# So x and y are not y given x u depend on u fy x v u 0

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Unformatted text preview: and then integrate it to ﬁnd the marginal pdf, fY . Basically we will be using the version of the law of total probability for pdfs in (4.7). The pdf of X is simple–X is uniformly distributed over the interval [0, 1]. The support of fX,Y is the triangular region, T = {(u, v ) : u ≥ 0, v ≥ 0, and u + v ≤ 1}. In particular, fX,Y (u, v ) = 0 if u ≤ 0 or if u ≥ 1. For 0 < u < 1, given X = u, the set of possible values of Y is the interval [0, 1 − u]. Since the second random point is uniformly distributed over [u, 1], and Y is the diﬀerence between that point and the constant u, the conditional distribution of Y is uniform over the interval [0, 1 − u]. That is, if 0 < u < 1: 1 1−u fY |X (v |u) = 0 0≤v ≤1−u else. For 0 < u < 1, fX,Y (u, v ) = fX (u)fY |X (v |u) = fX |Y (v |u), and fX,Y (u, v ) = 0 otherwise. Therefore, fX,Y (u, v ) = 1 1−u 0 (u, v ) ∈ T else. The support of the pdf of Y is [0, 1] so let vo ∈ [0, 1]. Then ∞ fY (vo ) = 1−v...
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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 Institute of Technology.

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