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Unformatted text preview: else,
u + v u, v ∈ [0, 1]
(b) fX,Y (u, v ) =
9u2 v 2 u, v ∈ [0, 1]
(c) fX,Y (u, v ) =
Solution: (a) No, X and Y are not independent. This one is a little tricky because the function
Cu2 v 2 does factor into a function of u times a function of v. However, fX,Y (u, v ) is only equal to
Cu2 v 2 over a portion of the plane. One reason (only one reason is needed!) X and Y are not
independent is because the support of fX,Y is not a product set. For example, fX,Y (0.3, 0.6) > 0
and fX,Y (0.6, 0.3) > 0, but fX,Y (0.6, 0.6) > 0, so the support is not a product set by Proposition
4.4.3. Another reason X and Y are not independent is that the conditional distribution of Y given
X = u depends on u: for u ﬁxed the conditional density of Y given X = u has support equal to
the interval [0, 1 − u]. Since the support of the conditional pdf of Y given X = u depends on u, the 4.5. DISTRIBUTION OF SUMS OF RANDOM VARIABLES 135 conditional pdf itself depends on u. So X and Y are not independent by Proposition 4.4.2.
(b) No, X and Y are not independent. One reason is that the function u + v on [0, 1]2 does not fac...
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- Spring '08
- The Land