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1. 1 f 1 ( x) = 2 e x 2 /2 and f 2 ( x) = 1 (1 + x 2 ) If x = 0 is observed the M.L.E of distribution of X is (i) f1 ( x) 2. (ii) f 2 ( x) If X i have mean 0 and variance 1 and Yi have mean 0 and variance 2 then ( X i + Yi ) n (i) N(0,1) 3. f ( x) = (i) 2x X 2 i converges to (ii) N(0,2) (iii) N(0,3) is sufficient for 6. (ii) X 2 i is sufficient for (iv) (max X i , min X i ) is sufficient for 1 If X 1 , X 2 , ..., X n be i.i.d distributed with F(.) and Y = 0 V(Y) is (i) 1/2 5. (iv) N(0,4) , 0 < x < , then (iii) max ( X i ) is sufficient for 4. (iv) can be f1 ( x) or f 2 ( x) (iii) does not exist (ii) 1/4 X1 X 2 > 0 otherwise (iii) cannot be predicted unless the value of F(.) given 0 1 / 2 0 1 / 2 0 2 / 3 0 1/ 3 find all stationary If transition matrix is P = 1 / 4 0 3 / 4 0 0 2 / 3 0 1/ 3 distributions of Markov Chain. Give three example of sequence (i) ( X n ) converges in probability but not in mean (ii) ( X n ) converges in distribution but not in probability then [ ] 7. 2 2 If X i ~ N (0,1) then lim P X 12 + X 2 + + X n < n + 2n a = P[X 1 a ] 8. X n Find the value of E exp it n , where X ~ P ( n) n 9. Prove that E (X k ) = 10. Suppose X is single observation from Beta ( ,1) . Let Y = ( log X ) 1 final convergence n 1 k 1 x (1 F ( x) ) dx where F(x) is distribution function. k 0 (i) proportional of interval estimate (Y / 2, Y ) for (ii) suggest a pivot based on random variable Y.

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