# $\min(X,Y)$ and $X-Y$ with $X,Y$ iid exponential distributions

Let :

• $$X$$ and $$Y$$ be two independent exponential distributions of parameter $$\lambda$$
• $$T=X-Y$$
• $$Z =\min(X,Y)$$

We want to show that $$T$$ and $$Z$$ are independant, without any memoryless property.

My attempt :

\begin{align*} f_{T}(s) & = \int_{ \mathbb{ R} }^ {} f_X(s-u) f_{-Y}( u) du \\ &= \int_{ \mathbb{ R} }^ {} e^{ - \lambda (s-u) } e^{ - \lambda u} \mathbb{1}_{ s-u \geq 0} \mathbb{1}_{ u \leq 0} \\ & = e^{ - \lambda s } \int_{ \mathbb{ R} }^ {} e^{ 2 \lambda u} \mathbb{1}_{ u \leq (0 \wedge s)} \\ & = \frac{1} {2} e^{ - \lambda (2 (0 \wedge s) -s )} \\ &= \frac{1} {2} e^{ - \lambda |s|} \\ \end{align*}

$$f_Z(z)= 2 e^{- 2 \lambda z}$$

$$\begin{cases} X-Y

For $$t>0$$ and $$z>0$$ \begin{align*} F_W(t,z)&= P(T

For $$t>0$$ and $$z>0$$, $$F_{Z,T}(z,t) =1 - \frac{1}{2} e^{-t}(1- e^{-2z}) - e^{ -2 z}$$

For $$z >0$$ and $$t<0$$, $$Y \geq X-t$$ and $$X \leq z$$, therefore,

\begin{align*} F_W(t,z) &= \int_{0}^{z} \lambda e^{ - \lambda x} ( \int_{x-t}^{ + \infty} \lambda e^{ - \lambda y } dy ) dx \\ &=\int_{0}^{z} \lambda e^{ - \lambda x} e^{ - \lambda (x-t)} \\ &=e^{ \lambda t} \frac{1}{2} (1- e^{ - 2 \lambda z })\\ \end{align*}

• "without any memoryless property" - why would you want to do this the hard, unenlightening way, when you can do it the quick, intuitive way? – Misha Lavrov Dec 23 '20 at 18:36
• Could you write down the intuitive way ? I have corrected my answer, that seems correct now. – zestiria Dec 23 '20 at 18:44
• The intuitive way is to use the memoryless property: $Z$ is the waiting time until the first event, $|T|$ is the waiting time until the second event (which is independent from the first waiting time by memorylessness), and the sign of $T$ tells us which of $X$ or $Y$ is larger, which is independent from both of the above by symmetry. – Misha Lavrov Dec 23 '20 at 18:48

If $$x\ge y$$, $$z=x$$ and $$z_x=1,\,z_y=0$$. In terms of Iverson brackets$$z_x=[x\ge y],\,z_y=1-[x\ge y],$$ so$$\frac{dtdz}{dxdy}=\left|\begin{array}{cc} 1 & -1\\ \left[x\ge y\right] & 1-\left[x\ge y\right] \end{array}\right|=1.$$Since $$x+y=|t|+2z$$, the infinitesimal probability is$$\lambda^2e^{-\lambda(x+y)}dxdy=\tfrac12\lambda e^{-\lambda|t|}dt\cdot2\lambda e^{-2\lambda z}dz.$$This also obtains the distributions of $$T\sim\operatorname{Laplace}(0,\,1/\lambda),\,Z\sim\operatorname{Exp}(2\lambda)$$.