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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 <t & \\ \min(X,Y) <z\\ \end{cases} \iff \begin{cases} X<t+Y & \\ \{ \min(X,Y) <z, Y \leq Z \} \cup \{ \min(X,Y) <z, Y \geq Z \} \\ \end{cases} \iff \begin{cases} \{ X< Y+t ,Y \leq Z \} \cup \\ \{ X<z, Y \geq z \} \\ \end{cases} $

For $t>0$ and $z>0$ $ \begin{align*} F_W(t,z)&= P(T<t,Z \leq z) \\ &= \int \int \mathbb{1}_{(X-Y) <t , \min(X,Y) <z } dP \\ &= \int_{0}^{z} \int_{0}^{y+t} f_X(x) dx f_Y(y) dy + \int_{z}^{ + \infty} \int_{0}^{z} f_X(x) dx ) f_Y(y) dy \\ \end{align*} $

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*} $

zestiria
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    "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
  • https://math.stackexchange.com/q/2240822/321264 – StubbornAtom Jan 10 '22 at 08:11

1 Answers1

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You only need to prove the PDF is separable.

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 $T\sim\operatorname{Laplace}(0,\,1/\lambda),\,Z\sim\operatorname{Exp}(2\lambda)$.

J.G.
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