Find marginal pdf given joint pdf (stats)

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Homework Help Overview

The discussion revolves around finding the marginal probability density function (pdf) from a given joint pdf in the context of statistics. The joint pdf is defined for x > 1 and y > 1, and participants are attempting to compute the marginal pdf fx(x) through integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to compute the marginal pdf by integrating the joint pdf with respect to y. There is a question about the correctness of the integral bounds used in the calculation. Some participants express confusion regarding the definition of the integral and the handling of constants during integration.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the integral setup. There is a suggestion that the original poster may not have utilized all available information, indicating a potential direction for further exploration. No consensus has been reached yet.

Contextual Notes

Participants are discussing the bounds of the integral and the implications of treating certain variables as constants during integration. There is an indication that the original problem may involve more complexity than initially perceived.

Phox
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Homework Statement



Given joint pdf:

f(x,y) = (2/(x2(x-1)))(y-(2x-1)/(x-1)) x>1, y>1

find marginal pdf fx(x)

Homework Equations



fx(x) = ∫f(x,y)dy, 0, ∞

The Attempt at a Solution



fx(x) = ∫(2/(x2(x-1)))(y-(2x-1)/(x-1)) dy, 1, ∞

= [-2y(x/(1-x))/x3],1,∞ = undefined. stuck here
 
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Phox said:

Homework Statement



Given joint pdf:

f(x,y) = (2/(x2(x-1)))(y-(2x-1)/(x-1)) x>1, y>1

find marginal pdf fx(x)

Homework Equations



fx(x) = ∫f(x,y)dy, 0, ∞

The Attempt at a Solution



fx(x) = ∫(2/(x2(x-1)))(y-(2x-1)/(x-1)) dy, 1, ∞

= [-2y(x/(1-x))/x3],1,∞ = undefined. stuck here

No, it is NOT undefined. You have not used all the available information.
 
Ray Vickson said:
No, it is NOT undefined. You have not used all the available information.

have I bounded the integral correctly?
 
Phox said:
have I bounded the integral correctly?

I don't know what you mean. However, why make things complicated? When we do the y-integral, x is a constant, so we might as well write ##f(x,y) = a y^{-b},## where ##a=2/[x^2(x-1)]## and ##b=(2x-1)/(x-1)## are 'constants' as far as y-variations are concerned. You need to use information about the magnitude of ##b##.
 

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