ESP%20I%20SOLUTIONS - ESP I EXAMPLES FOR STUDY AND PRACTICE...

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EXAMPLES FOR STUDY AND PRACTICE I ! STA261H " 1. Suppose we have obtained the following sample of 10 outcomes from a the continuous uniform distribution U ! unif[0,1] U : .189, .841, .994, .777, .281, .008, .310, .098, .892, .620, . .. a) What is the empirical relative frequency for the first 5 observations that .25 " U " .48 ? P 5 # .25 " U " .48 $ % 1/5 b) What is the empirical relative frequency for the first 10 observations that either .25 " U " .48 or .68 " U < .84 ? P 10 # .25 " U " .48 or .68 " U < .84 $ % 3/10 c) What is the emp. rel. freq. of the event (.68 " U < .84) over the first n observations, for each value of n = 1, . .. , 5 ; and, to what ultimate numerical value does this sequence eventually converge ? 0 , 0 , 1/3 , 1/4 , 1/5 , ... # .84 $ .68 % .16 c) What is the sample mean value, E 3 U 2 , of the random variable U 2 for the first 3 trial observations in the sequence of ten provided above, and, what is the simple numerical difference between this observed arithmetic average and its expected value, % E 3 U 2 - E U 2 % ? E 3 U 2 % !# .189 $ 2 .841 $ 2 .994 $ 2 " / 3 % .577 & EU 2 % 1 / 3 & thus % E 3 U 2 - E U 2 % % .244 ESP I $ 1 $
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2. Suppose that the probability function, p(x,y), for a bivariate random variable W = (X,Y) is given by the following table [cf.: p212, 229, 238] 9p(x,y): x 0 1 2 y 0 1 2 1 % 4 1 2 2 0 % 4 2 1 0 0 % 1 4 4 1 % 9 a) Let h( W ) = (X $ Y $ 2 and compute E h( W ) by direct appeal to the definition. E h( W ) % 1 0 2 2 1 2 1 2 2 2 ’# $ 1 $ 2 2 0 2 & 0 1 2 1 ’# $ 2 $ 2 & 0 ’# $ 1 $ 2 & 0 0 2 9 % 2 4 2 4 9 % 12 9 % 4 3 b) Set U = h( W ) and determine the ‘marginal’ probability function, p U (u) for the univariate random variable U. U ! ( ) * + 0 1 4 3/9 4/9 2/9 c) Use the marginal probability function to compute EU directly. Thus EU % 3 0 4 1 2 4 9 % 12 9 % 4 3 ! Another example of this nature may be found on p.224. Such examples are extremely simple (of course), but the principle being illustrated is very important: that, no matter how you compute it, EU = E h( W ). " ESP I $ 2 $
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3. For the same table as presented in question 1 above one might take the opportunity a) First compute EX, EY, varX, varY & cov(X,Y) EX % EY % 2/3 , EX 2 % EY 2 % 8/9 , EXY % 2/9 varX % varY % 4/9 , cov # X,,Y $ % $ 2/9 b) Use a) to determine each of the following: i) E(24X $ 16Y) % 16/3 E(X $ Y) % 0 E(X Y) % 4/3 ii) cov(3X $ 2Y, Y) % 3cov # X,Y $ $ 2varY % $ 14/9 cov( $ 2X+Y, 6X $ 3Y) % $ 12varX 12cov # X,Y $ $ 3varY % $ 84/9 cov(X $ Y, X Y) % varX $ varY % 0 iii) var(3X $ 5Y) % 9varX $ 30cov # X,Y $ 25varY % 76/9 var(X Y) % varX 2cov # X,Y $ varY % 4/9 var(X $ Y) % varX $ 2cov # X,Y $ varY % 12/9 c) Determine the constant a for which the linear combination Z = Y $ aX has smallest variance. What is this minimum variance ? g
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ESP%20I%20SOLUTIONS - ESP I EXAMPLES FOR STUDY AND PRACTICE...

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