Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: 3 fx.x(x, y) =(1+x²y)!(0,1)(x)/(0,2)(y) Note that fx(x) = (1+ x²) I0,1) (2) fy(y) : (1+ y) I(0,2)(y) Solve for fx|y(x | y).

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Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by:
3
fx.x(x, y) =(1+ a²y)!(0,1)(x)/(0.2)()
Note that
fx(x) = (1+ x²) I0,1) (2)
fy(y) :
(1+ y) I(0,2)(y)
Solve for fx|y(x | y).
Remark. It follows immediately from the definition of conditional probability density function that
fx,y(r,y) = fxjy(x | y) · fy (y),
fx.x (x, y) = fy|x(y | x) · fx(x),
-00 <x<
-00 < y <∞
Transcribed Image Text:Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: 3 fx.x(x, y) =(1+ a²y)!(0,1)(x)/(0.2)() Note that fx(x) = (1+ x²) I0,1) (2) fy(y) : (1+ y) I(0,2)(y) Solve for fx|y(x | y). Remark. It follows immediately from the definition of conditional probability density function that fx,y(r,y) = fxjy(x | y) · fy (y), fx.x (x, y) = fy|x(y | x) · fx(x), -00 <x< -00 < y <∞
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