hw4_soln

# hw4_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,2.4.2 2.4.3 2.5.4 2.5.10 2.6.1 2.6.3 2.7.4 2.8.1 2.8.5 and 2.8.11 Problem 2.4.2 (a) The given CDF is shown in the diagram below. FX(x) © ¤¨ ¤¨¥ ¥ ¦¤ § § ¢ £¡ FX x 0 x 1 02 1x0 07 0 x 1 1 x1 1 0.8 0.6 0.4 0.2 0 −2 −1 0 x 1 2 (b) The corresponding PMF of X is ¢ ¢ § § ¢ ¡ § PX x x 1 x0 x1 otherwise ¥ ¦¢ 02 05 03 0 Problem 2.4.3 (a) Similar to the previous problem, the graph of the CDF is shown below. FX(x) ¤ © ¤¨ ¨¥ ¥ ¤ § § ¢ ¡ FX x 0 x 3 04 3x5 08 5 x 7 1 x7 −3 0 57 x (b) The corresponding PMF of X is ¢ § § ¢ ¡ § 1 x 3 x5 x7 otherwise ¢ 04 04 02 0 ¥ ¦¢ PX x 1 0.8 0.6 0.4 0.2 0 Problem 2.5.4 the PMF of X is 02 05 03 0 § § x 1 x0 x1 otherwise 1 02 0 05 ¥ ¢ § ¢ ¢ ¢ ¡ PX x The expected value of X is 1 03 01 ¢ ¡ §  ¡ §  ¡ § x ¥¢ ¡ ¢   x § ∑ xPX EX Problem 2.5.10 By the deﬁnition of the expected value, 1! px ¡ #¡ ¥ ¥ ¥ ¡ ¥ ! 1 1 p n1 x1 1, we have n1 nx np ∑ PX x ! )( ' ¥ " 1 0 ( x p ¥ 0 px 1 ¢( " ¡ ∑ ¢ ' ¢   x 1 " n ¥ n1 np n1 0 x ¢ 1¡ ¢ E Xn ¥ x ¡ x1 x n 1! 1!n 1 x ¥ " ¢  ! n np ∑ nx & " %" " ¡ \$ x1 p ¥ With the substitution x nx p1 x "¡ n ∑x E Xn np ! )( The above sum is 1 because it is he sum of a binomial random variable for n 1 trials over all possible values. ¥ Problem 2.6.1 the PMF of Y is 14 y 1 14 y 2 12 y 3 0 otherwise 2 ¢ 2 ¢ ¢ PY u u . Hence, ¢ 1¡ PU 9 PY 3 ¢ £¡ 14 12 2 2 ¢ £¡ ¡ PY 2 ¢ ¡ 6 ¢ £¡ PU 4 u 1 4 9 . The PMF of U can be ¢ 0. The complete expression for the PMF of U is 2 2 ¢ 2 ¢ 1¡ 2 ¢ ¢ £¡ PU u 14 u 1 14 u 4 12 u 9 0 otherwise ¢ ¢  14 ¢ 2 ¢ £¡ ¢ £¡ For all other values of u, PU u PY 1 ¢ PU 1 PY PY 6 ¢ 5 4 4 ¢ 3 Since Y is never negative, PU u u ¢ P Y2 u  PU Y 2 is SU 6¥ 7¢ ¢ £¡ 1 2 3 , the range of U (a) Since Y has range SY found by observing that 5 4 4 ¢ 3 2 PY y (b) From the PMF, it is straighforward to write down the CDF. 4 9 ¤ 1 u u 9 ¤ ¤ ¨ ¨ 2 ¢ £¡ 2 © FU u 0 u 14 1 12 4 1 u (c) From Deﬁnition 2.14, the expected value of U is 414 912  ¡ 2 114 5 75 ¢ 8¡ 2  ¡ 2 ¢ £¡ u ¢  u § ∑ uPU EU From Theorem 2.10, we can calculate the expected value of U as 22 1 4 32 1 2  ¡ 2 12 1 4  ¡ 2 ¢ £¡ y ¢ y 5 75 ¢ 9¡ 2 ∑ y2 PY § E Y2 EU ¢  As we expect, both methods yield the same answer. Problem 2.6.3 Problem 2.4.3, the PMF of X is x 3 x5 x7 otherwise ¥ ¢ § ¢ § ¢ ¢ ¡ § PX x X satisﬁes w PX ¢  X w ¥ ¢ ¥ P ¡ ¢ ¡ ¥ ¢ PW w (a) The PMF of W 04 04 02 0 This implies 04 PW 3 § ¢ £¡ PX 5 ¢ ¡ 5 ¢ 1¡ ¥ PW PX 02 § ¢ £¡ ¢ £¡ ¥ The complete PMF for W is ¢ § § ¢ ¡ § PW w w 7 w 5 w3 otherwise ¥¢ ¥ ¢ 02 04 04 0 (b) From the PMF, the CDF of W is ¥ ¤ ¤ © ¨¥ ¨¥ ¥ ¦¤ § § ¢ ¡ FW w 0 w 7 02 7w 06 5w 1 w3 5 3 (c) From the PMF, W has expected value ¥ @¡ § 3 7 02 5 04  ¡ § ¥¢ ¦¡ w w 3 04 ¥¢ ¡ § ∑ wPW 22 § EW ¥ PX 7 3 ¢ ¡ 7 04 § PW ¢  Problem 2.7.4 Given the distributions of D, the waiting time in days and the resulting cost, C, we can answer the following questions. (a) The expected waiting time is simply the expected value of D. 2 04 3 03 4 01  ¡ §  ¡ § 1 02  ¡ § ¢ £¡ PD d 23 ¢ 8¡ § A ∑d ¢   d1 § 4 ED ! (b) The expected deviation from the waiting time is µD ¥ E µd ED µD 0 ¢ µD ¢   ¥ C  ED ¢ B ¥ (c) C can be expressed as a function of D in the following manner. ¢ ¢ ¢ D D D D ¢ ¢ £¡ CD 90 70 40 40 1 2 3 4 (d) The expected service charge is 40 0 3  ¡ § 70 0 4  ¡ § 90 0 2 40 0 1 62 dollars ¢ D¡ §  ¡ § EC ¢ Problem 2.8.1 Given the following PMF 02 07 01 0 n0 n1 n2 otherwise ¢ § ¢ § ¢ ¢ 1¡ § PN n § ¢ ¡§ ¥§ ¢ ¥ ¢   § ¢ ¡§  ¡§  ¡§ ¢ ¢ ¡§  ¡§  ¡§ ¢ %  02 0 07 1 (b) E N 2 0 2 02 (c) Var N E N2 0 7 12 2 EN Var N 09 0 1 22 11 41 0 24 4 12 2 09 0 11 0 29 0 29 § ¢ 6   ¢ (d) σN 01 2 § (a) E N Problem 2.8.5 (a) The expected value of X is 41 1 24 12 4 24 2 G  4 ¢ 1     F¢  ¢ ¡ x0 x 2 2 41 2 24  ∑ xPX 3 41 3 24  4 ! EX 4 41 4 24 ¢ E  The expected value of X 2 is  36 16 24 41 2 24 22 41 3 24 32 42 41 4 24 5 ¢ 24 41 1 24 12  2 G   ¢ ¡ x0 41 0 24  x 02 4 ∑ x2PX  4 !  F¢  ¢ E X2 The variance of X is 5 ¢ H  ¡ ¥ ¢ %  Var X 22 ¥ Thus, X has standard deviation σX 2 EX 1 ¢ E X2 Var X 1. ¢ %  ¢ (b) The probability that X is within one standard deviation of its expected value is 1 X 2 ¨ ¨ P2 1  ¥ ¢ B σX P1 ¢  µX ¨  X X ¨ ¨ σX 3  ¨ ¥ P µX  This calculation is easy using the PMF of X . PX 2 PX 3  I¡  ¡ PX 1 3 78 ¢ £¡ ¢  X 2 ¨ P1 ¨ Problem 2.8.11 The PMF of K is the Poisson PMF λ 2" k! k 0 1 otherwise ¢ λk e 0 §§§ ###4 4 ¢ £¡ PK k The mean of K is k2 λk e λ k 2! ¥ ∞ ¡ ¢ ! " ¡ ¥ 2, we obtain λ je j! 0 " ¥ ∞ λ2 ∑ ! ¢ P¡ j λ ¢ ¢ P¡ ! 1 ∑ λ ¥ ¢ ¥  EKK λk e k! λ " ¢   ! k ! By factoring out λ2 and substituting j k0 1 ¥ ∑k k ¡ " ¢ ∞ 1 k To ﬁnd E K 2 , we use the hint and ﬁrst ﬁnd EKK λk 1 e λ 1 k 1! λ∑ " k0 ∞ λ λk e k! " ∑k ¢ ∞ EK λ2 1  The above sum equals 1 because it is the sum of a Poisson PMF over all possible values. Since EK λ, the variance of K is ¥ ¥ λ ¥ ¢     λ2 EK K EK 1 λ2 5 2 ¢ ¡ G  ¥R  QP¡  ¡ G  ¢   E K2 Var K EK λ EK 2 ¢ ¢ ...
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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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