7 since the area of s is 34 the pdf is fxy u v find

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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 continuous-type 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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