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Unformatted text preview: Chapter 3 Discrete Random Variables 3.1 Random Variables Def 1 A random variable X of a probability space (Ω , A ,P ) is a realvalued function defined on Ω , i.e., X : Ω→ R Example 1 Tossing two fair dice. Then the sample space will be Ω = { ( a,b )  a,b ∈ { 1 , 2 , 3 , 4 , 5 , 6 }} For ω = ( a,b ) ∈ Ω , define 1. X ( ω ) = a + b 2. Y ( ω ) = max( a,b ) 3. Z ( ω ) = ( 1 if a = b if a 6 = b Explain the above random variables and discuss their ranges. 1 Example 2 Cut a wire of length l into two segments. The sample space will be Ω = { ( a,b )  a,b > ,a + b = l } For ω = ( a,b ) ∈ Ω , define 1. X ( ω ) = a 2. Y ( ω ) = min( a,b ) 3. Z ( ω ) = ab Explain the above random variables and discuss their ranges. Remark 1. Although random variables are defined as “functions”, but treating them as “variables” is easier for probability calculation. 2. For example, we will use following notations: P ( X = x ) ≡ P ( { ω  X ( ω ) = x,ω ∈ Ω } ) P ( X > x ) ≡ P ( { ω  X ( ω ) > x,ω ∈ Ω } ) P ( a < X ≤ b ) ≡ P ( { ω  a < X ( ω ) ≤ b,ω ∈ Ω } ) Def 2 A discrete random variable X is a random variable with range being a finite or count able infinite subset { x 1 ,x 2 ,... } of real numbers R . Example 3 All the random variables defined in Example 1 are discrete, while those in Example 2 are not. 3.2 The Probability Mass Functions (Discrete Density Functions) Def 3 Let X be a discrete random variable. The probability mass function (pmf) of X , denoted by p X ( x ) , is defined as p X ( x ) = P ( X = x ) = X X ( ω )= x P ( ω ) 2 Example 4 Determine the pmf’s for random variables X , Y and Z in Example 1, and plot their corresponding diagrams. sol ) 1. • coef ( z x ) of the following equation gives the number of possible outcomes for X = x . ( z + z 2 + z 3 + z 4 + z 5 + z 6 )( z + z 2 + z 3 + z 4 + z 5 + z 6 ) = z 2 (1 + z + z 2 + z 3 + z 4 + z 5 ) 2 = z 2 (1 z 6 ) 2 (1 z ) 2 = ( z 2 2 z 8 + z 14 ) ∞ X j =0 ˆ j + 1 j ! z j coef ( z x ) = ˆ x 1 x 2 ! 2 ˆ x 7 x 8 ! + ˆ x 13 x 14 ! = x 1 if x = 2 , 3 , 4 , 5 , 6 , 7 13 x if x = 8 , 9 , 10 , 11 , 12 otherwise • Therefore, p X ( x ) = ( x 1) / 36 x = 2 , 3 , 4 , 5 , 6 , 7 (13 x ) / 36 x = 8 , 9 , 10 , 11 , 12 otherwise 2. • Let X 1 ,X 2 be the r.v’s representing the outcomes of the 1st and 2nd dice, respec tively. Then p Y ( y ) = P ( Y = y ) = P [ max ( X 1 ,X 2 ) = y ] • Event analysis: [ max ( X 1 ,X 2 ) = y ] = [1 ≤ x 1 ≤ 6][1 ≤ x 2 ≤ 6][ x 1 = y ][ x 2 ≤ y ] + [1 ≤ x 1 ≤ 6][1 ≤ x 2 ≤ 6][ x 2 = y ][ x 1 < y ] = [1 ≤ x 2 ≤ y ][ x 1 = y ] + [1 ≤ x 1 < y ][ x 2 = y ] • Therefore, p Y ( y ) = P [ max ( X 1 ,X 2 ) = y ] = y X x 2 =1 P ( X 1 = y,X 2 = x 2 ) + y 1 X x 1 =1 P ( X 1 = x 1 ,X 2 = y ) = y 36 + y 1 36 = 2 y 1 36 y = 1 , 2 , 3 , 4 , 5 , 6 3 3. Clearly, p Z ( z ) = P ( Z = z ) = 5 6 z = 0 1 6 z = 1 0 otherwise Remark A pmf must satisfy...
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 Spring '11
 Teerana

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