Fundamentals of Probability and Statistics for Engineers

Functions of Random Variables 1215.1.1.1 Discrete Random VariablesLet us first dispose of the case when X is a discrete random variable, since itrequires only simple point-to-point mapping. Suppose that the possible valuestaken by X can be enumerated as x 1 ,x 2 ,.... Equation (5.2) shows that thecorresponding possible values of Y may be enumerated as y 1 ˆ g(x 1 ), y 2 ˆg(x 2 ), . . . . Let the pmf of X be given byThe pmf of y is simply determined asp X …x i †ˆp i ; i ˆ 1; 2; ...: …5:3†p Y …y i †ˆp Y ‰g…x i †Š ˆ p i ; i ˆ 1; 2; ...: …5:4†Ex ample 5. 1. Problem: the pmf of a random variable X is given as81; for x ˆ 1;2>< 1 ; for x ˆ 0;p X …x† ˆ41 ; for x ˆ 1;8>: 1 ; for x ˆ 2;8Determine the pmf of Y if Y is related to X by Y ˆ 2X ‡ 1.Answer: the corresponding values of Y are: g( 1) ˆ 2( 1) ‡ 1 ˆ 1;g(0) ˆ 1; g(1) ˆ 3; and g(2) ˆ 5. Hence, the pmf of Y is given by81 ; for y ˆ 1;2>< 1 ; for y ˆ 1;p Y …y† ˆ41; for y ˆ 3;8>: 1; for y ˆ 5.8Ex ample 5. 2. Problem: for the same X as given in Example 5.1, determine thepmf of Y if Y ˆ 2X 2 ‡ 1.Answer: in this case, the corresponding values of Y are: g( 1) ˆ 2( 1) 2 ‡1 ˆ 3; g(0) ˆ 1; g(1) ˆ 3; and g(2) ˆ 9, resulting in81 ; for y ˆ 1;>< 4 p Y …y† ˆ58 ˆ 12 ‡ 1 ; for y ˆ 3;8>: 1 ; for y ˆ 9:8TLFeBOOK