GIVEN
Let $A$,$B$ and $C$ be independent poisson processes with arrival rates $\lambda_1,\lambda_2$ and $ \lambda_3$ respectively.
Let $X,Y$ and $Z$ are times until the first arrival in processes A,B and C respectively.
$\Rightarrow X,Y $ and $Z$ are independent exponentials with parameters $\lambda_1,\lambda_2$ and $ \lambda_3$ respectively.
Also,
$ \min(Y,Z)$ is the time until the first arrival in merged process BC
and
$ \min(X,Y,Z)$ is the time until the first arrival in merged process ABC
$\Rightarrow \min(Y,Z)$ and $ \min(X,Y,Z)$ are exponential random variables with parameters $(\lambda_2+\lambda_3)$ and $(\lambda_1+\lambda_2+\lambda_3)$ respectively
event $X<Y<Z$ occurs when arrivals in merged process ABC follow the following order
- the very first arrival comes from process A.
- first arrival from process C, comes after an arrival from process B.
event $ \min(X,Y,Z)>t $ occurs when there are no arrivals in the merged process ABC until time $t$
two ways to prove $P(X<Y<Z \mid \min(X,Y,Z)>t) = P(X<Y<Z )----(1)$ are as follows:
- using the memorylessness property of exponential r.v's.
- using the time independence property of poisson processes.
we were able to use the time independence property of poisson processes(or the memorylessness property of exponential r.v's) to prove ---(1) because when we condition on $ \min(X,Y,Z)>t $, we are essentially told some information about the past. But the event $X<Y<Z$ occurs in the future.
TO PROVE
I have been asked to prove $P( \min(X,Y,Z)>t \mid X<Y<Z ) = P( \min(X,Y,Z)>t )----(2)$
without converting the LHS of (2) to LHS of (1) .
any hints to avoid the following computation (that involves integrals)?
WHAT I HAVE TRIED
$P( \min(X,Y,Z)>t \mid X<Y<Z ) = P( X>t \mid X<Y<Z ) = \int_t^{\infty}f_{X \mid X<Y<Z}(x)dx$
but I have been unable to simplify the pdf $f_{X \mid X<Y<Z}(x)$
ADDITIONAL QUESTION
how to simplify the pdf $f_{X \mid X<Y<Z}(x)$?
ps: i know the following formulas
and