A05ans - EXERCISES IN STATISTICS Series A, No. 5 1. Find...

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Unformatted text preview: EXERCISES IN STATISTICS Series A, No. 5 1. Find the moment generating function of x f ( x ) = 1, where 0 < x < 1, and thereby confirm that E ( x ) = 1 2 and V ( x ) = 1 12 . Answer: The moment generating function is M ( x, t ) = E ( e xt ) = Z 1 e xt dx = e xt t 1 = e t t- 1 t . But e t = t 0! + t 1! + t 2 2! + t 3 3! + , so M ( x, t ) = 1 t + 1 + t 2! + t 2 3! + t 3 4! + - 1 t = 1 + t 2 + t 2 6 + t 3 24 + . By the process of differentiating M ( x, t ) with respect to t and the setting t = 0, we get E ( x ) = M ( x, t ) t fl fl fl fl t =0 = 1 2 + 2 t 3! + 3 t 2 4! + t =0 = 1 2 , E ( x 2 ) = 2 M ( x, t ) t 2 fl fl fl fl t =0 = 2 3! + 6 t 4! + t =0 = 1 3 . Combining these results gives V ( x ) = E ( x 2 )- ' E ( x ) 2 = 1 3- 1 4 = 1 12 . 2. Find the moment generating function of x f ( x ) = ae- ax ; x 0....
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A05ans - EXERCISES IN STATISTICS Series A, No. 5 1. Find...

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