Mathematics Learning Centre Sigma notation Jackie Nicholas c 2005 University of Sydney
Mathematics Learning Centre, University of Sydney 1 1 Sigma Notation 1.1 Understanding Sigma Notation The symbol Σ (capital sigma) is often used as shorthand notation to indicate the sum of a number of similar terms. Sigma notation is used extensively in statistics. For example, suppose we weigh five children. We will denote their weights by x 1, x 2, x 3, x 4 and x 5. The sum of their weights x 1 + x 2 + x 3 + x 4 + x 5 is written more compactly as x j. j1 The symbol Σ means add up. Underneath Σ we see j 1 and on top of it 5. This means that j is replaced by whole numbers starting at the bottom number, 1, until the top number,5, is reached. Thus x j x 2 + x 3 + x 4 + x 5, and So the notation x j tells us: j1 a. to add the scores x j, b. where to start: x 1, j2 x j x 2 + x 3 + x 4. j2 c. where to stop: x n (where n is some number). Now take the weights of the children to be x 1 10kg, x 2 12kg, x 3 14kg, x 4 8kg and x 5 11kg. Then the total weight (in kilograms) is x i x 1 + x 2 + x 3 + x 4 + x 5 10+12+14+8+11 55. Notice that we have used i instead of j in the formula above. The j is what we call a dummy variable - any letter can be used, ie, x j x i. j1 Now let us find 2x i where x 1 2,x 2 3,x 3 2 and x 4 1.
Mathematics Learning Centre, University of Sydney 2 Again, starting with i 1 we replace the expression 2x i with its value and add up the terms until i 4 is reached. So, 2x i 2x 1 +2x 2 +2x 3 +2x 4 2(2) + 2(3) + 2( 2) + 2(1) 4+6 4+2 8. Similarly, let us find Here, (x k 4) where x 1,x 2 4,x 3 1. k1 (x k 4) (x 1 4) + (x 2 4)+(x 3 4) k1 ( 4) + (4 4) + (1 4) 3+0+( 3) 0. Notice that this is different from x k 4 where x 1,x 2 4,x 3 1. In this case, we have, x k 4 x 1 + x 2 + x 3 4 k1 k1 +4+1 4 8. We use brackets to indicate what should be included in the sum. In the previous example, there were no brackets, so the 4 was not included in the sum. Example: Write out in full: x k. k1 Solution: x 1 + x 2 + x 3 + x 4 + x 5. We also use sigma notation in the following way: j 2 1 2 +2 2 +3 2 +4 2 30. j1 This is the same principle: replace j in the expression (this time j 2 ) by whole numbers starting with 1 and ending with 4, and add.
Mathematics Learning Centre, University of Sydney 3 1.1.1 Exercises 1. Evaluate x i where x 1 5,x 2 2,x 3 3,x 4 8. 2. Evaluate 5x k where x 1 10, x 2 14, x 3 2, and n 3. k1 3. Find μ 1 x j where the x 1 10kg, x 2 12kg, x 3 14kg, x 4 8kg and x 5 11kg 5 j1 are the weights of 5 children. (μ is the mean weight of the children.) 4. Find the value of (x i μ) 2 where x 1 105, x 2 100, x 3 95, and μ 100. 1.2 Rules of summation We will prove three rules of summation. These rules will allow us to evaluate formulae containing sigma notation more easily and allow us to derive equivalent formulae. Rule 1: If c is a constant, then cx i c x i. To see why Rule 1 is true, let s start with the left hand side of this equation, as required. cx i cx 1 + cx 2 + cx 3 + + cx n c(x 1 + x 2 + x 3 + + x n ) c x i Suppose that x i 55 as in a previous example. Then 3x i 3 x i 3 55 165. Rule 2: If c is a constant, then c nc. This rule looks a bit strange as there is no x i. The left hand side of this formula means sum c, n times. That is, n {}}{ c c + c + + c n c nc.
Mathematics Learning Centre, University of Sydney 4 For example, 25 2 10. Rule 3: (x i + y i ) x i + y i. To prove this rule, let s start with the left hand side, (x i + y i ) (x 1 + y 1 )+(x 2 + y 2 )+(x 3 + y 3 )+ +(x n + y n ) (x 1 + x 2 + x 3 + + x n )+(y 1 + y 2 + y 3 + + y n ) x i + y i. For example, if x i 21 and y i 35 then (x i + y i ) x i + y i 21+3556. 1.2.1 Exercises In the following exercises you may assume that yi 2 50 and x i y i 105. Evaluate the following expressions: x i 3, y i 12, x 2 i 303, 1. 2y i 2. 3. x i 1 (x i 1) 4. ( x i ) 2 5. 6.. 8. (2x i + y i ) (2x i +3y i ) (2x i 5y i +3) (x i 2y i ) 2
Mathematics Learning Centre, University of Sydney 5 1.3 Using Sigma Notation in Statistics Here are some examples of how sigma notation is used in statistics: The formula for a mean of a group of N scores, is μ 1 N x i. A measure of how spread out the scores are, called the variance, has the following formula: σ 2 1 N (x i μ) 2. For example, the number of customers having lunch at a certain restaurant on weekdays were x 1 92, x 2 84, x 3 0, x 4 6, x 5 66, x 6 80, x 1. The mean is μ 1 N x i 1 (92 + 84 + 0 + 6 + 66 + 80 + 1) 539. Note: There are scores so N. The variance is σ 2 1 (x i ) 2 1 [(15)2 + () 2 +( ) 2 +( 1) 2 +( 11) 2 + (3) 2 +( 6) 2 ] 1 [225 + 49 + 49 + 1 + 121 + 9 + 36] 1 [490] 0. An alternative formula for variance is σ 2 1 N N ( x 2 i Nμ 2 )
Mathematics Learning Centre, University of Sydney 6 For the above example we get: σ 2 1 N [x2 1 + x 2 2 + x 2 3 + x 2 4 + x 2 5 + x 2 6 + x 2 Nμ 2 ] 1 [922 +84 2 +0 2 +6 2 +66 2 +80 2 +1 2 () 2 ] 1 [8464 + 056 + 4900 + 56 + 4356 + 6400 + 5041 (5929)] 1 [490] 0 as before. We can use the rules to show that two formulae for variance are equivalent, since (x i μ) 2 x 2 i Nμ 2. (x i μ) 2 (x 2 i 2μx i + μ 2 ) x 2 i 2μx i + μ 2 x 2 i 2μ x i + Nμ 2 x 2 i 2μ Nμ+ Nμ 2 x 2 i Nμ 2 since μ is a constant since N x i Nμ 1.3.1 Exercises 1. Find the variance of the weights of the five children (in Exercise 1.1.1 number 3), using each of the above formulae for σ 2. 2. During a 5 week period, a salespersons weekly income (in dollars) was x 1 400, x 2 250, x 3 15, x 4 300, x 5 35. Calculate μ 1 5 x 5 i and σ 2 1 5 ( 5 x 2 i 5μ 2 ). 3. An insurance company is concerned about the length of time required to process claims. The length of time, measured in days, taken to process claims produced the data x 1 23, x 2 20, x 3 22, x 4 25, x 5 24,x 6 23, x 21. Evaluate the mean μ and variance σ 2 for these data.
Mathematics Learning Centre, University of Sydney 1.4 Answers to Exercises Answers to Exercise 1.1.1 1. 18 2. 110 3. 11kg 4. 50 Answers to Exercise 1.2.1 1. 2. 3. 2y i 2 y i 24 x i 13 136 (x i 1) x i 13 5(1) 32 4. ( x i ) 2 (3) 2 1369 Note this is different from x 2 i 303. 5. 6.. 8. (2x i + y i )2 x i + y i 2(3) + 12 86 (2x i +3y i )2 x i +3 y i 2(3) + 3(12) 110 (2x i 5y i +3)2 x i 5 y i + 3 2(3) 5(12) + 5(3) 29 (x i 2y i ) 2 (x 2 i 4x i y i +4yi 2 ) x 2 i 4 x i y i +4 yi 2 303 420 + 200 83 Answers to Exercise 1.3.1 1. σ 2 4 2. μ 300, σ 2 650 3. μ 22.5 to two decimal places, σ 2 2.53 taking the mean as 22.5 and using the formula σ 2 1 N ΣN (x i μ) 2. If, however the formula σ 2 1 N [ΣN x 2 i Nμ 2 ] is used, then the answer σ 2 2.60 is obtained. This discrepancy is due to round off error and can be avoided by using μ 22.51429 in the above formula.