Unformatted text preview: + 0.1 + 0.2 = 0.6. 3/6 v = 1
2/6 v = 2
(e) pY X (v 2) = 1/6 v = 3.
The following two properties are necessary and suﬃcient for p to be a valid pmf:
Property pmf.1: p is nonnegative,
Property pmf.2: 4.3 (u,v )∈Z2 p(u, v ) = 1. Joint probability density functions The random variables X and Y are jointly continuoustype if there exists a function fX,Y , called
the joint probability density function (pdf), such that FX,Y (u0 , vo ) is obtained for any (u0 , vo ) ∈ R2
by integration over the shaded region in Figure 4.1:
uo vo −∞ −∞ FX,Y (uo , vo ) = fX,Y (u, v )dvdu. 4.3. JOINT PROBABILITY DENSITY FUNCTIONS 123 It follows from (4.1) that if R is the rectangular region (a, b] × (c, d] in the plane, shown in
Figure 4.2, then
P {(X, Y ) ∈ R} = fX,Y (u, v )dudv. (4.2) R More generally, (4.2) holds for any set R in the plane that has a piecewise diﬀerentiable boundary,
because such sets can be approximated by a ﬁnite or countably inﬁnite union of disjoint rectangular
regions. If g is a function on the...
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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 Tech.
 Spring '08
 Zahrn
 The Land

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