Suppose we have an infinite number of counters (numbered $0,1,2,\dots$) which fire (i.e. increment by $1$) independently according to a Poisson process of rate $1$. All counters are initially switched off. At time $0$ we turn on counter $0$. Each time counter $0$ fires, the next counter which is still off turns on (e.g. the first time counter $0$ fires, counter $1$ turns on, the second time counter $0$ fires, counter $2$ turns on, etc.). What is the expected number of fires over all counters in the interval $[1,2]$?
Here is my attempt:
Let $T_i$ denote the number of times counter $i$ fires in $[1,2]$, and let $T=T_0+T_1+T_2+\dots$. Then $$\mathbb{E}[T]=\sum_{i=0}^{\infty}\mathbb{E}[T_i]$$
Note $\mathbb{E}[T_0]=\lambda(2-1)=1$. $\mathbb{E}[T_1]$ is a little trickier, but we can compute it by conditioning on the first firing time $F_0$ of counter 0:
$$\mathbb{E}[T_1]=\int_0^{2}\mathbb{E}[T_1|F_0=t]p_{F_0}(t)dt=\int_0^{1}\mathbb{E}[T_1|F_0=t]p_{F_0}(t)dt+\int_1^{2}\mathbb{E}[T_1|F_0=t]p_{F_0}(t)dt=$$
$$\lambda(2-1)\cdot\mathbb{P}(F_0\leq 1)+\int_1^2\lambda(2-t)\cdot p_{F_0}(t)dt$$
Here $\mathbb{P}(F_0\leq t)=1-e^{-\lambda t}=1-e^{-t}\implies p_{F_0}(t)=\frac{d}{dt}\mathbb{P}(F_0\leq t)=e^{-t}$, so the above becomes
$$1-e^{-1}+\int_1^2(2-t)e^{-t}dt=1-e^{-1}+e^{-2}$$
Now we can proceed similarly for $\mathbb{E}[T_2]$, but it does seem to get a little messy. Namely, conditioning on the event $F_1=t$ seems trickier to do because now we need $p_{F_1}(t)=\int_0^tp_{F_0}(t_0)p_{F_0}(t-t_0)dt_0=\int_0^te^{-t}dt_0=te^{-t}$. If we compute this expectation and compare to the previous iteration, we can probably start to see some pattern emerge.
However, does anyone have a cleaner approach (without all this messy algebra)? Any ideas would be appreciated (even if it's just building on my attempt)! Thanks!
Best Answer
Addendum: this answer computes an expression for $E[T_i]$ that is quite complicated. However, it is possible to compute $E[T]$ directly without computing each $E[T_i]$; see Matthew H's answer.
Your work seems reasonable to me. In general, if $S_i$ is the time of the $i$th firing of counter zero (note that $S_1$ is your $F_0$, and $S_2$ is your $F_1$), you've essentially shown that
$$E[T_i] = E[T_i \mathbf{1}_{S_i \le 1}] + E[T_i \mathbf{1}_{1 < S_i < 2}] = P(S_i \le 1) + E[(2-S_i) \mathbf{1}_{1 < S_i < 2}].$$
Let $N_{[a,b]}$ denote the number of times counter zero fires in $[a,b]$. We know $$P(S_i \le 1) = 1 - P(N_{[0, 1]} \le i-1) = 1 - e^{-1} \sum_{j=0}^{i-1} \frac{1}{j!}.$$
For the second term, I think finding the density of $S_i$ is unavoidable, due to the necessary truncation onto the interval $[1, 2]$. Your induction idea would work, and would eventually lead you to $f_{S_i}(t) = \frac{1}{(i-1)!} t^{i-1} e^{-t}$, the density of the $\text{Gamma}(i,1)$ distribution.
In order to compute $E[(2-S_i) \mathbf{1}_{1 < S_i < 2}]$ it suffices to compute $E[\mathbf{1}_{1 < S_i < 2}]$ and $E[S_i \mathbf{1}_{1 < S_i < 2}]$. The first term is $$ E[\mathbf{1}_{1 < S_i < 2}] = P(1 < S_i < 2) = P(N_{[0,1]} \le i-1) - P(N_{[0, 2]} \le i-1) = e^{-1} \sum_{j=0}^{i-1} \frac{1}{j!} - e^{-2} \sum_{j=0}^{i-1} \frac{2^j}{j!}.$$
The second term can be rewritten into something resembling the first term. $$E[S_i \mathbf{1}_{1 < S_i < 2}] = \int_1^2 t \frac{1}{(i-1)!} t^{i-1} e^{-t} \, dt = i \int_1^2 \frac{1}{i!} t^i e^{-t} \, dt = i E[\mathbf{1}_{1 < S_{i+1} < 2}] = i \left(e^{-1} \sum_{j=0}^{i} \frac{1}{j!} - e^{-2} \sum_{j=0}^{i} \frac{2^j}{j!}\right).$$
Putting everything together, we have
\begin{align} E[T_i] &= 1 - e^{-1} \sum_{j=0}^{i-1} \frac{1}{j!} + 2\left(e^{-1} \sum_{j=0}^{i-1} \frac{1}{j!} - e^{-2} \sum_{j=0}^{i-1} \frac{2^j}{j!} \right) - i\left(e^{-1} \sum_{j=0}^{i} \frac{1}{j!} - e^{-2} \sum_{j=0}^{i} \frac{2^j}{j!}\right) \\ &= 1 - (i-1) e^{-1} \sum_{j=0}^{i-1} \frac{1}{j!} + (i-2) e^{-2} \sum_{j=0}^{i-1} \frac{2^j}{j!} + i\left(e^{-2} \frac{2^i}{i!} - e^{-1} \frac{1}{i!}\right) \end{align}
When $i=1$ we have $$E[T_1] = 1 - 0 + (-1) e^{-2} + (2e^{-2} - e^{-1}) = 1 - e^{-1} + e^{-2},$$ which corresponds with your result.