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Unformatted text preview: = {u : (u, v ) ∈ S for some v } and B = {v : (u, v ) ∈ S for some u}. To complete
the proof it will be shown that S = A × B. To prove this, it is shown that S ⊂ A × B and A × B ⊂ S.
On one hand, if (a, b) ∈ S then a ∈ A and b ∈ B so (a, b) ∈ A × B. Therefore, S ⊂ A × B. On the
other hand, if (a, b) ∈ A × B, then a ∈ A so (a, v ) ∈ S for some v , and b ∈ B so (u, b) ∈ S for some
u. But since (a, v ), (u, b) ∈ S, it follows that (a, b) ∈ S. So A × B ⊂ S. So S = A × B as claimed,
and the proposition is proved. 134 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES The next proposition gives a necessary condition for X and Y to be independent. While the
condition is not suﬃcient for independence, it is suﬃcient for an important special case, covered in
the corollary to the proposition.
Proposition 4.4.4 If X and Y are independent, jointly continuoustype random variables, then
the support of fX,Y is a product set.
Proof. By assumption, fX,Y (u, v ) = f...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
 Spring '08
 Zahrn
 The Land

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