Q.6.56 - A First Course in Probability

Question

If X and Y are independent and identically distributed uniform random variables on$(0,1)$, compute the joint density of

$$\begin{gathered} \left(a\right) U = X + Y, V = X/Y; \\ \left(b\right) U = X, V = X/Y; \\ \left(c\right) U = X + Y, V = X/(X + Y). \end{gathered}$$

Step-by-Step Solution

Answer

(a) $f_{U,V}(u,v)=\frac{u}{{(v+1)}^2}$

(b) $f_{U,V}(u,v)=\frac{u}{v^2}$

(c) $f_{U,V}(u,v)=u$

Step 1 : Definition

The joint probability density function (joint pdf) is a function that is used to characterize a continuous random vector's probability distribution.

Step 2: Explanation (part a)

X and Y are independent and are identically distributed on $\left[0,1\right]$.

Such that 

$U=X+Y$ and $V=X/Y$

The joint probability distribution function of random vector (X,Y),

$f(x,y)=f_X\left(x\right) f_Y\left(y\right)=1$

Apply the transformation,

$g:{(0,1)}^2\to R^2$

Such that 

$g(x,y)=(u,v)= \left(x+y, \frac{x}{y}\right)$

By using theorem, the density function of random vector $(U,V)=g(X,Y)$ as

$f_{U,V}(u,v)= f_{X,Y}(x,y)·{\left|det(\nabla g(x,y\left)\right)\right|}^{-1}$

Then calculate,

$\nabla g(x,y)=\left[\begin{array}{cc}1& 1\\ \frac{1}{y}& -\frac{x}{y^2}\end{array}\right]$

That yields

$$\begin{gathered} \left|det(\nabla g(x,y\left)\right)\right|=\frac{x}{y^2}+\frac{1}{y}=\frac{x+y}{y^2} \\ \text{Thus} \\ {f}_{U,V}(u,v)=f_{X,Y}(x,y)·\frac{y^2}{x+y}=\frac{y^2}{x+y} \end{gathered}$$

Now, write x, y in terms of u and v and substitute it in the last equality,

But we have

$u=x+y and y^2=\frac{u^2}{{(v+1)}^2}$

That yields,

$f_{U,V}(u,v)=\frac{{\displaystyle \frac{u^2}{{(v+1)}^2}}}{u}=\frac{u}{{(v+1)}^2} , 0<u v<1+v, 0<u<1+v$

Step 3: Explanation (part b)

X and Y are identically independent and identically distributed on $\left[0,1\right]$.

Such that 

$U=X V=X/Y$

The joint probability distribution function of random vector (X,Y),

$f(x,y)=f_X\left(x\right)f_Y\left(y\right)=1$

Now, apply transformation,

$g:{(0,1)}^2\to R^2$

Such that

$g(x,y)=(u,v)= \left(x, \frac{x}{y}\right)$

By using theorem, the density function of random vector $(U,V)=g(X,Y)$ as

$f_{U,V}(u,v)= f_{X,Y}(x,y)·{\left|det(\nabla g(x,y\left)\right)\right|}^{-1}$

Calculate,

$\nabla g(x,y)=\left[\begin{array}{cc}1& 0\\ \frac{1}{y}& -\frac{x}{y^2}\end{array}\right]$

$$\begin{gathered} \left|det(\nabla g(x,y\left)\right)\right|=\frac{x}{y^2} \\ {f}_{U,V}(u,v)=f_{X,Y}(x,y)·\frac{y^2}{x}=\frac{y^2}{x} \end{gathered}$$

Now, write x, y in terms of u and v and substitute it in the last equality,

But we have,

$x=u y=\frac{u}{v}$

That yields,

$f_{U,V}(u,v)=\frac{{\displaystyle \frac{u^2}{v^2}}}{u}=\frac{u}{v^2} , v\ge 1, 0<u<v$.

Step 4: Explanation (part c)

X and Y are identically independent and identically distributed on $\left[0,1\right]$.

Such that 

$U=X+Y and V=X/(X+Y)$

The joint probability distribution function of random vector (X, Y),

$f(x,y)=f_X\left(x\right)f_Y\left(y\right)=1$

Now, apply the transformation,

$g:{(0,1)}^2\to R^2$

Such that

$g(x,y)=(u,v)= \left(x+y, \frac{x}{x+y}\right)$

By using theorem, the density function of random vector $(U,V)=g(X,Y)$ as

$f_{U,V}(u,v)= f_{X,Y}(x,y)·{\left|det(\nabla g(x,y\left)\right)\right|}^{-1}$

Calculate,

$\nabla g(x,y)=\left[\begin{array}{cc}1& 1\\ \frac{1}{{\left(x+y\right)}^2}& -\frac{x}{{(x+y)}^2}\end{array}\right]$

$$\begin{gathered} \left|det(\nabla g(x,y\left)\right)\right|=\frac{x}{{(x+y)}^2}+\frac{y}{{(x+y)}^2}=\frac{1}{x+y} \\ \text{Thus,} \\ {f}_{U,V}(u,v)=f_{X,Y}(x,y)·(x+y) \\ =x+y \\ =u, 0<u<1 \end{gathered}$$