Question

if f(x, y) = 3x²-8xy+6y² (0<x<1, 0<y<1).find fx(x|y) and fy(y|x) & show that x and y are independent.
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Answer 1

Answer:

A

∫

−2010

2010

F(x)dx=∫

0

2011

F(x)dx

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Answer 2

SOLUTION

GIVEN

$\sf{f(x, y) = 3 {x}^{2} - 8xy + 6 {y}^{2} \: \: \: ( 0 < x < 1,0 < y < 1) }$

TO DETERMINE

$\sf{1. \: \: f_x(x|y) \: \: and \: \: f_y(y|x)}$

2. X and Y are dependent

EVALUATION

Here it is given that

$\sf{f(x, y) = 3 {x}^{2} - 8xy + 6 {y}^{2} \: \: \: ( 0 < x < 1,0 < y < 1) }$

Now

$\sf{f_x(x)}$

$\displaystyle \sf{ = \int\limits_{- \infty}^{\infty} f(x,y) \, dy }$

$\displaystyle \sf{ = \int\limits_{0}^{1} (3 {x}^{2} - 8xy + 6 {y}^{2} ) \, dy}$

$\displaystyle \sf{ =3 {x}^{2} - 4x +2 \: \: \: \: \: \: , \: 0 < x < 1}$

Again

$\sf{f_y(y)}$

$\displaystyle \sf{ = \int\limits_{- \infty}^{\infty} f(x,y) \, dx}$

$\displaystyle \sf{ = \int\limits_{0}^{1} (3 {x}^{2} - 8xy + 6 {y}^{2} ) \, dx}$

$\displaystyle \sf{ =1 - 4y + 6 {y}^{2} \: \: \: \: \: \: , \: 0 < y < 1}$

1.

$\sf{ f_x(x|y) }$

$\displaystyle \sf{ = \frac{f(x,y)}{f_y(y)} }$

$\displaystyle \sf{ = \frac{(3 {x}^{2} - 8xy + 6 {y}^{2} )}{(6 {y}^{2} - 4y + 1)} \: \: \: ( 0 < x < 1,0 < y < 1) }$

Again

$\sf{ f_y(y|x) }$

$\displaystyle \sf{ = \frac{f(x,y)}{f_x(x)} }$

$\displaystyle \sf{ = \frac{(3 {x}^{2} - 8xy + 6 {y}^{2} )}{(3 {x}^{2} - 4x + 2)} \: \: \: ( 0 < x < 1,0 < y < 1) }$

2. From above we observe that

$\sf{f(x, y) \ne \: f_x(x). f_y(y) }$

So X and Y are dependent

Hence proved

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