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Probability, EE351K Statistics and Random Processes Instructor:
Shakkottai/Vishwanath Homework 6: Solutions Spring 2008
{shakkott/sriram}@ece.utexas.edu
Problem 1: pX (k) = p e MX ()  k k! + (1  p) e X  k k! , for k 0. Hence, = E(e = ) ek + (1
 p) e  e k k=0 (p k! k k! ek ) = pe (e )k k=0 k! + (1  p)e (e )k k=0 k! = pe ee + (1 
p)e ee , = pe(e since ez = zk k=0 k! , for all complex numbers z 1) + (1  p)e(e 1) This
expression for MX () is valid for all complex numbers .
Problem 2: Part a: To find a and b, we use the following two properties of the MGF: 1.
MX (0) = 1 2. dMX () d =0 = E[X] MX () = ae + be4(e 1) . MX () = ae + be4(e 1) 4e . MX
(0) = a + 4b. Hence, 1. a + b = 1 2. a + 4b = 3 a = 1/3, b = 2/3 . Part b: MX () = 1 e + 2
e4(e 1) . 3 3 So, this MGF is a convex combination of two MGFs, namely M1 () = e and
M2 () = e4(e 1) . As a result, the PMF of X will be the corresponding convex
combination of the PMFs associated with the two MGFs. (This follows from the fact that
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This note was uploaded on 02/05/2012 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.
 Spring '07
 BARD

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