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
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.