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Calculating expected value, variance, and covariance from a sample (empirical data) Objectives: Practice calculating mean, variance, and covariance. Become more familiar with math symbols. Better understand the meaning of covariance. Note that individual answers (observations) are recorded in boxes and class answers (summary statistics) are recorded in circles. Step 1. Roll your die. Define X = the result of the first die roll (1-6) record your observation, Xi here: Step 2. Roll your die a second time. Define Y = the sum of both rolls (2-12) record your observation, Yi here: Step 3. We will calculate the sample mean of X and the sample mean of Y as a class. (Follow along to make sure you understand how the calculation is being made). Record these below: n X i 1 i Sample mean of X: X n = n n Y i 1 i Sample mean of Y: Yn = n Step 4: Calculate the squared deviation of your X and Y observations from the group means: record your ( X i X n ) here: 2 record your (Yi Yn ) here: 2 Step 5. We will calculate the sample variance of X and the sample variance of Y as a class. (Follow along to make sure you understand how the calculation is being made). Record these below: n (X i 1 i X n )2 Sample variance of X: s x 2 = n 1 n (Y i 1 i Yn ) 2 Sample variance of Y: s y 2 = n 1 Step 6. Calculate the product of X and Y deviations: record your ( Yi Yn )( X i X n ) here: Step 6. We will calculate the sample covariance of X and Y. (Follow along to make sure you understand how the calculation is being made). Record this below: n (Y i 1 i Yn )( X i X n ) Sample covariance of X and Y: s x , y = n 1 For class discussion: What do you note about the relationship between variance and covariance here? How does this make sense intuitively? RECORD CLASS OBSERVATIONS HERE. Observation Xi Yi (X i X n )2 (Yi Yn ) 2 ( Yi Yn )( X i X n ) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Calculating expected value, variance, and covariance of a random variable with a known (a priori) probability distribution: Mean (expected value): E ( x) x p(x ) all x i i Variance: x 2 Var ( x) E ( x x ) 2 [x all x i x ] 2 p( xi ) algebra E ( x 2 ) x E ( x 2 ) E ( x) 2 2 Covariance: xy Cov ( xy ) E[( x x )( y y )] (x all x i , y i i x )( y i y ) p( xi & y i ) algebra E ( xy ) x y E ( xy ) E ( x) E ( y ) The covariance of two random variables measures their tendency to vary together, i.e., to co-vary. Where the variance is the average squared deviation of a random variable from its mean, the covariance is the average of the products of the deviations of two random values from their respective means. In-Class Exercise: A Priori version of the dice problem: Fill in the following joint probabilities, P(X&Y) As before, X = the result of the first die roll Y = sum of both rolls X 1 2 3 4 5 6 1 P(X=1&Y=1)=0 P(X=2&Y=1)=0 P(X=3&Y=1)=0 2 P(X=1&Y=2)= P(X=2&Y=2)=0 1/36 3 4 5 Y 6 7 8 9 10 11 12 Next, using the above joint probability table, Calculate the a priori means of X and Y, variances of X and Y, and covariance of X and Y. X Marginal probability table: x x2 p(x) 1 1 p(x=1)=1/6 2 4 p(x=2)=1/6 3 9 p(x=3)=1/6 4 16 p(x=4)=1/6 5 25 p(x=5)=1/6 6 36 p(x=6)=1/6 1.0 E(x) = E(x2) = x 2 =Var(x) = E(x2) – [E(x)]2 = Y Marginal probability table: y y2 p(y) 1 1 0 2 4 1/36 3 9 2/36 4 16 3/36 5 25 4/36 6 36 5/36 7 49 6/36 8 64 5/36 9 81 4/36 10 100 3/36 11 121 2/36 12 144 1/36 E(y) = E(y2)= y 2 =Var(y) = E(y2) – [E(y)]2 = E(xy) = (refer to joint probability table) Cov(xy) = E(xy) – [E(x)][E(y)] = Answer: X 1 2 3 4 5 6 1 0 0 0 0 0 0 2 1/36 0 0 0 0 0 3 1/36 1/36 0 0 0 0 4 1/36 1/36 1/36 0 0 0 5 1/36 1/36 1/36 1/36 0 0 Y 6 1/36 1/36 1/36 1/36 1/36 0 7 1/36 1/36 1/36 1/36 1/36 1/36 8 0 1/36 1/36 1/36 1/36 1/36 9 0 0 1/36 1/36 1/36 1/36 10 0 0 0 1/36 1/36 1/36 11 0 0 0 0 1/36 1/36 12 0 0 0 0 0 1/36 X x x2 p(x) 1 1 p(x=1)=1/6 2 4 p(x=2)=1/6 3 9 p(x=3)=1/6 4 16 p(x=4)=1/6 5 25 p(x=5)=1/6 6 36 p(x=6)=1/6 1.0 x p(x ) (1)( 6 ) 2( 6 ) 3( 6 ) 4( 6 ) 5( 6 ) 6( 6 ) 1 1 1 1 21 1 1 E(x) = i i 3.5 all x 6 1 1 1 1 1 1 E(x2) = xi p(x i ) (1)( ) 4( ) 9( ) 16( ) 25( ) 36( ) 15.16666 2 all x 6 6 6 6 6 6 x 2 =Var(x) = E(x2) – [E(x)]2 = 15 - (3.5)2 = 2.916666 Y Marginal probability table: y y2 p(y) 1 1 0 2 4 1/36 3 9 2/36 4 16 3/36 5 25 4/36 6 36 5/36 7 49 6/36 8 64 5/36 9 81 4/36 10 100 3/36 11 121 2/36 12 144 1/36 y p(y 1 E(y) = i i )( )[ 2(1) 3(2) 4(3) 5(4) 6(5) 7(6) 8(5) 9(4) 10(3) 11(2) 12(1)] 7.0 all y 36 E(y )= y i 2 p(y i ) 1 [4(1) 9(2) 16(3) 25(4) 36(5) 49(6) 64(5) 81(4) 100(3) 121(2) 144(1)] 54.833 2 all y 36 y =Var(y) = E(y ) – [E(y)]2 = 54.8333-49 = 5.83333 2 2 Notice: the expected value and variance are exactly twice that of those for X!! Covariance of X and Y: 1 2 3 4 5 6 1 0 0 0 0 0 0 2 1/36 0 0 0 0 0 3 1/36 1/36 0 0 0 0 4 1/36 1/36 1/36 0 0 0 5 1/36 1/36 1/36 1/36 0 0 6 1/36 1/36 1/36 1/36 1/36 0 7 1/36 1/36 1/36 1/36 1/36 1/36 8 0 1/36 1/36 1/36 1/36 1/36 9 0 0 1/36 1/36 1/36 1/36 10 0 0 0 1/36 1/36 1/36 11 0 0 0 0 1/36 1/36 12 0 0 0 0 0 1/36 x y p(x 1 i , yi ) [2 3 4 5 6 7 6 8 10 12 14 16 12 15 18 21 24 27 20 E(xy) = all x, y i i 36 24 28 32 36 40 30 35 40 45 50 55 42 48 54 60 66 72] 987 / 36 27.41666 Cov(xy) = 27.41666 – (3.5) (7.0) = 2.91666 Notice: Covariance of X and Y = Variance of X (they share completely the variance of X).

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posted: | 10/28/2011 |

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