The distribution of ratio of two uniform variables

  • Thread starter gimmytang
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Hello,
Let X ~ U(0,1), Y ~U(0,1), and independent from each other. To calculate the density of U=Y/X, let V=X, then:
[tex]f_{U,V}(u,v)=f_{X,Y}(v,uv)|v|[/tex] by change of variables.
Then:
[tex]f_{U}(u)=\int_{0}^{1}{f_{X,Y}(v,uv)|v|dv}=\int_{0}^{1}{vdv}={1\over 2}, 0<u<\infty[/tex], which is not integrated to 1.
Where I am wrong?
gim :cry:
 
Last edited:
you only integrated with respect to fu(u). Now you have to integrate with respect to fv(v). Or you could have just used a double integral to start with...
 
Actually the marginal distribution of U, namely the distribution of the ratio of two uniform variables, is the only thing that I am interested. To be more clear:
[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 my result 1/2 is not a reasonable density since it's not integrated to 1.
gim :bugeye:
 
Question says Let X and Y be independent random variables with join cumulative distribution function (CDF) F subscript X,Y of (x,y)= P (X</= x, Y</=y). Show that the CDF Fz(z) of the random variable Z=min (X,Y) can be computed via
Fz(z)= Fx(z) + Fy(z) - Fx(z) . Fy(z) = 1 - (1 - Fx (z)) . (1- Fy (z))

Please reply to this asap. I need to submit this answer by Friday. Thanks!
 

statdad

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Question says Let X and Y be independent random variables with join cumulative distribution function (CDF) F subscript X,Y of (x,y)= P (X</= x, Y</=y). Show that the CDF Fz(z) of the random variable Z=min (X,Y) can be computed via
Fz(z)= Fx(z) + Fy(z) - Fx(z) . Fy(z) = 1 - (1 - Fx (z)) . (1- Fy (z))

Please reply to this asap. I need to submit this answer by Friday. Thanks!
Don't jump into the thread of another. What have you done so far?
 
Well, I'm new here and had problems starting a new thread.

I looked at PDF of an exponential function which is (1 - Fx (x)) and also since its also given its independent, we know it can be split into Fx (x) . Fy (y) but I can't put these things together.
 

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