This preview shows page 1. Sign up to view the full content.
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 < v < 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...
View
Full
Document
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.
 Spring '08
 Zahrn
 The Land

Click to edit the document details