# soln4 - 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,2.4.1 2.4.3 2.5.3 2.5.5 2.6.1 2.6.3 2.7.2 2.8.1 2.8.7 and 2.9.1 Problem 2.4.1 Using the CDF given in the problem statement we find that (g) From the staircase CDF of Problem 2.4.1, we see that Y is a discrete random variable. The jumps in the CDF occur at at the values that Y can take on. The height of each jump equals the probability of that value. The PMF of Y is 1 4 y 1 1 4 y 2 1 2 y 3 0 otherwise Problem 2.4.3 (a) Similar to the previous problem, the graph of the CDF is shown below. 0 x 3 04 3 x 5 08 5 x 7 1 x 7 1 0.8 0.6 0.4 0.2 0 -3 0 x 5 7 (b) The corresponding PMF of X is 1 ) ( ' & % PX x 0 1) ) ( ( \$ " FX x 04 04 02 0 x 3 x 5 x 7 otherwise FX(x) # # ! PY y " (f) P Y 3 1 2 (e) P Y 1 1 4 (d) P Y 2 1 PY 2 1 1 4 3 4 (c) P Y 2 1 PY (b) P Y 1 (a) P Y 1 0 1 4 2 1 1 2 1 2 Problem 2.5.3 the PMF of Y is 1 4 y 1 1 4 y 2 1 2 y 3 0 otherwise The expected value of Y is y Problem 2.5.5 Problem 2.4.3, the PMF of X is The expected value of X is x Problem 2.6.1 the PMF of Y is 1 4 y 1 1 4 y 2 1 2 y 3 0 otherwise 2 Q P F ID E PU u Q Q P P F D E For all other values of u, PU u 0. The complete expression for the PMF of U is 1 4 u 1 1 4 u 4 1 2 u 9 0 otherwise G F D E F ID E G F D E F HD E G F D E F D E PU 1 PY 1 1 4 PU 4 D B C9 A @ 9 Since Y is never negative, PU u PY u . Hence, PY 2 1 4 PU 9 PY 3 1 2 7 6 8!5 4 3 2 1 \$ ) \$ ( PU u 0 P Y2 u PY u ' & & \$ % \$ 1 2 3 , the range of U (a) Since Y has range SY found by observing that \$ # PY y Y 2 is SU 1 4 9 . The PMF of U can be PY " \$ \$ " # # ! ' & & \$ % EX xPX x 3 04 5 04 PX x 04 04 02 0 x 3 x 5 x 7 otherwise 7 02 22 EY yPY y 11 4 21 4 PY y 31 2 9 4 u (b) From the PMF, it is straighforward to write down the CDF. (c) From Definition 2.14, the expected value of U is u From Theorem 2.10, we can calculate the expected value of U as y As we expect, both methods yield the same answer. Problem 2.6.3 Problem 2.4.3, the PMF of X is This implies The complete PMF for W is (b) From the PMF, the CDF of W is (c) From the PMF, W has expected value w 3 C A H (SP C I Q P C I A Q RP C I A H B9P I H G F EW wPW w 7 02 5 04 @ C 5 98 7 FW w A (@ 0 w 7 7 w 02 06 5 w 1 w 3 E D A D A A B@ 5 4 C & 1 0 PW w 6)5 6 )5 02 04 04 0 w 7 w 5 w 3 otherwise 5 3 3 04 22 % & 1 '0 & 1 0 % & 1 0 4 4 & 1 '0 % & 1 0 & 1 '0 PW 7 PX 7 02 PW 5 PX 5 04 1 '0 & 3 & 2' & 1 0 ' )& (a) The PMF of W X satisfies PW w P X w PX w & % PX x ' (& & 04 04 02 0 x 3 x 5 x 7 otherwise PW 3 PX \$ # ! " # ! " % % ! EU y ! " ! E Y2 y2 PY 12 1 4 22 1 4 32 1 2 EU uPU u 11 4 41 4 FU u 0 u 1 4 1 1 2 4 1 u 1 u u 9 4 9 91 2 5 75 5 75 3 04 Problem 2.7.2 Whether a lottery ticket is a winner is a Bernoulli trial with a success probability of 0 001. If we buy one every day for 50 years for a total of 50 365 18250 tickets, then the number of winning tickets T is a binomial random variable with mean ET 18250 0 001 18 25 Since each winning ticket grosses \$1000, the revenue we collect over 50 years is R The expected revenue is ER 1000E T 18250 But buying a lottery ticket everyday for 50 years, at \$2.00 a pop isn't cheap and will cost us a total of 18250 2 \$36500. Our net profit is then Q R 36500 and the result of our loyal 50 year patronage of the lottery system, is disappointing expected loss of EQ Problem 2.8.1 Given the following PMF ER 36500 18250 Problem 2.8.7 For Y aX b, we wish to show that Var Y a2 Var X . We begin by noting that Theorem 2.12 says that E aX b aE X b. Hence, by the definition of variance. c Wb a ` Y V WU T S I PI E H G D "F E D 8 CA @ Var Y e E aX i 2 b aE X b EX Problem 2.9.1 From the solution to Problem 2.4.1, the PMF of Y is 1 4 y 1 1 4 y 2 1 2 y 3 0 otherwise 4 u u u t t t PY y r i s q p h Wg f Since E X d EX Var X , the assertion is proved. X R Q 2 E a2 X A @ 8 BA @ 9 A @ 7 6 4 53 2 8 5A 9 9 @ ( (d) N Var N 0 29 ) ' & (c) Var N EN 11 09 ( 1 ) "% ) 0 ( % E N2 2 2 "! "! (b) E N 2 0 2 02 0 7 12 0 1 22 (a) E N 02 0 07 1 01 2 09 11 0 29 PN n 02 07 01 0 n 0 n 1 n 2 otherwise 2 a2 E X EX 1000T dollars. \$ # 8 2 The conditional first and second moments of Y are y y The conditional variance of Y is 5 R R C R 2 \$A @ 9 Var Y B EY B 5 2 9 4 Q I D Q PH G F EC E Y2 B 2 1 4 6 2 85 6 4 7 )5 6 4 2 5 4 3 B 2 E Y2 B y2 PY B y 12 1 2 22 1 2 0' & ( )' & ' & % \$# " ! EY B yPY B y 11 2 0 21 2 PY B y PY y PB y B otherwise 1 2 y 1 1 2 y 2 0 otherwise 3 2 5 2 The probability of the event B conditional PMF of Y given B is Y 3 is P B 1 PY 3 1 2. From Theorem 2.19, the 1 ...
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