X y x dy dx Therefor E Y X x R yf Y X y x dy 1 Problem 2 Suppose X and Y are

# X y x dy dx therefor e y x x r yf y x y x dy 1

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| X ( y | x ) dy dx. Therefor, E [ Y | X = x ] = R -∞ yf Y | X ( y | x ) dy . 1

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Problem 2 Suppose X and Y are jointly continuous random variables with joint pdf given by f XY ( x, y ) = e - y if x > 0 and y > x 0 otherwise i) Show that f XY is a legitimate joint pdf. f XY ( x, y ) is obviously nonnegative. Also, Z -∞ Z -∞ f XY dxdy = Z 0 Z x e - y dydx = Z 0 e - x dx = 1 . Thus, f XY ( x, y ) is a legitimate joint pdf. ii) Find the Marginal pdf’s X and Y . f Y ( y ) = Z -∞ f XY ( x, y ) dx = 0 for y 0 ye - y for y > 0 f X ( x ) = Z -∞ f XY ( x, y ) dy = 0 for x 0 e - x for x > 0 iii) Find E [ Y | X = x ] for x > 0. E [ Y | X = x ] = Z -∞ y f XY ( x, y ) f X ( x ) dy = x + 1 , x > 0 Problem 3 Suppose X n i.p. --→ X and that there is a constant c such that | X n | ≤ c for all n . Show that X n m.s. ---→ X . | X n | ≤ c n F X n = 1 for x c ? for - c < x < c 0 for x < - c Therefore F X ( x ) = F X n ( x ) for | x | > c Now, given > 0, let A n = { ω Ω : | X n ( ω ) - X ( ω ) | ≥ } . If we use the indicator function I A n and for I A c n , for each n we have: E[ | X n - X | 2 ] = E[( I A n + I A c n ) | X n - X | 2 ] = E[ I A n | X n - X | 2 ] + E[ I A c n | X n - X | 2 ] (1) Using triangle inequality, it can be proven that | X n - X | < 2 c ⇒ | X n - X | 2 < (2 c ) 2 . Using this result, we have the first part of equation ( ?? ) as: E[ I A n | X n - X | 2 ] 4 c 2 P( | X n - X | > ) , > 0
• Spring '14
• Aazhang,Behnaam
• Calculus, dy, lim P

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