Jtdylktuy

So this is the problem

Let $X_1$ and $X_2$ be $i.i.d.$ random variables with a common density function $chi_2^2$. Let $Y_1 = frac{1}{2}(X_1-X_2)$ and $Y_2 = X_2$. Show that $Y_1$ follows a Laplace distribution with density function given by
$$f(y) = frac{1}{2}e^{-|y|} $$
for $y in mathbb{R}.$

Now, here is my reasoning:

Since $X_1,X_2 sim chi_2^2$ (Chi-squared distribution) , then each of these random variables have their density function given by
$$f(x)= frac{1}{2} e^{-x/2}$$
for $x >0.$ Now,we have the transformations:

$$y_1 =g_1(x_1,x_2) = frac{1}{2}(x_1-x_2), y_2 =g_2 (x_1,x_2)= x_2$$
and their inverses
$$x_1 = g_1^{-1}(y_1,y_2) = 2y_1+y_2, x_2 = g_2^{-1} (y_1,y_2)= y_2. $$

Now we obtain the Jacobian of the transformation, which is
$$J =begin{vmatrix}
2 & 1 \ 0 & 1
end{vmatrix} = 2$$
Thus the joint density of $Y_1,Y_2$ is given by
$$f_{Y_1,Y_2}(y_1,y_2) = f_{X_1,X_2}(x_1,x_2) cdot |J|$$
and since $X_1$ and $X_2$ are independent we have that
begin{align*}
f_{Y_1,Y_2}(y_1,y_2) &= f_{X_1}(x_1) cdot f_{X_2}(x_2) cdot 2\
&= 2 cdot left(frac{e^{-(2y_1+y_2)/2}}{2} right)cdot left(frac{e^{-y_2/2}}{2} right)\
&= frac{1}{2}cdot e^{-(y_1+y_2)}.
end{align*}
Now, we are asked to find the distribution of $Y_1$, so
begin{align*}
f_{Y_1}(y_1) &= int_{0}^{infty} f_{Y_1,Y_2}(y_1,y_2) dy_{2}\
&= int_{0}^{infty} frac{1}{2}cdot e^{-(y_1+y_2)}dy_{2}\
&=frac{1}{2}cdot e^{-y_1} cdot int_{0}^{infty}e^{-y_2}dy_{2}
end{align*}
and since $ int_{0}^{infty}e^{-y_2}dy_{2} = 1$ we can conclude that
$$f_{Y_1}(y_1)= frac{1}{2}cdot e^{-y_1}.$$

Clearly this is not entirely correct since the $y$ exponent should be in absolute value. If anyone could point out my mistake and/or provide some intuition as to how the absolute value plays a role in this problem, I would appreciate it very much!

Do not forget to include the proper supports of the distributions. Use indicator functions for brevity.

Joint density of $(X_1,X_2)$ is

begin{align}
f_{X_1,X_2}(x_1,x_2)&=frac{1}{2}e^{-x_1/2}mathbf1_{x_1>0},frac{1}{2}e^{-x_2/2}mathbf1_{x_2>0}
\&=frac{1}{4}e^{-(x_1+x_2)/2}mathbf1_{x_1,x_2>0}
end{align}

For the transformation $(X_1,X_2,)to (Y_1,Y_2)$ such that $$Y_1=frac{1}{2}(X_1-X_2)quad,quad Y_2=X_2,,$$

we have $$x_1=2y_1+y_2quad,quad x_2=y_2$$

So, $$x_1,x_2>0implies 2y_1+y_2>0,y_2>0$$

In other words, $$y_2>max(0,-2y_1)quadtext{ where }quad y_1inmathbb R$$

Absolute value of the jacobian as you say is $$|J|=2$$

So the joint density of $(Y_1,Y_2)$ is

begin{align}
f_{Y_1,Y_2}(y_1,y_2)&=f_{X_1,X_2}left(2y_1+y_2,y_2right)|J|,mathbf1_{2y_1+y_2>0,y_2>0}
\&=frac{1}{2}e^{-(y_1+y_2)}mathbf1_{2y_1+y_2>0,y_2>0}
end{align}

Hence the marginal density of $Y_1$ for $y_1inmathbb R$ is given by

begin{align}
f_{Y_1}(y_1)&=int_{max(0,-2y_1)}^infty f_{Y_1,Y_2}(y_1,y_2),mathrm{d}y_2
\\&=frac{e^{-y_1}}{2}lim_{Atoinfty}left[-e^{-y_2}right]_{max(0,-2y_1)}^A
\\&=frac{e^{-(y_1+max(0,-2y_1))}}{2}
\\&=begin{cases}frac{1}{2}e^{-y_1}&,text{ if }y_1ge 0\\frac{1}{2}e^{y_1}&,text{ if }y_1<0end{cases}
end{align}