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

  • Thread starter gimmytang
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In summary, the conversation discusses the method of finding the density of Y/X, using the joint distribution of two i.i.d uniform random variables X and Y. The marginal density of U is needed, and it is found to be 1/2. However, there is a question about the correctness of this result since it does not integrate to 1. The expert suggests that the limit should be a function of the other variable, and provides a sketch of the joint distribution to illustrate this.
  • #1
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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  • #2
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
 
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  • #3
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
 

1. What is the density of Y/X?

The density of Y/X is the probability distribution that describes the relative likelihood of Y and X being equal to a specific value. In this case, X and Y are both uniformly distributed between 0 and 1, so the density of Y/X would be a straight line with a slope of 1 and an intercept of 0, representing equal probability for all values of Y/X.

2. How is the density of Y/X calculated?

The density of Y/X is calculated by taking the ratio of the probability density functions of Y and X. In this case, both Y and X have a uniform distribution, so their probability density functions are both constant over the range of 0 to 1. Therefore, the density of Y/X would be a constant value of 1 over the same range.

3. What is the range of possible values for the density of Y/X?

Since the density of Y/X is calculated as a ratio of two uniform distributions, it will always be a constant value of 1 over the range of 0 to 1. This means that the range of possible values for the density of Y/X is also 0 to 1.

4. How does the density of Y/X relate to the individual densities of Y and X?

The density of Y/X is related to the individual densities of Y and X by taking the ratio of these two distributions. This can be thought of as a way to compare the likelihood of a specific value of Y/X occurring, given the individual probabilities of Y and X. Since Y and X are uniformly distributed, their individual densities have no effect on the overall shape of the density of Y/X, which will always be a straight line with a slope of 1.

5. How can the density of Y/X be used in scientific research?

The density of Y/X can be used in scientific research to analyze the relationship between two variables that are both uniformly distributed. It can help researchers understand the relative likelihood of certain values of Y/X occurring and can be used to make predictions or in statistical analyses. Additionally, the density of Y/X can be used to generate simulations or models that mimic real-world scenarios where Y and X are both uniformly distributed.

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