probability theory - Equality involving a sequence of independent exponentially distributed variables

I'm trying to prove the following statement: Let $\left( {{T_n}:n \geqslant 1} \right)$ be a sequence of independent, exponentially distributed random variables with ${T_n} \sim Exp\left( {{q_n}} \right)$, let $q: = \sum\limits_{n = 1}^\infty {{q_n} < \infty } $ and $T: = \mathop {\inf }\limits_{n \in \mathbb{N}} {T_n}$. Then there exists a random variable $N$ with positive integer values such that $\mathbb{P}\left( {T = {T_N}} \right) = 1$. If $N = n$ then ${T_n} < {T_j},\forall j \ne n$. Furthermore, $T$ and $N$ are independent, $T \sim Exp\left( q \right)$ and $\mathbb{P}\left( {N = n} \right) = \frac{{{q_n}}}{q}$.

I'm having problems with this part: Let $N = n$ if ${T_n} < {T_j}$ for all $j \ne n$ and not defined otherwise. Then for each $t \geqslant 0$ we have $\mathbb{P}\left( {N = n,T \geqslant t} \right) = \mathbb{P}\left( {{T_j} > {T_n} \geqslant t,\forall j \ne n} \right) = \int_t^\infty {\mathbb{P}\left( {{T_n} \geqslant s} \right)\mathbb{P}\left( {{T_j} > s,\forall j \ne n} \right)ds} $.

How do I get the last equality?

EDIT: $T \sim {\text{Exp}}\left( \lambda \right)$ if its PDF is as defined here. $\mathbb{E}\left[ T \right] = \frac{1}{\lambda }$.

EDIT 2: I've noticed a problem given later, it goes like this:

Let $T$ and $S$ be non-negative, independent, absolutely continuous random variables with PDFs ${f_T}$ and ${f_S}$, respectively. Then for each $t \geqslant 0$ we have

$\mathbb{P}\left( {T < t \leqslant T + S} \right) = \int_0^t {{f_T}\left( u \right)\left( {\int_{t - u}^\infty {{f_S}\left( v \right)dv} } \right)du} $. I guess I should be able to solve the initial problem if I knew how to show that.

Answer

Let $A=[N=n,T\geqslant t]=[\forall j\ne n,T_j\gt T_n\geqslant t]$. You write:

Then (...) $\mathbb{P}(A) = \int_t^\infty {\mathbb{P}\left( {{T_n} \geqslant s} \right)\mathbb{P}\left( {{T_j} > s,\forall j \ne n} \right)ds} $.
How do I get the last equality?

This equality does not hold. Actually, $P(T_j\geqslant s)=\mathrm e^{-q_js}$ for every $s\geqslant0$ hence, by independence,
$$
P(A\mid T_n)=\mathbf 1_{T_n\geqslant t}\cdot\prod_{j\ne n}\mathrm e^{-q_jT_n}=\mathrm e^{-(q-q_n)T_n}\cdot\mathbf 1_{T_n\geqslant t},
$$
Now, $P(A)=E(P(A\mid T_n))$ hence
$$
P(A)=E(\mathrm e^{-(q-q_n)T_n};T_n\geqslant t)=\int_t^\infty\mathrm e^{-(q-q_n)s}q_n\mathrm e^{-q_ns}\mathrm ds,
$$
that is,

$$
P(A)=q_n\int_t^\infty\mathrm e^{-qs}\mathrm ds=\frac{q_n}q\mathrm e^{-qt}.
$$
This identity shows the property you state at the beginning of your post.