0
$\begingroup$

A random variable $(T_1, T_2)$ is called two-dimensional exponentially distributed with parameters $(\lambda_1, \lambda_2, \lambda_3)$, if the distribution function has the form

$$F(t_1,t_2) = \left\{\begin{matrix} 1-e^{-(\lambda_1+\lambda_3)t_1}-e^{-(\lambda_2+\lambda_3)t_2}+e^{-\lambda_1t_1-\lambda_2t_2-\lambda_3 \text{max}(t_1,t_2)}, \text{ if } t_1>0, t_2>0\\ 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ else } \end{matrix}\right.$$

Show that: $T_1$ and $T_2$ are independent and exponentially distributed, if and only if $\lambda_3 = 0$

This is from exam of last year and I like to know how to do this correct. From my other question I'm very happy I almost do it correct and now know better how it work: Determine the marginal distributions of $(T_1, T_2)$

So we already know $$\lim_{t_1 \rightarrow \infty}f(t_1,t_2) = 1-e^{-\infty}-e^{-(\lambda_2+\lambda_3)t_2}+e^{-\infty} = 1+0-e^{-(\lambda_2+\lambda_3)t_2}+0=1-e^{-(\lambda_2+\lambda_3)t_2}$$

$$\lim_{t_2 \rightarrow \infty}f(t_1,t_2) = 1-e^{-(\lambda_1+\lambda_3)t_1}-e^{-\infty}+e^{-\infty} = 1-e^{-(\lambda_1+\lambda_3)t_1}-0+0 = 1-e^{-(\lambda_1+\lambda_3)t_1}$$

But I think this is not really useful for answer the question.. I try insert the given $\lambda_3=0$ and see maybe we can form it then:

$$F(t_1,t_2) = \left\{\begin{matrix} 1-e^{-\lambda_1t_1}-e^{-\lambda_2t_2}+e^{-\lambda_1t_1-\lambda_2t_2 }, \text{ if } t_1>0, t_2>0\\ 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ else } \end{matrix}\right.$$

$$\Leftrightarrow F(t_1,t_2) = \left\{\begin{matrix} 1-e^{-\lambda_1t_1}-e^{-\lambda_2t_2}+\frac{e^{-\lambda_1t_1}}{e^{\lambda_2t_2}}, \text{ if } t_1>0, t_2>0\\ 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ else } \end{matrix}\right.$$

$$\Leftrightarrow F(t_1,t_2) = \left\{\begin{matrix} 1-e^{\lambda_2t_2} \left(\frac{e^{-\lambda_1t_1}}{e^{\lambda_2t_2}}+1-e^{-\lambda_1t_1}\right), \text{ if } t_1>0, t_2>0\\ 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ else } \end{matrix}\right.$$

$\endgroup$
  • 1
    $\begingroup$ Maybe you misread the last exponent, when $\lambda_3 = 0$, the ugly $\max$ term vanished and you can factorize the rest with ease. $\endgroup$ – BGM Jan 14 '18 at 15:45
  • $\begingroup$ @BGM Thank you I editted it now! :) $\endgroup$ – roblind Jan 14 '18 at 16:44
4
$\begingroup$

We note that in general two random variables $X,Y$ with joint CDF $F_{X,Y}$ are independent if and only if there exist CDFs $G, H$ such that

$$F_{X,Y}(x,y) = G(x)H(y), \qquad \forall \, x,\,y \in \mathbf{R},$$

in which case $G$ is the CDF of $X$, and $H$ is the CDF of $Y$.


Suppose that such a factorisation exists for your example $F$ the CDF of $(T_1,T_2)$, i.e.

$$F(t_1,t_2) = G(t_1)H(t_2)$$

Then in particular (and using your previous question) we would have

\begin{align*} G(t_1) & = \lim_{t_2 \rightarrow \infty} G(t_1)H(t_2)\\ & = \lim_{t_2 \rightarrow \infty} F(t_1,t_2) \\ & = 1 - e^{-(\lambda_1 + \lambda_3)t_1} \\ H(t_2) & = \lim_{t_1 \rightarrow \infty} G(t_1)H(t_2)\\ & = \lim_{t_1 \rightarrow \infty} F(t_1,t_2) \\ & = 1 - e^{-(\lambda_2 + \lambda_3)t_2}. \end{align*}

But taking the product of $G,H$ we see

\begin{align*} G(t_1)H(t_2) & = \left( 1 - e^{-(\lambda_1 + \lambda_3)t_1} \right) \left( 1 - e^{-(\lambda_2 + \lambda_3)t_2} \right) \\ & = 1 - e^{-(\lambda_1 + \lambda_3)t_1} - e^{-(\lambda_2 + \lambda_3)t_2} + e^{-(\lambda_1t_1 + \lambda_2 t_2 - \lambda_3(t_1 + t_2))} \end{align*}

Case $\lambda_3 \neq 0$ In this case we see from the above that $$ G(t_1)H(t_2) \neq F(t_1,t_2),$$ but this contradicts the assumption that was made the $F = GH$. Therefore $T_1,T_2$ are not independent.

Case $\lambda_3 = 0$ In this case we see that $G(t_1)H(t_2) = F(t_1,t_2)$, so that the factorization holds. Further, since both $G,H$ are themselves CDFs (of Exponential distributions), we have that $T_1,T_2$ are independent.

$\endgroup$
  • $\begingroup$ Just a very little thing but I will say it anyway. Before your first "Case" you forgot a closing bracket in the exponent. $\endgroup$ – cnmesr Jan 14 '18 at 21:34
  • $\begingroup$ Corrected, thanks. $\endgroup$ – owen88 Jan 14 '18 at 22:08
4
$\begingroup$

$F_{T_1}(t_1)=\lim_{t_2\to\infty}F(t_1,t_2)=(1-e^{-(\lambda_1+\lambda_3)t_1})1_{(0,\infty)}(t_1)$.

$F_{T_2}(t_2)=\lim_{t_1\to\infty}F(t_1,t_2)=(1-e^{-(\lambda_2+\lambda_3)t_2})1_{(0,\infty)}(t_2)$.

Then $F_{T_1}(t_1)F_{T_2}(t_2)=[1-e^{-(\lambda_1+\lambda_3)t_1}-e^{-(\lambda_2+\lambda_3)t_2}+e^{-(\lambda_1+\lambda_3)t_1-(\lambda_2+\lambda_3)t_2}]1_{(0,\infty)}(t_1)1_{(0,\infty)}(t_2)$

$T_1$ and $T_2$ are independent iff $F_{T_1}(t_1)F_{T_2}(t_2)=F(t_1,t_2)$ for every pair $(t_1,t_2)$, or equivalently iff: $$\forall t_1,t_2>0[\lambda_3(t_1+t_2)=\lambda_3 \text{max}(t_1,t_2)]$$

For this it is necessary and sufficient that $\lambda_3=0$.

In that case $F_{T_i}(t)=(1-e^{-\lambda_1t})1_{(0,\infty)}(t)$ for $i=1,2$.

So then $T_1,T_2$ have exponential distribution with parameter $\lambda_1,\lambda_2$ respectively.

$\endgroup$

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.