Dr. Hackney STA Solutions pg 22

Dr. Hackney STA Solutions pg 22 - Second Edition 2-7 3 4 2-...

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Second Edition 2-7 b. E X = 2 β/ π E X 2 = 3 β 2 / 2 Var X = β 2 ± 3 2 - 4 π ² . 2.23 a. Use Theorem 2.1.8 with A 0 = { 0 } , A 1 = ( - 1 , 0) and A 2 = (0 , 1). Then g 1 ( x ) = x 2 on A 1 and g 2 ( x ) = x 2 on A 2 . Then f Y ( y ) = 1 2 y - 1 / 2 , 0 < y < 1 . b. E Y = R 1 0 yf Y ( y ) dy = 1 3 E Y 2 = R 1 0 y 2 f Y ( y ) dy = 1 5 Var Y = 1 5 - ( 1 3 ) 2 = 4 45 . 2.24 a. E X = R 1 0 xax a - 1 dx = R 1 0 ax a dx = ax a +1 a +1 ³ ³ ³ 1 0 = a a +1 . E X 2 = R 1 0 x 2 ax a - 1 dx = R 1 0 ax a +1 dx = ax a +2 a +2 ³ ³ ³ 1 0 = a a +2 . Var X = a a +2 - ´ a a +1 µ 2 = a ( a +2)( a +1) 2 . b. E X = n x =1 x n = 1 n n x =1 x = 1 n n ( n +1) 2 = n +1 2 . E X 2 = n i =1 x 2 n = 1 n n i =1 x 2 = 1 n n ( n +1)(2 n +1) 6 = ( n +1)(2 n +1) 6 . Var X = ( n +1)(2 n +1) 6 - ( n +1 2 ) 2 = 2 n 2 +3 n +1 6 - n 2 +2 n +1 4 = n 2 +1 12 . c. E X = R 2 0 x 3 2 ( x - 1) 2 dx = 3 2 R 2 0 ( x 3 - 2 x 2 + x ) dx = 1 . E X 2 = R 2 0 x 2 3 2 ( x - 1) 2 dx = 3 2 R 2 0 ( x 4 - 2 x 3 + x 2 ) dx = 8 5 . Var X = 8 5 - 1 2 = 3 5 . 2.25 a. Y = - X and g - 1 ( y ) = - y . Thus f Y ( y ) = f X ( g - 1 ( y )) | d dy g - 1 ( y ) | = f X ( - y ) | - 1 | = f X ( y ) for every y . b. To show that M X ( t ) is symmetric about 0 we must show that M X (0 + ± ) = M X (0 - ± ) for all ± > 0. M X (0 + ± ) = Z -∞ e (0+ ± ) x f X ( x ) dx = Z 0 -∞ e ±x f X ( x ) dx + Z 0 e ±x f X ( x ) dx = Z 0 e ± ( - x ) f X ( - x ) dx + Z 0 -∞ e ± ( - x ) f X ( - x ) dx = Z -∞ e - ±x f X ( x ) dx = Z -∞ e (0 - ± ) x f X ( x ) dx = M X (0 - ± ) . 2.26 a. There are many examples; here are three. The standard normal pdf (Example 2.1.9) is symmetric about a = 0 because (0 - ±
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