- #1

_joey

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I have a question I need to answer.

[tex]f\left(x,y\right)= 6x^2y [/tex] such that [tex] 0<x<y[/tex] and [tex]x+y<2[/tex] where [tex] f\left(x,y\right)[/tex] is a probability density function for two random variables: [tex]X,\;Y[/tex]

I need to find marginal density distribution for the random variables [tex]X[/tex] and [tex]Y[/tex]

It appears to be a straight forward question except that when I integrate twice, in instance with the marginal density for [tex]Y,\;f_Y[/tex], it does not integrate to 1.

Here are my calculations:

[tex]

0&<&x<y[/tex] and [tex]x+y<2 \implies 0<x<1[/tex] and [tex] x<2-y[/tex]. Hence, marginal density for [tex]X[/tex] is

[tex]f_x\left(x,y\right)=\int\limits_{x}^{2-x}6x^2y\,dy =\[\left. 3{{x}^{2}}{{y}^{2}} \right|_{x}^{2-x}\]=12x^2-12x^3,\;0<x<1[/tex]

If I integrate the above function again over [tex] \left(0,1\right)[/tex]I will obtain 1. This is a property of marginal density distribution (?!)

Things start falling apart with marginal density for [tex]Y[/tex] variable

[tex]f_y\left(x,y\right)=\int\limits_{0}^{1}6x^2y\,dx =\[\left. 2{{x}^{3}}{{y}} \right|_{0}^{1}\]=2y,\;x<y<2-x[/tex]

If I integrate [tex]f_Y(y)=2y[/tex] again over [tex]x<y<2-x[/tex] I will obtain [tex]4-4x[/tex].

Any help and suggestions will be much appreciated.

Thanks!

[tex]f\left(x,y\right)= 6x^2y [/tex] such that [tex] 0<x<y[/tex] and [tex]x+y<2[/tex] where [tex] f\left(x,y\right)[/tex] is a probability density function for two random variables: [tex]X,\;Y[/tex]

I need to find marginal density distribution for the random variables [tex]X[/tex] and [tex]Y[/tex]

It appears to be a straight forward question except that when I integrate twice, in instance with the marginal density for [tex]Y,\;f_Y[/tex], it does not integrate to 1.

Here are my calculations:

[tex]

0&<&x<y[/tex] and [tex]x+y<2 \implies 0<x<1[/tex] and [tex] x<2-y[/tex]. Hence, marginal density for [tex]X[/tex] is

[tex]f_x\left(x,y\right)=\int\limits_{x}^{2-x}6x^2y\,dy =\[\left. 3{{x}^{2}}{{y}^{2}} \right|_{x}^{2-x}\]=12x^2-12x^3,\;0<x<1[/tex]

If I integrate the above function again over [tex] \left(0,1\right)[/tex]I will obtain 1. This is a property of marginal density distribution (?!)

Things start falling apart with marginal density for [tex]Y[/tex] variable

[tex]f_y\left(x,y\right)=\int\limits_{0}^{1}6x^2y\,dx =\[\left. 2{{x}^{3}}{{y}} \right|_{0}^{1}\]=2y,\;x<y<2-x[/tex]

If I integrate [tex]f_Y(y)=2y[/tex] again over [tex]x<y<2-x[/tex] I will obtain [tex]4-4x[/tex].

Any help and suggestions will be much appreciated.

Thanks!

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