hw5_soln

hw5_soln - Probability and Stochastic Processes: A Friendly...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,3.1.2 3.1.4 3.2.3 3.2.5 3.3.3 3.3.5 3.4.1 and 3.4.6 Problem 3.1.2 On the X Y plane, the joint PMF is y £ c ¥ ¨ ¨ ¨ 3c 2c 2c 3c ¨ ¨ 1 ¡ 1 c ¢ § c ¨ ¦ PX Y x y 2 x c ¨ ¨ ¤ (a) To find c, we sum the PMF over all possible values of X and Y . We choose c so the sum equals one. ∑ cx 101  # $ 1 PX Y 2 ¦ 1 # $ 1 2 0 2y ¥ ¥  ¥ ¥    © § x y 6c ©  ∑ xy  ¦ ¥ 2c 6c 14c 1 14.  © Thus c y  x © ∑ ∑ PX Y (b) # $ 12  © ¥ 7c  &§ © 3c ¥  %§  ¦ 2c ¦  ¥ c PX Y 2 1  %§  © "! c PX Y 2 0 ¥ PX Y 0 ¦ X § PY  © (c) © #¥ ¦ 7c 12    #¥ ¦ © ! © ¦ ' (d) From the sketch of PX Y x y given above, P X PX Y Y 0. © "! ©  %§ c  &§ c 20 #¥ ¦ 2c PX Y 21  %§ 3c 2 PX Y 0 1 ¥ PX Y ¦ X § PY § © ¥ (e) 21  )§ #¥ ¦  %§ 1 PX Y PX Y 0 ¥ 20 ¦ #¥ ¦ PX Y # $  %§ # $ © #¥ ¦ 8 14 1 1  %§ 8c 2 PX Y 0 1 ¥ PX Y ¦ 1 § © © (! PX  Problem 3.1.4 hh ht th tt and each sample point has probability 1 4. Each samThe sample space is the set S ple outcome specifies the values of X and Y as given in the following table 2  0 1© outcome hh ht th tt XY 01 10 11 20 The joint PMF can represented by the table 0y 1 0 14 14 14 14 0   ¦ §   ¥ © © © Problem 3.2.3 We recognize that the given joint PMF is written as the product of two marginal PMFs PN n and PK k where ¦ # © 7§ © 3§  ¥ © 3§ ¦ ¦ ¦ 0 100 k § ¦ ∑ PN K n k p k 01 otherwise © ¦ ¥ n0 pk 1  ¦ © 4§ 100 k nk n 01 otherwise 666 )$$$ 5 ∑ PN K 0 © ¦ PK k k0 ∞ 100 100n e n! ∑ PN K n k 666 $$$ § 100 PN n  Problem 3.2.5 N is    1 ∞ 100n 1 e 100 ∑ n 1! 100 n k 8 1 §   ¦ ¦  k 100  9! ' © © w. That is, x ∑   © "! # 2 x ∞ PX Y x w ¥ w # xY ¦ ∞ © PX x § © ©  PN § 1∞ ∑ PN n 100 n k  666 $$$ ¦ © © ∞ § © 4§ ∑ 100 100n e 100 ∑ n k n 1! ¦   x 100n e n!  © 4§ © 7§ ¦ PW w ∞  ¥ x in order for W § ¦ © 7§ ∞ PK k # Problem 3.3.3 have Y w ¦ , the marginal PMF of K is  01 100n e 100 ∑ k 0 n 1! © For k k n  PN n 100 § y © xy 0 1 2 © PX Y x x x © Problem 3.3.5 A GE "FD The x y pairs with nonzero probability are shown in the figure at right. From the figure, we observe that for w 01 10, © §# ! ' ' 7§ wY 2 w 2 1 2w w 10 # ¦  # #  6 ¦ 0 01 21 PW w 2 !§ ' 0 01 10 1 # w § #§ # ¦ #! © PW 10,  6  © 6 ¦ © © 7§ PW w 1 ' ¦ ' ' To find the PMF of W , we observe that for w ! w 666 )$$$ © © (! 0 01 10 w B  PX w @ ¦ P min X Y ! 666 )$$$ w C PW © The complete expression for the PMF of W is w 12 otherwise § 2w © # ¦ 0 01 21 0 666 )$$$ 6 © 7§ ¦ PW w 10 Problem 3.4.1 In Problem 3.2.1, we found that the joint PMF of X and Y is y ¦ £ PX Y x y § ¥ 4 3 28 6 28 12 28 1 28 2 28 4 28 ¨H ¨H ¨H 3 2 ¨H ¨H 0 0 1 2 3 4 ¢ ¨H 1 x The expected values and variances were found to be EX 3 Var X 10 7 EY 52 Var Y 34  © "!  © I!   3  ©(!  © "!  We will need these results in the solution to this problem. Y X has expected value  y PX Y x y 13x 36 2 28 14 4 28   © "!   ¥  ¥¥  11 33 12 1 28 1 28 2 28 30 28 15 14 3 12 4 28  ¥ x 1 2 4y ¦ ∑∑ EY X §    © © (a) Random variable W ©  © (b) The correlation of X and Y is xyPX Y x y 236 28 PP  212 28  PP 123 28 414 28 PP   PP PP ¥  ¥¥  111 28 207 28  © "! x 1 2 4y 1 3 4 3 12 28 PP ¥ ∑∑ ¦  E XY §  rX Y ©¥ © © (c) The covariance of X and Y is EXEY © (!  !  #! E XY # σX Y 3 28  ©¥ (d) The correlation coefficient is ¦ © 1 16 ¨¨ H W4 V4   ¦ PX Y x y § 1 12 1 16  ¥ £ W3 V3 W3 V4 18 1 12 1 16 W2 V2 W2 V3 W2 V4   ¨¨ H  ¨H   ¨¨ H  ¨¨H   2 ¨H 3  14 18 1 12 1 16 W1 V1  W1 V2  W1 V3  W1 V4 1 2 3 4    ¨H H¨  0 0 4 ¢  ¨H 1 ¨H 4 x § ©¥ y © min X Y and V ¦ Q ! ! !  Problem 3.4.6 problem is for each possible pair x y, we identify the values of W max X Y . § © Var X Var Y 3 2 210 # Cov X Y ρX Y We observe that W Y and V X . This is a consequence of the statement of the underlying experx are equally likely. Hence Y X iment in which given X x, each value of Y in the set 1 2 and this implies min X Y Y and max X Y X . Using the results from Problem 3.4.4, we have the following answers. © 2 %$$$ 0 666 R © ¦ © 7§ © © 7§ ¦ (a) The expected values are 74 EV EX 52 © "!  © (!  EY   © (!  EW © (!  (b) The variances are 41 48 Var V Var X © "!  © !  Var Y  © (!  © (!  (c) The correlation is E XY © "!  © (! E WV rX Y 5 ©¥  rW V ©¥ (d) The covariance of W and V is Cov X Y 10 16  © "!  Cov W V © "!  (e) The correlation coefficient is § ¦ §   ©¥ ¦ 5 10 16 41 48 5 4 S ρX Y 0 605 6 ρW V 54  Var W ©¥ ...
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This note was uploaded on 02/11/2012 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.

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