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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!

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    $\begingroup$ When you plug in $f_{X_1}(x_1) = f_{X_1}(2y_1 + y_2)$, you forget to check the support of $X_1$ that requires $2y_1 + y_2 > 0$, which in turns give $y_2 > -2y_1$. Combine with $y_2 > 0$, it gives the joint support of $(Y_1, Y_2)$ to be $y_1 \in \mathbb{R}, y_2 > \max\{0, -2y_1\}$. Once you replace your integration lower limit by $\max\{0, -2y_1\}$ and you should get your desired result. $\endgroup$
    – BGM
    Nov 20, 2018 at 7:30

1 Answer 1

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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 $$x_1=2y_1+y_2\quad,\quad x_2=y_2$$

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

In other words, $$y_2>\max(0,-2y_1)\quad\text{ where }\quad y_1\in\mathbb R$$

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_2\right)|J| \\&=\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 $y_1\in\mathbb R$ 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_{A\to\infty}\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_1\ge 0\\\\\frac{1}{2}e^{y_1}&,\text{ if }y_1<0\end{cases} \end{align}

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  • $\begingroup$ Ahh I forgot to check the supports, now it makes sense. Thank you very much! $\endgroup$
    – Bergson
    Nov 20, 2018 at 21:39

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