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How does one read the $1_{[T_n,T_{n+1})} (t)$ in exploding random process?

I.e. in $X_t = \sum_{n=1}^{\infty} n 1_{[T_n,T_{n+1})} (t)$.

I read that $t$ should belong to $[T_n,T_{n+1})$ for it to take value $1$.

But since $T_n=S_1+...+S_n$, $S_i \sim Exp(\lambda_i)$.

Then I don't understand what it means for some $t$ to belong "between sums of random exponential variables".

mavavilj
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  • Are you saying that you do not understand what $1_{[T_n, T_{n+1})}(t)$ means for arbitrary random variables $T_n$ and $T_{n+1}$? Or is there something about the particular case $T_n = S_1 + \cdots +S_n$ that confuses you? – angryavian Nov 18 '19 at 19:39
  • @angryavian I don't understand where $t$ comes from and how one can know, whether it's "between sums of $S_i$". Or perhaps $t$ and $T_n$ correlate somehow? – mavavilj Nov 18 '19 at 19:40
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    For $\omega$ in the sample space $\Omega$, ${T_n(\omega)}_{n\ge 1}$ is a fixed non-decreasing sequence and

    $$ X_t(\omega) = \sum_{n=1}^{\infty} n 1_{[T_n(\omega),T_{n+1}(\omega))} (t) $$

     –  Nov 18 '19 at 19:41
  • Perhaps the $t$ serves as a index? Since elsewhere I see defs. $X_t=\sum_{n \geq 1} 1_{t \geq T_n}$. So perhaps the idea is to merely say that "as $t$ grows, then so will $T_n$". And then each $t$ is such that is "larger than some $T_n$". – mavavilj Nov 18 '19 at 19:53

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If it helps, think of $t$ as a fixed number for now (say, $t=4$). It is not a random quantity.

$1_{[T_n, T_{n+1})}(t)$ is a random variable that depends on the random variables $T_n$ and $T_{n+1}$. It equals $1$ when $T_n \le t < T_{n+1}$, and equals zero otherwise. So in the case $t=4$, this random variable equals $1$ when the event $T_n \le 4 < T_{n+1}$ holds, and equal zero otherwise.

Then the whole sum $\sum_{n\ge 1} n 1_{[T_n, T_{n+1})}(t)$ is a sum of random variables. Different choices of $t$ will give different random variables.

In your particular case, you can think of $T_n$ as the arrival time of the $n$th customer to your store (and $S_i$ as the time between the arrival times of the $(i-1)$st and $i$th customers). Then $1_{[T_n, T_{n+1})}(t)$ equals $1$ if the $n$th customer arrives before or at time $t$ and if the $(n+1)$st customer arrives after time $t$. All terms in the sum $\sum_{n \ge 1} n 1_{[T_n, T_{n +1})}(t)$ will equal zero except for one (since for a fixed $t$, there is only one $n$ such that the event $\{T_n \le t < T_{n+1}\}$ occurs). In particular, $X_t$ will be the number of customers before/at time $t$.

angryavian
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  • "number of the last customer before/at time $t$". What does this mean? – mavavilj Nov 18 '19 at 19:58
  • @mavavilj Edited for clarity. What I meant was "if the last customer before/at time $t$ was the $10$th one, then $X_t = 10$," but it's equivalent to "the number of customers before/at time $t$." – angryavian Nov 18 '19 at 20:02
  • Couldn't it be something else than $10$? Since the sum involves $n$. Is the $n$ some kind of weight? – mavavilj Nov 18 '19 at 20:03