# RP-HW1-sol - Kyung Hee University Department of Electronics...

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1 Kyung Hee University Department of Electronics and Radio Engineering C1002900 Random Processing Homework 1 Solutions Spring 2010 Professor Hyundong Shin Issued: March 22, 2010 Due: March 31, 2010 Reading: Course textbook Chapters 1–6 HW 1.1 A random variable X has a cumulative distribution function (CDF)   ,. x X Fx e x  2 10 (a) By definition    X Fx PXx    X P XF e  2 11 1 .  . X PX Fe e      4 4 21 2 12 1 1 2 Note that  22 because   X is continuous at . x 2  20 because   X is continuous at x 2 as we have already noticed. (b)   X x X dF x d f xe x dx dx   2 1u     . x X f x  2 2u Because   X is continuous,   X f x has no impulses. (c) Since y can take on only the values 0 or 1 ,   Y fy will have the form           Y XX fy PY y PY y y PX y Fy F y     01 1 1 2 1         Y e y e y  44

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2   Y f y has impulses   Y f y will be discontinuous. HW 1.2 Let X and Y be independent identically distributed (i.i.d.) random variables with common density function  , , f   10 1 0o t h e r w i s e . Let SXY  . (a) X and Y are independent     SX Y f sff s  . X and Y are identically distributed     S f sf f s . That is,      S f f s d   . 0 1 1   f s 1 s 1   fs  Convolve graphically by flipping and shifting as shown. 0 1 2 s 1   S f s , , , S ss s s     01 21 2 0e l s e Notice that   S f s integrates to 1 as it must.
3 (b) Find and sketch   XS f xs versus x with s viewed as a known parameter. Intuitively, we expect to be distributed like Y but with a shift by S since XSY  . Formally we must be a bit cauful since ,, XYS only take values over a finite interval with non-zero probability:      X SX S f sx f x fx s fs . We are given   X f x and   S f s is a normalizing constant which we have computed for all s in part (a). Thus we must compute   f sx . Because Y is uniform on  , 01 , is uni- form on , xx 1 : , , , , xsx x   11 0e l s e 1a n d1 l s e Using our result from part (a) we obtain: x s 1   f , , s x s s s  For 0 1 1 0 l s e s x s  1 2   f , , s s s    For 1 2 1 2 l s e s 1 1 So we see that is distributed like Y but with a shift and a renormalization depending on S .

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## This note was uploaded on 06/10/2010 for the course ELECTRONIC C1002900 taught by Professor Hyungdongshin during the Spring '10 term at Kyung Hee.

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RP-HW1-sol - Kyung Hee University Department of Electronics...

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