solution_set2 - X x 1/2 1 1 y 1/2 1 1 F (y) Y 7. F X ( α )...

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EE 511 Solutions to Problem Set 2 1. P ( AB | A ) = P ( AB ) P ( A ) and P ( AB | ( A + B )) = P ( AB ( A + B )) P ( A + B ) = P ( AB ) P ( A + B ) . Since P ( A + B ) P ( A ), P ( AB | ( A + B )) P ( AB | A ). 2. Since A and B are mutually exclusive, AB = φ and P ( AB ) = 0. For A and B to be independent, we need P ( AB ) = P ( A ) P ( B ). This is possible only if P ( A ) = 0 or P ( B ) = 0 or both. 3. (a) True. (b) False. (c) False, in general. True only if X ( s ) = a s S . (d) True. 4. (i) For f X ( x ) to be a valid pdf, we need Z 1 - 1 f X ( x ) dx = 1 i.e., Z 1 - 1 c (1 - x 2 ) dx = 1 Therefore, we have c x - x 3 3 ± ± ± ± 1 - 1 = 1 2 c - 2 c 3 = 1 Therefore, c = 3 / 4. (ii) P [ X > 0] = Z 1 0 3 4 (1 - x 2 ) dx = 1 2 . P [ X < 1 2 ] = Z 0 . 5 - 1 3 4 (1 - x 2 ) dx = 27 32 . P [ | X | > 0 . 75] = 2 Z 1 0 . 75 3 4 (1 - x 2 ) dx = 11 128 . 5. (i) This function cannot be a pdf since f ( x ) < 0 for - 1 x < 0. (ii) This function can be a pdf since f ( x ) 0 for all x and Z -∞ f ( x ) dx = 1 . 6. F X ( x ) = F X,Y ( x, ) = 1 x > 1 1 2 0 x 1 0 x < 0 Similarly, we have F Y ( y ) = F X,Y ( , y ) = 1 y > 1 1 2 0 y 1 0 y < 0 1
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Unformatted text preview: X x 1/2 1 1 y 1/2 1 1 F (y) Y 7. F X ( α ) = P [ X ≤ α ] and F Y ( α ) = P [ Y ≤ α ]. We know that X ( s ) ≤ Y ( s ) for all s ∈ S . Therefore, we can say that Y ( s ) ≤ α implies X ( s ) ≤ α , i.e., the set of all elements s ∈ S such that Y ( s ) ≤ α is a subset of the set of all elements s ∈ S such that X ( s ) ≤ α . { s ∈ S : Y ( s ) ≤ α } ⊂ { s ∈ S : X ( s ) ≤ α } Therefore, we have P { s ∈ S : Y ( s ) ≤ α } ≤ P { s ∈ S : X ( s ) ≤ α } which implies F Y ( α ) ≤ F X ( α ). 2...
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solution_set2 - X x 1/2 1 1 y 1/2 1 1 F (y) Y 7. F X ( α )...

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