Linear Combination of Random Variables

The random variables have the countable values for their probability in the statistics. The sum of the probability for the random variable is one, then the random variable is said to be discrete random variable. The random variable is the measurable function. Here the mean, variance and standard deviation for the random variables are calculated. This article contains the information about the linear combination of random variables in the probability theory and statistics.

Formula used for linear combination of random variables:

The formulas used to determine the mean, variance and the standard deviation in the random variable are

Mean = `sum x p(x)`

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

In this above mentioned formula the `(sum x p(x)) ^2` refers the squared value of the mean in the statistics.

Standard deviation = `sqrt (variance)`

In the above mentioned formula x refers the given set of discrete random values and p(x) refers the probability value for the random variables in the statistics.

Examples for linear combination of random variables:

Example 1 to linear combination of random variables:

Predict the mean, variance and standard deviation for the random variables.

x	20	40	60	80	90
P(x)	0.11	0.12	0.34	0.24	0.19

Solution:

Mean = `sum x p(x)`  

Mean = 20 (0.11) +40 (0.12) +60 (0.34) +80 (0.24) + 90(0.19)

Mean = 2.2+ 4.8 + 20.4 + 19.2 + 17.1                   

Mean = 63.7

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

Variance = ((0.11) (20) (20) + (0.12) (40) (40) + (0.34) (60) (60) + (0.24) (80) (80) + (0.19) (90) (90)) - (63.7)² 

Variance = (44+ 192+ 1224 + 1536+ 1539) -4057.69

Variance = 4535 -4057.69

Variance = 477.31

Standard deviation = `sqrt (variance)`

Standard deviation = `sqrt (477.31)`

Standard deviation = 21.847

The mean value is 63.7, variance is 477.31 and the standard deviation is 21.847 for the given random variables.  

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Example 2 to linear combination of random variables:         

Predict the mean, variance and standard deviation of random variables.  

  

x	10	20	30	40	50
P(x)	0.20	0.12	0.22	0.13	0.33

Solution:

Mean = `sum x p(x)`

Mean = 10(0.20) +20(0.12) + 30(0.22) +40(0.13) + 50(0.33)

Mean = 2.0 +2.4 +6.6+ 5.2+ 16.5

Mean = 32.7

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

Variance = ((0.20) (10) (10) + (0.12) (20) (20) + (0.22) (30) (30) + (0.13) (40) (40) + (0.33) (50) (50)) - (32.7)² 

Variance = (20+48+ 132 + 208 +825) -1069.29

Variance = 1233 -1069.29                                     

Variance = 163.71

Standard deviation = `sqrt (variance)`

Standard deviation = `sqrt (163.71)`

Standard deviation = 12.79

The mean is 32.7, variance is 163.71 and the standard deviation is 12.79 for the given random variables.