ST102_2017_LT0_commentary.pdf

Others attempted to derive separate means and

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Others attempted to derive separate means and variances for each component of the pdf. There were some instances of Var( X ) = E( X 2 ), which is only true if E( X ) = 0, and in some cases a negative answer was arrived at, demonstrating a lack of understanding about what a variance is! Full solutions are as follows. i. We must have R f ( x ) d x = 1. Hence: 1 = Z k 0 3 16 x 2 d x + Z 2 k k 3 16 ( x - 2 k ) 2 d x = x 3 16 k 0 + ( x - 2 k ) 3 16 2 k k = k 3 8 giving k = 2. ii. We have: E( X ) = Z 2 0 3 x 3 16 d x + Z 4 2 3 x ( x - 4) 2 16 d x = 3 x 4 64 2 0 + 3 x 4 64 - x 3 2 + 3 x 2 2 4 2 = 3 4 + 12 - 32 + 24 - 3 4 + 4 - 6 = 2 and: E( X 2 ) = Z 2 0 3 x 4 16 d x + Z 4 2 3 x 2 ( x - 4) 2 16 d x = 3 x 5 80 2 0 + 3 x 5 80 - 3 x 4 8 + x 3 4 2 = 6 5 + 192 5 - 96 + 64 - 6 5 + 6 - 8 = 22 5 . Hence Var( X ) = E( X 2 ) - (E( X )) 2 = 2 / 5. (12 marks) (b) Feedback on this question: Many successful attempts at this question, for which the starting point is remembering that M X ( t ) = E(e tX ). Common errors included a lower limit of summation of x = 0 instead of x = 1, and confusing a geometric series with the geometric distribution. Full solutions are as follows. We have: M X ( t ) = E(e tX ) = X x =1 e tx (1 - π ) x - 1 π = X x =1 π e t (e t (1 - π )) x - 1 = π e t 1 - e t (1 - π ) where the final step recognises that we have a sum to infinity of a geometric series with first term a = π e t and common ratio r = e t (1 - π ). 5
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ST102 Elementary Statistical Theory (LT0 examination) Differentiating with respect to t , we have: M 0 X ( t ) = (1 - e t (1 - π )) π e t - π e t (e t ( π - 1)) (1 - e t (1 - π )) 2 = π e t (1 - e t (1 - π )) 2 . Hence, since e 0 = 1, we have: E( X ) = M 0 X (0) = π (1 - (1 - π )) 2 = π π 2 = 1 π . (13 marks) 6
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  • Fall '15
  • Probability theory, Elementary Statistical Theory

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