Chapt 4d - 4.10 Uniform Distribution 141 4.10 Uniform...

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4.10 Uniform Distribution 141 4.10 Uniform Distribution A continuous random variable X is said to have a uniform distribution over the interval [ a , b ] if its PDF is given by f X ( x ) = 1 b a a x b 0 otherwise It is used to model events that are equally likely to occur at any time within a given time interval. The plot of the PDF is shown in Figure 4.6. The CDF of X is given by F X ( x ) = P [ X x ] = 0 x < a x a b a a x b 1 x b The expected value of X is given by E [ X ] = integraldisplay −∞ xf X ( x ) dx = integraldisplay b a x b a dx = bracketleftbigg x 2 2 ( b a ) bracketrightbigg b a = b 2 a 2 2 ( b a ) = ( b a )( b + a ) 2 ( b a ) = b + a 2 The second moment of X is given by Figure 4.6 PDF of the Uniform Random Variable
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142 Chapter 4 Special Probability Distributions E [ X 2 ] = integraldisplay −∞ x 2 f X ( x ) dx = integraldisplay b a x 2 b a dx = bracketleftbigg x 3 3 ( b a ) bracketrightbigg b a = b 3 a 3 3 ( b a ) = ( b a )( b 2 + ab + a 2 ) 3 ( b a ) = b 2 + ab + a 2 3 Thus, the variance of X is given by σ 2 X = E bracketleftbig X 2 bracketrightbig ( E [ X ] ) 2 = b 2 + ab + a 2 3 b 2 + 2 ab + a 2 4 = b 2 2 ab + a 2 12 = ( b a ) 2 12 Example 4.21 The time that Joe, the teaching assistant, takes to grade a paper is uniformly distributed between 5 minutes and 10 minutes. Find the mean and variance of the time he takes to grade a paper. Solution Let X be a random variable that denotes the time it takes Joe to grade a paper. Since X is uniformly distributed, we find that the mean and variance are as follows: E [ X ] = 10 + 5 2 = 7 . 5 σ 2 X = ( 10 5 ) 2 12 = 25 12 trianglesolid 4.10.1 The Discrete Uniform Distribution A discrete random variable K is said to have a uniform distribution in the range k = a , a + 1 , a + 2 ,..., a + N 1 if it has the PMF p K ( k ) = braceleftBigg 1 N k = a , a + 1 ,..., a + N 1 0 otherwise
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4.10 Uniform Distribution 143 The mean of K is given by E [ K ] = a + N 1 summationdisplay k = a k N = 1 N braceleftBigg a + N 1 summationdisplay k = 1 k a 1 summationdisplay k = 1 k bracerightBigg = 1 N braceleftbigg ( a + N 1 )( a + N ) 2 a ( a 1 ) 2 bracerightbigg = 1 2 { N + 2 a 1 } = 1 2 { ( a + N 1 ) + a } where the second line follows from the fact that n summationdisplay k = 1 k = n ( n + 1 ) 2 Thus, the expected value is the arithmetic average of the lowest and highest values of the random variable as in the case of the continuous uniform random variable. The second moment is given by E bracketleftbig K 2 bracketrightbig = a + N 1 summationdisplay k = a k 2 N = 1 N braceleftBigg a + N 1 summationdisplay k = 1 k 2 a 1 summationdisplay k = 1 k 2 bracerightBigg Using the fact that n summationdisplay k = 1 k 2 = n ( n + 1 )( 2 n + 1 ) 6 we obtain E bracketleftbig K 2 bracketrightbig = 2 N 2 + 6 aN + 6 a 2 6 a 3 N + 1 6 Thus, the variance is given by σ 2 K = N 2 1 12 Example 4.22 Let X be the random variable that denotes the outcome of the roll of a fair die. We know that the PMF of X is given by p X ( x ) = 1 6 x = 1 , 2 ,..., 6
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144 Chapter 4 Special Probability Distributions We can compute the mean and variance of X either directly or via the above formulas. First, by the direct method,
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