0
$\begingroup$

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*} $

$\endgroup$
3
  • 1
    $\begingroup$ "without any memoryless property" - why would you want to do this the hard, unenlightening way, when you can do it the quick, intuitive way? $\endgroup$ – Misha Lavrov Dec 23 '20 at 18:36
  • $\begingroup$ Could you write down the intuitive way ? I have corrected my answer, that seems correct now. $\endgroup$ – zestiria Dec 23 '20 at 18:44
  • $\begingroup$ 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. $\endgroup$ – Misha Lavrov Dec 23 '20 at 18:48
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.