# 451 sums of integer valued random variables suppose s

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Unformatted text preview: o fX,Y (u, vo )du = −∞ 0 1 du = − ln(1 − u) 1−u 1−vo = − ln(vo ). 0 4.4. INDEPENDENCE OF RANDOM VARIABLES 131 This is inﬁnite if vo = 0, but the density at a single point doesn’t make a diﬀerence, so we set fY (0) = 0. Thus, we have: − ln(v ) 0 &lt; v &lt; 1 fY (v ) = 0 else. Example 4.3.6 Let Z = Y , X2 such that X and Y have joint pdf given by 1 if 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 0 else. fX,Y (u, v ) = (a) Find the numerical value of P {Z ≤ 0.5}. Also, sketch a region in the plane so that the area of the region is P {Z ≤ 0.5}. (b) Find the numerical value of P {Z ≤ 4}. Also, sketch a region in the plane so that the area of the region is P {Z ≤ 4}. Solution. 1 2 0 (0.5)u du (a) P {Z ≤ 0.5} = P {Y ≤ (0.5)X 2 } = = 1. 6 v 1 v=(0.5)u 2 u 1 (b) P {Z ≤ 4} = P {Y ≤ 4X 2 } = 0 .5 2 0 4u du 2 + 0.5 = 3 . v 1 v=4u 2 u 0.5 4.4 1 Independence of random variables Independence of events is discussed in Section 2.4. Recall that two events, A and B , are independent, if and only if P (AB ) = P (A)P (B ). Even...
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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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