Let $(X_i)$ and $(Y_i)$ be independent sequences of i.i.d. exponential random variables with common parameter $\lambda > 0$. Further, let $S_n = \sum_{i=1}^n (X_i + Y_i)$ and $N_t = \sup_{S_n \leq t} n = \sup \{n\in\mathbb N \vert S_n \leq t \} $.

I want to compute $\mathrm E[N_t]$.

[Edit: Keeping the original questions for reference, see below for my attempt to a solution.]

Currently I'm still trying to make sense of this problem conceptually. Couple of questions:

  1. What difference does it make whether we have two processes $(X_i)$ and $(Y_i)$, instead of just one? How does $S_n$ differ from $\sum_{i=1}^n X_i$ and $X_i\sim$Exp$(\lambda/2)$? I don't see the difference since all the $X_i$'s and $Y_i$'s should be i.i.d.?
  2. $S_n$ should have some kind of gamma distribution, but not just Gamma$(n,\lambda/2)$, I suppose?
  3. Since the exponential describes (memoryless) waiting times and the Poisson counts appearences, is $N_t$ just a Poisson random variable with parameter $2\lambda$? (Of course this comes down to question 1.)
  4. In case 3. is correct, I think I can compute $\mathrm E[N_t]$ by the renewal function. Is there another neat way to do it in this case without this theorem?

Thanks in advance for every help!

I think I figured out how to solve the problem.

$T_i = X_i + Y_i$ has a Gamma$(2,\lambda)$ distribution. The sum of gamma distributions is itself a gamma distribution, so $S_n$ is distributed according to Gamma$(2n,\lambda)$. To compute $N_t$, I need to compute $\Pr(S_n \leq t, S_n + T_{n+1} > t)=\Pr(N_t = n)$, like discussed in this previous question: Supremum of sum of exponentially distributed random variables

According to this, $$\Pr(N_t = n) = \int_{x=0}^t f_{S_n}(x) G_{Z_{n+1}}(t-x) \, dx\, ,$$ where $G(x)=\Pr(T_n > x)$ is the renewal function of $Z_n$ and $f_{S_n}$ is the, Gamma$ (2n,\lambda)$, PDF of $S_n$.

However, can someone explain to me how exactly to arrive at that specific integral?

I haven't done the computation yet, but I should get that $N_t$ is Poisson$(\lambda t)$. From there I can work with the renewal equation and solve anotehr integral to arrive at the solution.

  • 1
    $\begingroup$ I think you mean $\sum_{i=\color{blue}{1}}^n$. But isn't $N_t$ trivially $S_n$? $\endgroup$
    – J.G.
    Jan 18, 2021 at 20:36
  • $\begingroup$ What I wrote was, indeed, wrong. My thought was that I have twice as many $X_i$, but of course they would come in pairs, so I guess that would mean that we double the expected waiting time to $2/\lambda$, hence $X_i$ should then be Exp($\lambda/2$). Also: Shoot, I switched terms in the definition of $N_t$, it's supposed to be the Supremum over all $n$ such that $S_n \leq t$. Thanks for pointing out my mistake! $\endgroup$
    – Hölderlin
    Jan 18, 2021 at 21:45
  • $\begingroup$ The $N_t$ comment was based on an equation you've since edited, but $\sum_\color{red}{i=i}^n$ doesn't make sense. $\endgroup$
    – J.G.
    Jan 18, 2021 at 21:49
  • $\begingroup$ That's $i=1$ of course. $\endgroup$
    – Hölderlin
    Jan 20, 2021 at 21:05


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