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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

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 70EQ
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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(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∞
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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