Lets derive the solution again this time using the

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Unformatted text preview: at can often be used to help determine whether X and Y are independent. To begin, the following proposition gives a condition equivalent to independence, based on conditional pdfs. 4.4. INDEPENDENCE OF RANDOM VARIABLES 133 Proposition 4.4.2 X and Y are independent if and only if the following condition holds: For all u ∈ R, either fX (u) = 0 or fY |X (v |u) = fY (v ) for all v ∈ R. Proof. (if part) If the condition holds, then for any (u, v ) ∈ R2 there are two cases: (i) fX (u) = 0 so fX,Y (u, v ) = 0 so fX,Y (u, v ) = fX (u)fY (v ). (ii) fX (u) > 0 and then fX,Y (u, v ) = fX (u)fY |X (v |u) = fX (u)fY (v ). So in either case, fX,Y (u, v ) = fX (u)fY (v ). Since this is true for all (u, v ) ∈ R2 , X and Y are independent. (only if part) If X and Y are independent, then fX,Y (u, v ) = fX (u)fY (v ) for all (u, v ) ∈ R2 , so if f (u,v ) fX (u) > 0, then fY |X (v |u) = X,Y (u) = fY (v ), for all v ∈ R. fX The remaining propositions have to do with the notion of product sets. Suppose A and B each consist of a finite union of disjoint fin...
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