The distribution of ratio of two uniform variables

1. Jun 18, 2005

gimmytang

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:
$$f_{U,V}(u,v)=f_{X,Y}(v,uv)|v|$$ by change of variables.
Then:
$$f_{U}(u)=\int_{0}^{1}{f_{X,Y}(v,uv)|v|dv}=\int_{0}^{1}{vdv}={1\over 2}, 0<u<\infty$$, which is not integrated to 1.
Where I am wrong?
gim

Last edited: Jun 18, 2005
2. Jun 18, 2005

akito458

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...

3. Jun 19, 2005

gimmytang

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:
$$f_{U}(u)={\int_{-\infty}^{\infty}f_{U,V}(u,v)dv}={\int_{0}^{1}f_{X,Y}(u,uv)|v|dv}={\int_{0}^{1}vdv}=1/2$$
Now the question is my result 1/2 is not a reasonable density since it's not integrated to 1.
gim

4. May 4, 2010

electroissues

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))

5. May 4, 2010