For the second part of equation we have E I A c n X n X 2 Z I A c n X n X 2 d F

For the second part of equation we have e i a c n x n

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For the second part of equation ( ?? ) we have E[ I A c n | X n - X | 2 ] = Z + -∞ I A c n | X n - X | 2 d F | X n - X | 2 Z + -∞ d F | X n - X | = 2 , > 0 2

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Which means that : E[ I A c n | X n - X | 2 ] = 0 ( small). Thus, we have: lim n + E [ | X n - X | 2 ] = lim n + E[ I A n | X n - X | 2 ] + lim n + E[ I A c n | X n - X | 2 ] = 4 c 2 lim n + P( | X n - X | > ) + 0 And we can conclude that lim n + E[ | X n - X | 2 ] = 0. Problem 4 Show that if X n D -→ X and P ( X = c ) = 1 for some constant c then X n i.p. --→ X . lim n + F X n = 0 if x < c 1 if x c For > 0 and n 1 we have P ( | X n - c | ≥ ) = P ( X n - c ) + P ( X n - c < - ) = P ( X n + c ) + P ( X n < c - ) = 1 - F X n ( c + ) + F X n ( c - ) + P ( X n = c + ) (2) Now lim n + P ( | X n - c | ≥ ) = 1 - 1 + 0 + 0 = 0 X n i.p. --→ c. Problem 5 We investigate the case of | X n | < Y, n . Using triangle inequality we have: | X | = | X - X n + X n | ≤ | X - X n | + | X n | , and thus P ( | X | > Y + ) P ( | X - X n | + | X n | - Y > ) P ( | X - X n | > ) Applying the limit it becomes: lim n + P ( | X | > Y + ) lim n + P ( | X - X n | > ) Because X n i.p. --→ X , lim n + P ( | X - X n | > ) = 0, and we have no dependence on n on the left hand side so: P ( | X | > Y + ) 0 P ( X > Y + ) = 0 So, now we can use exploit the fact that | X | ≤ Y : | X n | ≤ | X n | + | X | ≤ 2 Y ⇒ | X n - X | 2 2 Y 2 .
• Spring '14
• Aazhang,Behnaam
• Calculus, dy, lim P

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