Let $X_1,X_2,X_3$ be independent with a common exponential distribution. I am trying to find the density of $(X_2-X_1 ,X_3-X_1 )$.

Assuming $X_1,X_2,X_3\sim Exponential(λ)$

If $x > 0$ $$f_{X_2-X_1}(x) =f_{X_3-X_1}(x) = {\dfrac {\lambda } {2}e^{-\lambda x}} $$ If $x < 0$ $$f_{X_2-X_1}(x) =f_{X_3-X_1}(x) = {\dfrac {\lambda } {2}e^{\lambda x}} $$

For simplicity let $X_2-X_1 =X$ and $X_3-X_1= Y$. The Joint density can be calculated as $$f_{X,Y}( x,y) = f_{Y\mid X}( y\mid x) f( x) = f_{X\mid Y}( x\mid y) f(y)$$

I am unsure here how to Calculate the conditional pdf. Any help would be much appreciated.


You start with the joint PDF, as you have done $$ f : \mathbb{R}_{\geq 0}^3 \rightarrow \mathbb{R} : \mathbf{x} \rightarrow \alpha^3 e^{-\alpha(x_1+x_2+x_3)} $$

Then you pick an appropriate (invertible!) transform $T$, also as you have done, and find its inverse $$ \begin{eqnarray} T(\mathbf{x}) &=& \begin{pmatrix} -1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\;\mathbf{x} &\text{,}& T^{-1}(\mathbf{x}) &=& \begin{pmatrix} -1 & 0 & 1 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{pmatrix}\;\mathbf{x} \text{.} \end{eqnarray} $$

The transformed PDF $g$ (i.e. $g$ such that $g(T(\mathbf{x})) = f(\mathbf{x})$) is thus $$ g(\mathbf{y}) = f(T^{-1}(\mathbf{y}))\frac{1}{\det(T)} = \alpha^3 e^{-\alpha(3y_3 + y_2 - 2y_1)} $$

Now we have to figure out $\text{dom}\, g \subset \mathbb{R}^3$ such that $T^{-1}(\text{dom}\, g) = \text{dom}\, f$. Requiring $T^{-1}(\mathbf{y}) \in \mathbb{R}_{\geq 0}^3$ for all $y \in \text{dom}\, g$ produces the conditions $$ \begin{eqnarray} y_3 - y_1 &\geq& 0 \text{,}\\ y_3 &\geq& 0 \text{,}\\ y_3 + y_2 - y_1 &\geq& 0 \text{.} \end{eqnarray} $$ This yields $y_3 \geq \max\{0,y_1, y_1 - y_2\}$ as necessary and sufficient, i.e. you get $$ \text{dom}\, g = \left\{\mathbf{y} \,:\, \mathbf{y} \in \mathbb{R}^3, y_3 \geq \max\{0,y_1, y_1 - y_2\}\right\} \text{.} $$

To find the PDF $h$ of $(X_2-X_1,X_3-X_1)=(Y_1,Y_2)$ you must now integrate $g$ over $y_3$, but without leaving the domain of $g$, i.e. compute $$ \begin{eqnarray} h(y_1,y_2) &=& \int_{c}^\infty \alpha^3 e^{-\alpha(3y_3 + y_2 - 2y_1)} dy_3 \:\text{where}\: c = \max\{0,y_1, y_1 - y_2\} \\ &=& \frac{\alpha^2}{3} e^{-\alpha(y_2 - 2y_1)} \int_{c}^\infty 3\alpha e^{-3\alpha y_3} dy_3 \\ &=& \frac{\alpha^2}{3} e^{-\alpha(y_2 - 2y_1)} e^{-3\alpha c} \\ &=& \frac{\alpha^2}{3} e^{-\alpha(3c - 2y_1 + y_2)} \end{eqnarray} $$

Note that if $y_1,y_2\geq 0$ then $c=y_1$, if $y_1 \leq 0,y_1 \leq y_2$ then $c=0$ and if $y_2 \leq 0,y_2 \leq y_1$ then $c=y_1-y_2$. This gives $$ h(y_1,y_2) = \frac{\alpha^2}{3}\begin{cases} e^{-\alpha(y_1 + y_2)} & \text{if } y_1,y_2 \geq 0 \\ e^{-\alpha(2|y_1| + y_2)} & \text{if } y_1 \leq 0, y_1 \leq y_2 \\ e^{-\alpha(y_1 + 2|y_2|)} & \text{if } y_2 \leq 0, y_2 \leq y_1 \end{cases} $$

  • $\begingroup$ Firstly sorry for a late reply, sleeping was way overdue.This is an excellent answer and i completely agree that i had made a big boo-boo with determining the new domain.I do have a suggestion though i think in your final answer one does not need those absolute values brackets around $y_1$ and $y_2$.These should just be negative as per Feller's provided answer.I suppose some how he is not considering to adapt the density $h(y_1,y_2)$ so it's distribution function would always converge.I am going to take down my post instead of improving mine to match yours.Thank you for this. Welcome to MSE:-) $\endgroup$ – Comic Book Guy Oct 9 '12 at 19:19
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    $\begingroup$ @Hardy I actually just wrote $-2y_1$ respectively $-2y_2$ initially, and later changed it to the absolute value. Mostly because that how the laplace distribution is usually written, and I wanted to make the connection more obvious. Though I can see your point too, guess the absolute value could make things look slightly more complex than they are. Anyway, glad you like the answer, I certainly had fun solving this! $\endgroup$ – fgp Oct 9 '12 at 19:35

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