# What distribution has the sum of two finite independent sequences of i.i.d. exponential random variables?

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.

• I think you mean $\sum_{i=\color{blue}{1}}^n$. But isn't $N_t$ trivially $S_n$?
– J.G.
Jan 18, 2021 at 20:36
• 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! Jan 18, 2021 at 21:45
• The $N_t$ comment was based on an equation you've since edited, but $\sum_\color{red}{i=i}^n$ doesn't make sense.
– J.G.
Jan 18, 2021 at 21:49
• That's $i=1$ of course. Jan 20, 2021 at 21:05