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fX,Y (x, y) = 24xy, 0 < x < 1, 0 < y < 1, 0 < x + y < 1

(a) Is fX,Y (x, y) a valid probability density function?

using the integral 1→0

12x(1 − x)^2dx

12x^3 − 24x^2 + 12xdx =

3-8+6 =1 correct?

kurtk3
  • 149

1 Answers1

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In order to be a valid p.d.f., the function f(x, y) must satisfy two conditions: (i) f(x, y) ≥ 0, ∀x, y and

f(x, y)dydx = 1.Clearly f(x, y) is non-negative as it is zero outside of the range indicated above, and positive inside. In order to check the second condition we need to integrate the p.d.f. over the shaded region: that is, the region where y < 1.x for any x that lies between 0 and1. so that's correct

purugin
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