Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: Note that fx(x) =÷(1+ x²) I0,1)(7) 3 fr (y) =÷ (1 + y) I0,2)(y) Solve for fxjr(r | y). Remark. It follows immediately from the definition of conditional probability density function that fx,y(r, y) = fxjy(x | y) · fy (y), -0 < # < 0 fxy(x, y) = fyjx(y| a) fx(x), -00 < y
Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: Note that fx(x) =÷(1+ x²) I0,1)(7) 3 fr (y) =÷ (1 + y) I0,2)(y) Solve for fxjr(r | y). Remark. It follows immediately from the definition of conditional probability density function that fx,y(r, y) = fxjy(x | y) · fy (y), -0 < # < 0 fxy(x, y) = fyjx(y| a) fx(x), -00 < y
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by:
3
fx,x(r, y) = (1 + a°y)[(0,1)(=)[(0,2) (1)
Note that
fx(x) =÷ (1 + a²) I0,1)(x)
fr(y) =-
(1+y) I(0,2) (y)
Solve for fxjy(æ | y).
Remark. It follows immediately from the definition of conditional probability density function that
fx,x(x, y) = fxjy(x | y) · fy(y), -o∞ < x < ∞
fx,y (x, y) = fy|x(y | æ) · fx(x), - <y <o∞
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