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:

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

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

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

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    $\begingroup$ 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

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.

  • $\begingroup$ 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



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.


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