Help on the density of Y/X, where X,Y~U(0,1)

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SUMMARY

The discussion focuses on determining the density of the ratio Y/X, where X and Y are independent and identically distributed (i.i.d) uniform random variables on the interval (0,1). The initial approach led to an incorrect marginal density of 1/2, which does not integrate to 1. The correct marginal density is derived as f(u) = 0.5 for u in [0,1) and f(u) = 0.5 * u^(-2) for u in [1,∞), ensuring that the distribution integrates to 1.

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gimmytang
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Hi,
There are two i.i.d uniform random variables X and Y. Now I need to know the density of Y/X. My method is like this:
Let U=Y/X, V=X. Then the marginal density of U is what I need.

[tex]f_{U}(u)={\int_{-\infty}^{\infty}f_{U,V}(u,v)dv}={\int_{0}^{1}f_{X,Y}(u,uv)|v|dv}={\int_{0}^{1}vdv}=1/2[/tex]
Now the question is that my result 1/2 is not a reasonable density since it's not integrated to 1. Can anyone point out where I am wrong?
gim :bugeye:
 
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I'm really rusty on this stuff, too rusty too even formalize a proper answer without brushing up on notation etc. So excuse me if this answer is a little vague but I think I know roughly what your problem is.

When you integrate out one of the variables of the joint distribution the limit is not just a simple "1", its actually function of the other variable. It helps if you try to sketch the joint distribution, the way I'm picturing it you should end up with something like,

f(u) = int(v,dv,0..1) : u in [0..1)
and
f(u) = int(v,dv,0..1/u) : u in [1..infinity)

This gives

f(u) = 0.5 : u in [0..1)
and
f(u)=0.5 u^(-2) : u in [1..infinity)

I'm not 100% sure it's correct but it looks reasonable
 
Last edited:
Thank you for your answer. It's correct since the distribution function is integrated to 1. I always have some confusion about the right domain of the variable since it's a function of the other variable.
gim
 

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