Repeat Problem $6.56$ when X and Y are independent exponential random variables, each with parameter $\lambda =1$.

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

X and Y are independent exponential random variable.

With parameter, $\lambda =1$

Such that

$U=X+YandV=X/Y$

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

$f(x,y)=f_X\left(x\right)f_Y\left(y\right)={\lambda }^2{e}^{-\lambda (x+y)}={\left(1\right)}^2\times e^{-1\times (x+y)}=e^{-\left(x+y\right)}$

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,

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

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

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

But we have

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

That yields,

localid="1648186727302" $f_{U,V}(u,v)=e^{-u}·\frac{{\displaystyle \frac{u^2}{{(v+1)}^2}}}{u}=e^{-u}\frac{u}{{(v+1)}^2},0<uv<1+v,0<u<1+v$.

X and Y are independent exponential random variable.

With parameter, $\lambda =1$

Such that

$U=XandV=X/Y$

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

$f(x,y)=f_X\left(x\right)f_Y\left(y\right)={\lambda }^2{e}^{-\lambda (x+y)}={\left(1\right)}^2\times e^{-1\times (x+y)}=e^{-\left(x+y\right)}$

Apply the 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,

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

Then calculate,

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

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

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

But we have,

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

That yields,

localid="1648186831071" $f_{U,V}(u,v)=e^{-\left(u+\frac{u}{v}\right)}\frac{{\displaystyle \frac{u^2}{v^2}}}{u}=e^{-\left(u+\frac{u}{v}\right)}\frac{u}{v^2},v\ge 1,0<u<v$

X and Y are independent exponential random variable.

With parameter, $\lambda =1$

Such that

$U=X+YandV=X/(X+Y)$

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

$f(x,y)=f_X\left(x\right)f_Y\left(y\right)={\lambda }^2{e}^{-\lambda (x+y)}={\left(1\right)}^2\times e^{-1\times (x+y)}=e^{-\left(x+y\right)}$

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,

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

Then calculate,

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

localid="1648186869853" $$\begin{gathered} \left|det(\nabla g(x,y\left)\right)\right|=\frac{x}{{(x+y)}^2}+\frac{y}{{(x+y)}^2}=\frac{1}{x+y} \\ {f}_{U,V}(u,v)=f_{X,Y}(x,y)·(x+y) \\ =e^{-(x+y)}(x+y) \\ =u{e}^{-u},0<u<1 \end{gathered}$$