for determnaton of the pdf of Y,.e f Y (Y). Formula for determnng the pdf of Y,.e., f Y (Y): Let, X be a contnuous random varable wth pdf f x (X) and g(x) be a dfferentable functon of x. [ e.. g( x) 0]. We wsh to establsh a general expresson for the pdf of Y = g(x). Note that, an event { < Y + d} can be wrtten as a unon of several dsjont elementar events {E }. Let, the equaton = g( x) have n real roots x, x,, x n,.e. g(x ) = 0, for =,, n. Then, the dsjont events are of the forms: E = x < X < x f g x s ve { }, ( ) or E = x < X < x + f g x s + ve { }, ( ) In ether case, we can wrte (followng the basc defnton of pdf), that, Pr. of an event = (pdf at x = x). So, for the above dsjont events {E }, we ma, approxmatel wrte, PE { } = Probablt of event E = fx( x) As we have consdered the events E s dsjont, we ma now wrte that, Prob.{ < Y ( + d) } = f Y (). d = f x (x ). + f x (x ) f x (x n ). n

6 n = f x (x ). = The above expresson can equvalentl be wrtten as, n fy( ) = fx( x). d = n = f = X d ( x). Let us note that, at the -th root of = g(x), wth respect to x, evaluated at x = x. Usng the above convenent notaton, we fnall get, n Y( ) X( )/ ( ), = d = g ( x ). = value of the dervatve of g(x) f = f x g x.6.3 Here, x s the -th real root of = g(x) and g ( x ) 0. If, for a gven, = g(x) has no real root, then fy ( ) = 0 as X beng a random varable and x beng real, t can not take magnar values wth non-zero probablt. Let us take up a small example before concludng ths lesson. Example #.6.6 Let X be a random varable known to follow unform dstrbuton between -π and +π. So, the mean of X s 0 and ts probablt denst functon [pdf] s:, π < x π fx ( x) = π 0, otherwse Now consder a new random varable Y whch s a functon of X and the functonal relatonshp s, Y = g(x) = snx. So, we can wrte, = g(x) = sn x. Further, one can easl observe that, the pdf of Y exsts for -.0 <.0. Let us frst consder the nterval 0 <.0: The roots of sn x = 0 for >0 are, x = sn - () and x = π - sn - (). dg( x) Further, = cosx whle dg( x) = cos( sn ) and x= x

7 dg( x) x= x We see that, dg( x) dg( x) ( π sn ) = cos ( cos π.cos sn ) sn π.sn ( sn = + ) = cos( sn ) x x = f X( x ) f ( x ) fy ( ) = + g ( x ) ( ) X g x f f = + X(sn ) X( π sn ) = =., 0 < π π Followng smlar procedure for the range - < 0, t can ultmatel be shown that,., < fy ( ) = π 0, otherwse Problems Q.6.) Q.6.) Let, =x + 3x+. If pdf s x s f X (x), determne an expresson for pdf of. Sketch the pdf of of problem.6., f X has u form dstrbuton between - and +.