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Supose that $X_1$ is a continuous and positive (real) random variable with exponential distribution, namely $$P(X_1>t)=e^{-\lambda t}\quad t>0$$ Now suppose that $X_2$ is another continuous and positive (real) random variable such that for every $s>0$ we have $$P(X_2>t\mid X_1=s)=P(X_1>t)=e^{-\lambda t}$$

Now, why can I conclude that $X_2$ has exponential distribution and moreover that $X_1$ and $X_2$ are independent?

The claim comes from the book "S. Ross - Stochastic Processes (second edition)" at page 64 when the author proves that the interrarival times in a Poisson process are i.i.d. random variables.

Thanks in advance

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Since $e^{-\lambda t}$ does not depend on $s$, we have for all $t$ $$ P(X_2 > t) = \int_0^\infty \Pr(X_2 > t \mid X_1 = s) \lambda e^{-\lambda s}\,ds = e^{-\lambda t}. $$ Thus $X_2$ has exponential distribution with parameter $\lambda$ and the identity $$ \Pr(X_2 > t \mid X_1 = s) = \Pr(X_1 > t) = \Pr(X_2 > t), \qquad t,s > 0 $$ shows that $X_1$ and $X_2$ are independent.

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  • $\begingroup$ Dubious: Are you actually able to prove that the identity implies the independence of $X_1$ and $X_2$? $\endgroup$ Commented Aug 26, 2014 at 22:54
  • $\begingroup$ @Did: I guess I am missing something, but cannot we say that $\Pr(X_2 > t \mid X_1) = \Pr(X_2 > t)$, hence $\Pr(X_1 > t_1, X_2 > t_2) = E[1_{X_1 > t_1} \Pr(X_2 > t_2)] = \Pr(X_1 > t_1)\Pr(X_2 > t_2)$? $\endgroup$ Commented Aug 27, 2014 at 8:01
  • $\begingroup$ Sure we can, but this argument relies on properties of conditional expectations in the full sense, which is why I asked @Dubious (not you) if they were able to complete the proof (with no reaction to this date, which I guess is a kind of answer to my question). $\endgroup$ Commented Aug 27, 2014 at 8:06
  • $\begingroup$ @Did: ok! I thought that you were dubious about my statement... $\endgroup$ Commented Aug 27, 2014 at 8:09
  • $\begingroup$ Of course not. Sorry that I was not more explicit about this. $\endgroup$ Commented Aug 27, 2014 at 8:10

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