# Transformation of uniform random variables: Stuck on a question about density functio

• Dec 12th 2011, 08:30 AM
celestial
Transformation of uniform random variables: Stuck on a question about density functio
Given: X and Y, independent, continuous uniform random variables on the interval (0,1)

Goal: Find the density function of$\displaystyle \frac{X}{X+Y}$

My approach:
Use the transformation technique (as in point 5 of this wikipedia page Integration by substitution - Wikipedia, the free encyclopedia).

Define $\displaystyle$Z_{1} = \frac{X}{X+Y}$$. Define \displaystyle Z_{2} = X$$.
I find that the joint density of$\displaystyle$Z_{1}$$and \displaystyle Z_{2}$$ is

$\displaystyle$f_{Z_{1},Z_{2}}(z_{1},z_{2}) = \frac{Z_{2}}{{Z_{1}}^{2}}$$And this is where i get stuck... I tried integrating \displaystyle Z_{2} out of it but nothing reasonable comes out (not a density function anyway). So two questions: 1. Did I find the joint density correct? 2. I tried integrating\displaystyle Z_{2} over the interval\displaystyle Z_{1} to 1, is this correct? 3. What should be next step? Thanks for any help! • Dec 12th 2011, 10:28 AM chisigma Re: Transformation of uniform random variables: Stuck on a question about density fun Quote: Originally Posted by celestial Given: X and Y, independent, continuous uniform random variables on the interval (0,1) Goal: Find the density function of\displaystyle \frac{X}{X+Y} My approach: Use the transformation technique (as in point 5 of this wikipedia page Integration by substitution - Wikipedia, the free encyclopedia). Define \displaystyle Z_{1} = \frac{X}{X+Y}$$. Define $\displaystyle$Z_{2} = X$$. I find that the joint density of\displaystyle Z_{1}$$ and $\displaystyle$Z_{2}$$is \displaystyle f_{Z_{1},Z_{2}}(z_{1},z_{2}) = \frac{Z_{2}}{{Z_{1}}^{2}}$$

And this is where i get stuck... I tried integrating $\displaystyle Z_{2}$ out of it but nothing reasonable comes out (not a density function anyway). So two questions:

1. Did I find the joint density correct?
2. I tried integrating$\displaystyle Z_{2}$ over the interval$\displaystyle Z_{1}$ to 1, is this correct?
3. What should be next step?

Thanks for any help!

May be that it is useful to devide the procedure into two steps. The first step is to find the p.d.f. of the random variable $\displaystyle Z=\frac{X}{Y}$ being X and Y independent random variables uniformely distributed in $\displaystyle [0,1]$. Proceeding as in...

http://www.mathhelpforum.com/math-he...es-193906.html

... we find for $\displaystyle 0<z<1$...

$\displaystyle p_{z}(z)= \int_{0}^{1} y\ dy =\frac{1}{2}$ (1)

... for $\displaystyle z>1$...

$\displaystyle p_{z}(z)= \int_{0}^{\frac{1}{z}} y\ dy =\frac{1}{2\ z^{2}}$ (2)

... and for $\displaystyle z<0$...

$\displaystyle p_{z}(z)= 0$ (3)

The second step is to find the p.d.f. of the random variable $\displaystyle U=\frac{1}{1+Z}$... how to proceed?...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 12th 2011, 11:44 AM
celestial
Re: Transformation of uniform random variables: Stuck on a question about density fun
Quote:

Originally Posted by chisigma
May be that it is useful to devide the procedure into two steps. The first step is to find the p.d.f. of the random variable $\displaystyle Z=\frac{X}{Y}$ being X and Y independent random variables uniformely distributed in $\displaystyle [0,1]$. Proceeding as in...

http://www.mathhelpforum.com/math-he...es-193906.html

... we find for $\displaystyle 0<z<1$...

$\displaystyle p_{z}(z)= \int_{0}^{1} y\ dy =\frac{1}{2}$ (1)

... for $\displaystyle z>1$...

$\displaystyle p_{z}(z)= \int_{0}^{\frac{1}{z}} y\ dy =\frac{1}{2\ z^{2}}$ (2)

... and for $\displaystyle z<0$...

$\displaystyle p_{z}(z)= 0$ (3)

The second step is to find the p.d.f. of the random variable $\displaystyle U=\frac{1}{1+Z}$... how to proceed?...

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$

Nice trick with the U! Thanks for that. I knew the ratio distribution of two standard uniforms but couldn't get any use if it. So let me see if Mary and Jesus have answered my prayers and blessed me with knowledge:

So using the ratio of two uniform variables, let's call $\displaystyle g(z) = M = \frac{1}{1+Z}$. Assume we can prove all the desirable properties like g being a bijective function, and its inverse having a continuous derivative (I haven't proven them yet), then I have $\displaystyle g^{-1}(M) = \frac{1-M}{M}$. Hence I have for $\displaystyle f_{M}(m) = \frac{ \partial g^{-1}(m)}{\partial m}f_{z}(g^{-1}(m))$
So then I'll have

for $\displaystyle 0 < \frac{1-m}{m} < 1$
• $\displaystyle f_{M}(m) = - \frac{1}{2 \cdot m^{2}}$

for $\displaystyle \frac{1-m}{m} > 1$

• $\displaystyle f_{M}(m) = - \frac{1}{2 \cdot m^{2} \cdot \left(\frac{1-m}{m}\right)}$

and 0 otherwise?

Of course I need to simplify the conditions on m (the intervals) , but is that the right direction?
• Dec 13th 2011, 12:07 AM
chisigma
Re: Transformation of uniform random variables: Stuck on a question about density fun
A preminary investigation on the functional equation...

$\displaystyle U=\frac{1}{1+Z},\ Z \ge 0$ (1)

... may be useful. From (1) it is clear that for $\displaystyle u<0$ is $\displaystyle P\{U<u\}=0$ and for $\displaystyle u>0$ is $\displaystyle P\{U<u\}=1$, so that we have to investigate what happens for $\displaystyle 0<u<1$. From (1) we derive...

$\displaystyle P\{U<u\} = P\{Z>\frac{1-U}{U}\}= \int_{\frac{1-u}{u}}^{\infty} p_{z}(z)\ dz$ (2)

The function $\displaystyle p_{z}(*)$ has been computed in a previous post...

$\displaystyle p_{z}(z)=\begin{cases}\frac{1}{2} &\text{if}\ 0<z<1\\ \frac{1}{2\ z^{2}} &\text{if}\ z>1\\ 0 &\text{elsewhere}\end{cases}$ (3)

Now deriving (2) and taking into account (3) we obtain...

$\displaystyle p_{u}(u)=\begin{cases}\frac{1}{2\ (1-u)} &\text{if}\ 0<z<\frac{1}{2}\\ \frac{1}{2\ u^{2}} &\text{if}\ \frac{1}{2}<z<1\\ 0 &\text{elsewhere}\end{cases}$ (4)

http://www.sv-luka.org/ikone/ikone180a.jpg

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$