Linear Combination of Random Variables Linear Combination of Random Variables For independent random variables and . A linear combination can be written as + or + . When independent object are taken, and independent object are taken: + ⋯ + + + ⋯+ Mean / Expected Value ( (
+ )= ( )+ + )= ( )+
(
( )
+ ⋯+
+
+ ⋯+
)=
( )+
( )
Variance Var( Var(
+ ) = Var( ) + ) = Var( ) +
)
Var( + ⋯ + + +⋯+ = Var( ) + Var( )
Var( )
Standard Deviation sd( sd(
+ ) = sd( ) + )= sd( )
+
sd( )
sd(
Normal Distribution For independent random variables distribution
and
+ ⋯+
+
+ ⋯+
with normal distributions then
)=
+
sd( )
+
sd( )
also has a normal
Example A coffee machine dispenses volumes of coffee that are normally distributed with a mean 240 mL and standard deviation 8 mL. The machine also has the option of adding milk to a cup of coffee, where the volume of milk dispensed is also normally distributed with mean 10 mL and standard deviation 2 mL. Let the random variable represent the volume of coffee the machine dispenses and let the random variable represent the volume of milk the machine dispenses. and are independent random variables. The mean of the volume of the combined drink, that is, a cup of coffee with milk is ( + ) = ( ) + ( ) = 240 + 10 = 250 mL The mean of the volume of the combined drink, that is, a cup of coffee with milk is Var( + ) = Var( ) + Var( ) = 8 + 2 = 10 mL Example Given that is a normal random variable with mean 10 and standard deviation 8, and that is a normal random variable with mean 3 and standard deviation 2, and and are independent random variables, the random variable defined by = − 3 will have mean and standard deviation given by = ( − 3 ) = ( ) − 3 ( ) = 10 − 3 × 3 = 1 = sd( − 3 ) = sd( ) + 3sd( ) = 8 + (3 × 2) = √100 = 10 Example Oranges grown on a citrus farm have a mean mass of 204 grams with a standard deviation of 9 grams. Lemons grown on the same farm have a mean mass of 76 grams with a standard deviation of 3 grams. The masses of the lemons are independent of the masses of the oranges. The mean mass and standard deviation, in grams, respectively of a set of three of these oranges and two of these lemons are Assuming each of the oranges and lemons are selected randomly, they are all independent of each other. ( + + + + ) = 3 ( ) + 2 ( ) = 3 × 204 + 2 × 76 = 764 Var( sd(
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replaced by the combinations when a dominance check is invoked. In order to keep the combinations for each rule encountered, we use a matrix M with MBRs as rows and cardinalities columns. As the search order shown in Fig. 5, pattern p1 comes before p
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