Any number of ranges can be found that is a variety

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Any number of ranges can be found. That is, a variety of values 0 x and 1 x with 0 x < 380 and 1 x > 380 will satisfy: 80 . 0 ) x X x ( P 1 0 = < < The shortest range is centered at the mean $380. To calculate this range, find a number a such that: 80 . 0 ) a 380 X a 380 ( P = + < < - This is illustrated with a graph: PDF of ) , ( N ~ X 2 50 380 380+a 380 380-a f(x) x Upper Tail Area = 0.10 Lower Tail Area = 0.10 Area = 0.80
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Econ 325 – Chapter 5 29 By inspecting the graph, it can be seen that an equivalent statement of the problem is: find a number a such that: 90 . 0 ) a 380 X ( P = + < To work with the standard normal distribution consider: = < = - + < σ μ - = + < 50 a F 50 a Z P 50 380 ) a 380 ( X P ) a 380 X ( P The Appendix Table gives 90 . 0 ) ( F = 1.28 Therefore, 1.28 = 50 a and 64 50 1.28 = = ) )( ( a The range centered at $380 is: [ $380 – 64, $380 + 64] = [ $316, $444] As a check on the calculations, the upper limit can be calculated with Microsoft Excel by using the function: NORMINV(0.9, 380, 50) Econ 325 – Chapter 5 30 Chapter 5.6 Jointly Distributed Continuous Random Variables Results stated earlier for jointly distributed discrete random variables can be extended to work with continuous random variables. Let X and Y be two continuous random variables that take numeric values denoted by x and y , respectively. The joint cumulative distribution function (CDF) is: ) y Y x X ( P ) y , x ( F Y , X < < = and The marginal distribution functions are: ) x X ( P ) x ( F X < = and ) y Y ( P ) y ( F Y < = X and Y are statistically independent if and only if: ) y ( F ) x ( F ) y , x ( F Y X Y , X =
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Econ 325 – Chapter 5 31 A measure of linear association is covariance: Y X Y X ) XY ( E )] Y ( ) X [( E ) Y , X ( Cov μ μ - = μ - μ - = where ) X ( E X = μ and ) Y ( E Y = μ If X and Y are independent then 0 = ) Y , X ( Cov . However, zero covariance does not guarantee independence. X and Y may have some complicated non-linear relationship. head2right Special Case: If X and Y are joint normally distributed random variables then zero covariance also gives the result that X and Y are independent. Econ 325 – Chapter 5 32 xrhombus Linear Combinations of Random Variables For constant fixed numbers a and b , a linear combination of random variables X and Y is: Y b X a W + = The mean of the random variable W is: ) Y ( E b ) X ( E a ) W ( E W + = = μ The variance of W is: ) Y , X ( Cov b a 2 ) Y ( Var b ) X ( Var a ) W ( Var σ 2 2 2 W + + = = head2right Special Case: If X and Y are joint normally distributed random variables then Y b X a W + = is also normally distributed with mean and variance as given above. That is, ) σ , ( N ~ W 2 W W μ
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Econ 325 – Chapter 5 33 Now consider three random variables 1 X , 2 X and 3 X with means 1 μ , 2 μ and 3 μ and variances 2 1 σ , 2 σ 2 and 2 σ 3 . The sum of these random variables has the properties: 3 2 3 2 ) X X X ( E μ + μ + μ = + + 1 1 and ) X , X ( Cov 2 ) X , X ( Cov 2 ) X , X ( Cov 2 ) X X X ( Var 3 2 3 1 2 1 3 2 1 3 2 + + + σ + σ + σ = + + 2 2 2 1 With independence the covariance between every pair of these random variables is zero to give a simpler result for the variance of the sum: 2 2 2 1 σ + σ + σ = + + 3 2 1 3 2 ) X X X ( Var Econ 325 – Chapter 5 34 Example: Portfolio Analysis The random variables X and Y are the share prices of two companies trading on the stock market such that ) , ( N ~ X 81 25 and ) , ( N ~ Y 121 40 X μ 2 X σ Y μ 2 Y σ The correlation between the two stock prices is: 0.4 - = ρ XY A portfolio is the random variable: Y X W 30 20 + = Find the probability that the portfolio value exceeds 2,000.
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