LINEAR COMBINATIONS OF RANDOM VARIABLES in .NET

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LINEAR COMBINATIONS OF RANDOM VARIABLES
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A random variable is sometimes de ned as a function of one or more random variables. The CD material presents methods to determine the distributions of general functions of random variables. Furthermore, moment-generating functions are introduced on the CD
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5-7 LINEAR COMBINATIONS OF RANDOM VARIABLES
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and used to determine the distribution of a sum of random variables. In this section, results for linear functions are highlighted because of their importance in the remainder of the book. References are made to the CD material as needed. For example, if the random variables X1 and X2 denote the length and width, respectively, of a manufactured part, Y 2X1 2X2 is a random variable that represents the perimeter of the part. As another example, recall that the negative binomial random variable was represented as the sum of several geometric random variables. In this section, we develop results for random variables that are linear combinations of random variables.
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De nition Given random variables X1, X2, p , Xp and constants c1, c2, p , cp, Y c1 X1 c2 X2 p cp Xp (5-36)
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is a linear combination of X1, X2, p , Xp.
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Now, E(Y) can be found from the joint probability distribution of X1, X2, p , Xp as follows. Assume X1, X2, p , Xp are continuous random variables. An analogous calculation can be used for discrete random variables. E1Y2 1c1x1 c2 x2 p cp xp 2 fX1 X2 p Xp 1x1, x2, p , xp 2 dx1 dx2 p dxp
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x1 fX1 X2 p Xp 1x1, x2, p , xp 2 dx1 dx2 p dxp x2 fX1 X2 p Xp 1x1, x2, p , xp 2 dx1 dx2 p dxp xp fX1 X2 p Xp 1x1, x2, p , xp 2 dx1 dx2 p dxp
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By using Equation 5-24 for each of the terms in this expression, we obtain the following.
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Mean of a Linear Combination
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If Y
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c1 X1
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c2 X2 E1Y2
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cp Xp, c2E 1X2 2 p cp E 1Xp 2 (5-37)
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c1E 1X1 2
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Furthermore, it is left as an exercise to show the following.
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CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS
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Variance of a Linear Combination
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If X1, X2, p , Xp are random variables, and Y general V1Y 2 c2V1X1 2 1 c2V1X2 2 2 p c2V1Xp 2 p
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c1 X1
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c2 X2
cp Xp, then in (5-38)
2 a a cicj cov1Xi, Xj 2
If X1, X2, p , Xp are independent, V1Y 2 c2V1X1 2 1 c2V1X2 2 2 p c2V1Xp 2 p (5-39)
Note that the result for the variance in Equation 5-39 requires the random variables to be independent. To see why the independence is important, consider the following simple examX1. Clearly, X1 and X2 are not indeple. Let X1 denote any random variable and de ne X2 pendent. In fact, XY 1. Now, Y X1 X2 is 0 with probability 1. Therefore, V(Y) 0, regardless of the variances of X1 and X2. EXAMPLE 5-35 In 3, we found that if Y is a negative binomial random variable with parameters p and r, Y X1 X2 p Xr, where each Xi is a geometric random variable with parameter p and they are independent. Therefore, E1Xi 2 1 p and E1Xi 2 11 p2 p2 . From Equation 2 5-37, E1Y 2 r p and from Equation 5-39, V 1Y 2 r11 p2 p . An approach similar to the one applied in the above example can be used to verify the formulas for the mean and variance of an Erlang random variable in 4. EXAMPLE 5-36 Suppose the random variables X1 and X2 denote the length and width, respectively, of a manufactured part. Assume E(X1) 2 centimeters with standard deviation 0.1 centimeter and E(X2) 5 centimeters with standard deviation 0.2 centimeter. Also, assume that the covariance between X1 and X2 is 0.005. Then, Y 2X1 2X2 is a random variable that represents the perimeter of the part. From Equation 5-36, E1Y 2 and from Equation 5-38 V1Y2 22 10.12 2 22 10.22 2 0.04 0.16 0.04 2 2 21 0.0052 0.16 centimeters squared 0.4 centimeters. 2122 2152 14 centimeters
Therefore, the standard deviation of Y is 0.161 2
The particular linear combination that represents the average of p random variables, with identical means and variances, is used quite often in the subsequent chapters. We highlight the results for this special case.