Week3SolutionsPart1 - EENG_6629_81 RANDOM PROCESSES IN...

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EENG_6629_81 RANDOM PROCESSES IN COMMUNICATION Instructor: Gloria B. Reinish ASSIGNMENT #3 SOLUTIONS PROBLEM 2.12 PX X X n xx x PP P K K x K () / / / (, ,) ! ! !.......... ! ........ ! !!! 1 2 3 123 12 111 14 3 1 2 1 4 1 4 3 16 = = = === = = = PROBLEM 2.13 To prove: µσ XX np npq where q p == = , 2 1 µ X k n kn k k n k EX kPx k Px k n k pq wh e r e n k n knk k n =  = = = = {} ( ) ! ! ( ) ! ! ! ( ) ! 0 0 For k k n soEX k n n k k k n k k n k = = −− = = 00 1 1 1 , ! ! ( ) ! ,{ } ! ! ( ) ! ! ! ( ) ! Let kk ' =− 1 n kn k nn p np n n p n k np because n k k n k k n k n k n ! !( ' )! ! !( ' )! ! !( ' )! ' ' ' '' ' ' ' = = −= = = = = +− = = = ∑∑ 1 1 1 1 1 1 1 0 1 11 0 1 1 0 1 1 0 1 1 = = k n ' 0 1 1 1
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σµ XX EX 22 2 =− {} kPx k k n kn k pq k n p q it equals for k k n kn k Let k k n k k k n k k n nk k k n k k n k k n ( ) ! !( )! ! !( )! () ! ! ( ) ! ! ! ! ( ) ! 0 2 0 2 1 1 2 1 00 1 11 1 1 1 == = = = = −− =−+ = −+ = = = = ∑∑ n k n k Note that k for k and n k np k nn n k k n k k n k k n X k k n ! ! ( ) ! ! ! ( ) ! (( ) , ! ! ( ) ! ! ! ! ( ) ! ( ) ! ! ( + −= = = += = = = = 10 1 1 1 1 12 2 21 1 2 2 µ k pp q n p k k n )! 2 = + Let kk ' 2 p n k n p n k n k Therefore E X n p np np k n k n k n ( ) ! ! ( ) ! ! ! ( ) ! ' ,{ } '' ' ' ' 0 2 2 0 2 2 0 2 2 1 2 2 2 2 2 1 + = = + = = = E X n p np np np np np np p npq 2 2 2 2 2 2 1 = + = ( )
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Week3SolutionsPart1 - EENG_6629_81 RANDOM PROCESSES IN...

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