Find marginal pdf given joint pdf (stats)

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SUMMARY

The discussion focuses on finding the marginal probability density function (pdf) fx(x) from a given joint pdf f(x,y) = (2/(x²(x-1)))(y-(2x-1)/(x-1)) for x > 1 and y > 1. The correct approach involves integrating fx(x) = ∫f(x,y)dy from 1 to ∞. A key insight is that while attempting the integration, the variables must be treated appropriately, and the constants a and b should be identified to simplify the process. The integral is not undefined; rather, it requires proper handling of the constants involved.

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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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