Cars, of random length $L$, arrive at a gate. The first car parks against the gate. The other arriving cars park behind at a distance uniformly distributed on $[0,1]$. Let $N(t)$ be the number of cars parked at a distance $t$ from the gate. Find: $$ \lim_{t\to \infty} E[N(t)]/t $$
I have not been given a distribution for $L$.

If there was no space in between cars I could just use the continuous renewal equation. I'm not sure how to incorporate the space in between cars here.

I think the expected value of the spaces between cars is $1/2(N(t)-1)$. Assuming I have found this correctly, could I just add this expected value to the expected value of the general continuous renewal process?


Denote the distance between two cars as $U$. The first car takes $L + U$ of space. The second car takes $L + U$ of space, etc.

Using the notation of the Wikipedia page: the holding times $S_i$ are i.i.d. and distributed as $L + U$. The number of cars at distance $t$ is $N(t)$. $N(\cdot)$ is a renewal process. By the elementary renewal theorem

\begin{equation} \lim_{t \to \infty} \frac{\mathbb{E}[N(t)]}{t} = \frac{1}{\mathbb{E}[S_1]} = \frac{1}{\mathbb{E}[L] + \mathbb{E}[U]}. \end{equation}

  • $\begingroup$ I continued: E(U) = 1/2; E(L) = 1/(integral 0 to infinity of (k f(k)dk)), where f(k) is the p.d.f. of L. Is there a way to simplify further? $\endgroup$ – ak87 Mar 9 '16 at 17:15
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    $\begingroup$ @ak87 Your $1/\ldots$ in the expression for $\mathbb{E}[L]$ is incorrect. Assuming $L$ is non-negative (which it should be) $\mathbb{E}[L] = \int_0^\infty x f_L(x) \, \mathrm{d}x$ with $f_L(\cdot)$ the probability density function of $L$ and $\mathbb{E}[U] = 1/2$. I would actually leave $\mathbb{E}[L]$ as it is in my answer, this is already very clear. $\endgroup$ – Ritz Mar 9 '16 at 20:08
  • $\begingroup$ Thanks Ritz for the answers, I did it as you suggested, I just don't know how to write in the pretty math notation yet :) $\endgroup$ – ak87 Mar 9 '16 at 21:31
  • $\begingroup$ Good! As long as you understand the answer now. If you are satisfied with the answer, you can accept it. $\endgroup$ – Ritz Mar 10 '16 at 7:58
  • $\begingroup$ i don't think I'm allowed to vote or accept answers yet.. when I vote up, it says I have to wait till i get some on-site cred :) $\endgroup$ – ak87 Mar 10 '16 at 8:48

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