Expected Value (10D)
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Expected Value Suppose random variable X can take value x 1 with probability p 1, value x 2 with probability p 2, and so on, up to value x k with probability p k. Then the expectation of this random variable X is defined as Since all probabilities pi add up to one ( p 1 + p 2 +... + p k = 1), the expected value can be viewed as the weighted average, with p i s being the weights: Expected Value (10D) 3
Arithmetic Mean 2 elements {a, b} 3 elements {a, b, c} A= a b 2 A= a b c 3 n elements {a 1, a 2,..., a n } A= a 1 a 2 a n = 1 n n n i=1 a i Expected Value (10D)
Arithmetic Mean - Example 2 3 3 2 = 3 3 =2 3=2 A same length Arithmetic Mean: A = 3 2 6 same length 2 6 = =3 =3 A Arithmetic Mean: A = Expected Value (10D) 5
Geometric Mean 2 elements {a, b} G= a b a 0, b 0 3 elements {a, b, c} G= 3 a b c a 0, b 0, c 0 n elements {a 1, a 2,..., a n } n G= a n 1 a 2 a n = i=1 1 n a i a i 0 Expected Value (10D) 6
Geometric Mean - Example 2 2 = = 2 =G 2 same area Geometric Mean: G = 2 same volume 2 = = 3 =G 3 Geometric Mean: G = Expected Value (10D) 7
Harmonic Mean 2 elements a, b 3 elements a, b, c H= 2 1 a 1 b = 2ab a b H= 3 1 a 1 b 1 = 3abc c ab bc ca n elements a 1, a 2,..., a n n H= 1 1 1 = n a 1 a 2 a n n 1 i=1 a i Expected Value (10D)
Harmonic Mean same distance = D same duration = T A 60 km/hr B 60 km/hr 0 km/hr 0 km/hr A B C Average Speed total distance = total time = 2 D D 60 D = 2 60 0 60 0 =km/hr 0 Average Speed total distance = total time = 60 T 0 T 2T =50km/hr Harmonic Mean: H = Arithmetic Mean: A = 50 Expected Value (10D) 9
Absolute Frequency # of students 0 ~ 50 1 50 ~ 60 60 ~ 70 70 ~ 0 5 0 ~ 90 2 Total 20 16 12 0 50 60 70 0 90 Average Score? 5 1 55 65 75 5 5 2 1 5 2 = 66.5 Expected Value (10D) 10
Absolute Frequency & Average - Rectangles 16 16 12 12 10 20 30 0 50 60 70 0 90 100 10 20 30 0 50 60 70 0 90 100 5 55 65 75 5 5 55 65 75 5 20 5 x 1 1 student cumulative 16 12 55 x 65 x students students 75 x 5 5 students 5 x 2 2 students 10 20 30 0 50 60 70 0 90 100 Expected Value (10D) 11
Absolute Frequency & Average Absolute Frequency 16 16 12 12 0 50 60 70 0 90 10 20 30 0 50 60 70 0 90 100 5 55 65 75 5 20 20 5 x 1 1 student cumulative 16 12 66.5 x 20 cumulative 16 12 55 x 65 x students students 75 x 5 5 students 10 20 30 0 50 60 70 0 90 100 Avg = 66.5 5 x 2 2 students 10 20 30 0 50 60 70 0 90 100 Expected Value (10D) 12
Comparison of Histograms Absolute Frequency Group A # of students 0 ~ 50 1 50 ~ 60 60 ~ 70 70 ~ 0 5 0 ~ 90 2 absolute 16 12 Group A Total 20 0 50 60 70 0 90 Group B # of students 0 ~ 50 1 50 ~ 60 2 60 ~ 70 70 ~ 0 9 absolute 16 12 Group B 0 ~ 90 5 Total 25 0 50 60 70 0 90 Expected Value (10D) 13
Normalization Group A (20 students) 1 5 2 Total 20 student counts 1.0 Normalization A common scale makes comparison convenient f i = N = k n k f i N relative count (absolute) count total counts = j n j 1.0 1 2 9 5 Total 25 student counts Group B (25 students) Expected Value (10D) 1
Relative Frequency Normalization f i = N = i Group A f i Group B f i absolute relative 0 ~ 50 1 1/20 = 0.05 50 ~ 60 /20 = 0.20 60 ~ 70 /20 = 0.0 70 ~ 0 5 5/20 = 0.25 0 ~ 90 2 2/20 = 0.10 Total 20 20/20 = 1.00 N = i = 20 absolute relative 0 ~ 50 1 1/25 = 0.0 50 ~ 60 2 2/25 = 0.0 60 ~ 70 /25 = 0.32 70 ~ 0 9 9/25 = 0.36 0 ~ 90 5 5/25 = 0.20 Total 25 25/25 = 1.00 N = i = 25.05.20.0.25.10.0.0.32.36.20 i f i = 1.00 i f i = 1.00 Expected Value (10D) 15
Comparison of Histograms Relative Frequency Group A f i 0 ~ 50 1 1/20 = 0.05 50 ~ 60 /20 = 0.20 60 ~ 70 /20 = 0.0 70 ~ 0 5 5/20 = 0.25 0 ~ 90 2 2/20 = 0.10 Total 20 20/20 = 1.00 Group B absolute absolute N = i = 20 f i relative relative 0 ~ 50 1 1/25 = 0.0 50 ~ 60 2 2/25 = 0.0 60 ~ 70 /25 = 0.32 70 ~ 0 9 9/25 = 0.36 0 ~ 90 5 5/25 = 0.20 Total 25 25/25 = 1.00 N = i = 25 relative relative 1.0..6..2 1.0..6..2 Group A.20.05 0 50 60 70 0 90 Group B 0 50 60 70 0 90 Expected Value (10D) 16
Relative Frequency & Average Relative Frequency Group A relative 1.0..6. relative 1.0..6..2.2 0 50 60 70 0 90 10 20 30 0 50 60 70 0 90 100 5 55 65 75 5 cumulative relative 1.0..6. 66.5 cumulative relative 1.0..6. 5 x.05 55 x.20 65 x.0 1/20 /20 /20.2.2 75 x.25 5/20 10 20 30 0 50 60 70 0 90 100 Avg = 66.5 5 x.10 2/20 10 20 30 0 50 60 70 0 90 100 Expected Value (10D) 17
Average Group A absolute f i relative 0 ~ 50 1 1/20 = 0.05 50 ~ 60 /20 = 0.20 60 ~ 70 /20 = 0.0 70 ~ 0 5 5/20 = 0.25 0 ~ 90 2 2/20 = 0.10 Total 20 20/20 = 1.00 N = i = 20 absolute 16 12 Group A relative 0 50 60 70 0 90 1.0..6..2 Group A.20.05 0 50 60 70 0 90 Average Score? 5 1 55 65 75 5 5 2 1 5 2 = 5 1 20 55 20 65 20 75 5 20 5 2 20 = 5 1 20 55 20 65 20 75 5 20 5 2 20 = 66.5 Expected Value (10D) 1
Accumulative Frequency Group A absolute accumulative absolute 0 ~ 50 1 1 50 ~ 60 5 60 ~ 70 13 70 ~ 0 5 1 0 ~ 90 2 20 Total 20 20 N = i = 20 accumulative absolute 20 16 12 Group A 5 1 1 0 50 60 70 0 90 2 20 1 13 5 Group A f i relative accumulative relative 0 ~ 50 0.05 0.05 50 ~ 60 0.20 0.25 60 ~ 70 0.0 0.65 70 ~ 0 0.25 0.90 0 ~ 90 0.10 1.00 Total 1.00 1.00 1.0 = i f i = 25 accumulative relative 1.0..6..2 Group A.05.20.0.25.10 1.0.90.65.25.05 0 50 60 70 0 90 Expected Value (10D) 19
Relative Frequency & Ratio Absolute Group A f i 2 student absolute relative 0 ~ 50 1 1/20 = 0.05 50 ~ 60 /20 = 0.20 60 ~ 70 /20 = 0.0 70 ~ 0 5 5/20 = 0.25 0 ~ 90 2 2/20 = 0.10 Total 20 20/20 = 1.00 5 students N = i = 20 students Relative.10..25.6.0 students..2.20 1 students.05.1.2.3..5.6.7..9 1.0 20 Expected Value (10D) 20
Normalization Equations (1) f i = N N = f i = N f i relative = count total count total count = count relative freq count = total count relative fraction = part total total = part fraction part = total part 0.xxx = = 0.xxx = 0.xxx Expected Value (10D) 21
Normalization Equations (2) f n i N f i = N n N = i f i = N f i 1. 0 f i = slope N i n n f i = slope i f i = slope i 1. 0 N 1. 0 N Expected Value (10D) 22
Binomial Probability Distribution n k n k p p p p p q q q q q C (n, k) Expected Value (10D) 23
Expected Value of the Binomial Distribution k k P(X=k) ( n k) = n! k!(n k )! ( n 1 k 1) = = n (n 1)! (n k)!k! (n k) = (n 1) (k 1) (n 1)! (k 1)!((n 1) (k 1))! m ( m ) l =0 l pl (1 p) m l = ( p+(1 p)) m Expected Value (10D) 2
Expected Value (10D) 25
References [1] http://en.wikipedia.org/ [2] https://upload.wikimedia.org/wikiversity/en/6/60/sequence.1.a.pdf [3] https://upload.wikimedia.org/wikiversity/en/b/bf/histogram.1.a.pdf