If X, Y, and Z are independent random variables having identical density functions $f\left(x\right)=e^{-x},0<x<\infty$ derive the joint distribution of $U=X+Y,V=X+Z,W=Y+Z$.

The probability density function is defined as the integral of the variable density density over a certain range. It is represented by the letter f(x).

Independent random variables X,Y and Z.

With identical density functions

$f\left(x\right)=e^{-x},0<x<\infty$

Where, $U=X+Y,V=X+ZandW=Y+Z$

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

$f_{X,Y,Z}(x,y,z)=f_X\left(x\right)f_Y\left(y\right)f_Z\left(z\right)=e^{-(x+y+z)}$

Apply the transformation,

$g:{(0,\infty )}^3\to R^3$

Such that

$g(x,y,z)=(u,v,w)=(x+y,x+z,y+z)$

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

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

Then calculate

$\nabla g(x,y,z)=\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 0& 1& 1\end{array}\right]$

That yields

$\left|det(\nabla g(x,y,z\left)\right)\right|=2$

$f_{U,V,Z}(u,v,w)=\frac{1}{2}{e}^{-\left(x+y+z\right)}$

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

But we have $u+v+w=2(x+y+z)$

That yields

$f_{U,V,W}(u,v,w)=\frac{1}{2}{e}^{-\frac{u+v+w}{2}}$.