How to compute $P[T1 \le T2 \le t]$ for T1, T2 is independent random variable with exponential distribution in terms of cmf, pdf of T1 and T2?
Similarly for $P[T1 \le T2 \le T3.. \le t]$ ?
I tried this:
$P[T1 \le T2 \le t] = P [T2 \le t] P[T1 \le T2]$
$= P [T2 \le t] \int_{x=0}^\infty P[T1 \le x ]P[x = T2] dx$
$= F_{T2}(t) \int_{x=0}^\infty F_{T1}(x) f_{T2}(x)dx $
which does not seem to give me the right result.. I think I made mistake at the first expansion, but could not think of any other way.
Thanks
If the sequence $(T_k)$ is i.i.d., then, for each integer $n\geqslant1$ and each permutation $\pi$ on $\{1,2,\ldots,n\}$, by symmetry of the distribution of $(T_k)_{1\leqslant k\leqslant n}$, the event $$A_\pi=[T_{\pi(1)}\lt T_{\pi(2)}\lt\cdots\lt T_{\pi(n)}\lt t]$$ has the same probability. Furthermore, the union of these $n!$ events is $A=\bigcap\limits_{k=1}^n[T_k\lt t]$, hence, for every $\pi$, considering $T$ distributed like every $T_k$, one gets$$P(A_\pi)=\frac1{n!}P(A)=\frac1{n!}P(T\lt t)^n=\frac1{n!}F_T(t)^n.$$