This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE302 Spring 2006 HW3 Solutions February 2, 2006 1 Solutions to HW3 Note: Most of these solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics where I thought more detail was appropriate. I have also largely rewritten the solutions to problems 2.10.1, 2.10.2, and 2.10.3. Problem 2.2.1 The random variable N has PMF P N ( n ) = braceleftbigg c (1 / 2) n n = 0 , 1 , 2 , otherwise . (a) What is the value of the constant c ? (b) What is P [ N 1]? Problem 2.2.1 Solution (a) We wish to find the value of c that makes the PMF sum up to one. P N ( n ) = braceleftbigg c (1 / 2) n n = 0 , 1 , 2 otherwise (1) Therefore, 2 n =0 P N ( n ) = c + c/ 2 + c/ 4 = 1, implying c = 4 / 7. (b) The probability that N 1 is P [ N 1] = P [ N = 0] + P [ N = 1] = 4 / 7 + 2 / 7 = 6 / 7 (2) Problem 2.2.2 For random variables X and R defined in Example 2.5, find P X ( x ) and P R ( r ). In addition, find the following probabilities: (a) P [ X = 0] (b) P [ X < 3] (c) P [ R > 1] Problem 2.2.2 Solution From Example 2.5, we can write the PMF of X and the PMF of R as P X ( x ) = 1 / 8 x = 0 3 / 8 x = 1 3 / 8 x = 2 1 / 8 x = 3 otherwise P R ( r ) = 1 / 4 r = 0 3 / 4 r = 2 otherwise (1) From the PMFs P X ( x ) and P R ( r ), we can calculate the requested probabilities ECE302 Spring 2006 HW3 Solutions February 2, 2006 2 (a) P [ X = 0] = P X (0) = 1 / 8. (b) P [ X < 3] = P X (0) + P X (1) + P X (2) = 7 / 8. (c) P [ R > 1] = P R (2) = 3 / 4. Problem 2.2.3 The random variable V has PMF P V ( v ) = braceleftbigg cv 2 v = 1 , 2 , 3 , 4 , otherwise . (a) Find the value of the constant c . (b) Find P [ V { u 2  u = 1 , 2 , 3 , } ]. (c) Find the probability that V is an even number. (d) Find P [ V > 2]. Problem 2.2.3 Solution (a) We must choose c to make the PMF of V sum to one. 4 summationdisplay v =1 P V ( v ) = c (1 2 + 2 2 + 3 2 + 4 2 ) = 30 c = 1 (1) Hence c = 1 / 30. (b) Let U = { u 2  u = 1 , 2 ,... } so that P [ V U ] = P V (1) + P V (4) = 1 30 + 4 2 30 = 17 30 (2) (c) The probability that V is even is P [ V is even] = P V (2) + P V (4) = 2 2 30 + 4 2 30 = 2 3 (3) (d) The probability that V > 2 is P [ V > 2] = P V (3) + P V (4) = 3 2 30 + 4 2 30 = 5 6 (4) ECE302 Spring 2006 HW3 Solutions February 2, 2006 3 Problem 2.2.4 The random variable X has PMF P X ( x ) = braceleftbigg c/x x = 2 , 4 , 8 , otherwise . (a) What is the value of the constant c ? (b) What is P [ X = 4]? (c) What is P [ X < 4]? (d) What is P [3 X 9]? Problem 2.2.4 Solution (a) We choose c so that the PMF sums to one. summationdisplay x P X ( x ) = c 2 + c 4 + c 8 = 7 c 8 = 1 (1) Thus c = 8 / 7....
View
Full
Document
This note was uploaded on 10/19/2008 for the course ECSE 4500 taught by Professor Woods during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 WOODS

Click to edit the document details