Probability: Store opening time

Question

Smith has a small booth where he sells lottery tickets. Customers arrive according to a Poisson process of rate $\lambda$= 1 per minute. He will close the shop on the 1st occasion that $a$ minutes have elapsed since the last customer arrived.

let $X$ denote the first time when the gap between the arrival of two consecutive customers is less than a. Find $E(X)$

Attempt:

I think $X$ is just exponential distributed, but $X=0$ for $x>a$. So $E[X]=\int_0^a\lambda x e^{-\lambda x}dx$.

What do you guys think?

Answer 1

$X$ is not the length of the gap, $X$ is the time when the length of the gap is first of greater than $a$. Call it a "long-gap", while a lesser length is a "short-gap".

To clarify:  Suppose some customers have arrived, all with a short-gaps before each and its previous, and then another customer arrives after a long-gap.   $X$ is the time of arrival of that customer.

Thus $X$ is the expected time a long-gap plus the expect count of short-gaps before the first long-gap times the expected time of any short-gap (since each is iid, and expectation is linear).

$$\begin{align}\mathsf E(X) ~ = & ~ \mathsf E(\mathsf E(\Delta T_{N+1}\mid \Delta T_{N+1}\geq a) + \mathsf E(\sum_{k=1}^N\Delta T_k\mid\forall k\leq N:\Delta T_k<a)) \\ = & ~ \mathsf E(\Delta T_{\ast}\mid \Delta T_{\ast}\geq a) +\mathsf E(N)\cdot\mathsf E(\Delta T_\ast\mid\Delta T_\ast<a) \end{align}$$

Now the lengths of each gaps is iid exponentially distributed.

$$\Delta T_\ast \mathop{\sim}\limits^{iid} \mathcal {Exp}(1)$$

The count of short-gaps before the first long-gap is geometrically distributed, with success rate being the probability of a long-gap. $$N\sim\mathcal{Geo}_0(\mathsf P(\Delta T_\ast\geq a))$$

Can you complete?

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(NB: "count of fails before success" is the $k\in\{0,1,...\}$ version of geometric distribution.)

Answer 2

We're computing $E X$ for $X$ taken as the time Smith closes his store (this is because the question seemed to ask that - recent comment shows differently, but the proof is similar).

The interarrival times are indeed IID exponential with rate $1$. Denote them by $T_1,T_2,\dots$ ($T_1$ time the first customer arrives, $T_2$ time between first and second customer, etc.).

Let $M=\min\{n: T_n \ge a\}$. Then

1. $M$ is geometric with parameter $ \int_a^{\infty} e^{-t} dt = e^{-a}$.
2. $X = \sum_{i<M} T_i + a$.

Now compute $EX$. We do this by conditioning on $M$:

$$ E X = \sum_{j=1}^\infty E[ X |M=j] P(M=j)= a + \sum_{j=1}^{\infty} (j-1) E [T_1| M=j] P(M=j).$$

Where the last equality is due to the fact that conditioned on $\{M=j\}$, $T_1,\dots ,T_{j-1}$ are IID. Now

$$\{M=j\}=\{T_1 <a\}\cap \{T_2 < a\}\dots \cap \{T_{j-1}<a\}\cap \{T_j\ge a\}.$$

All but the first event in the intersection on the RHS but the first are independent of $T_1$. Therefore for $j>1$ (note that $j=1$ does not contribute in sum above):

$$ E [ T_1 | M=j] = E [ T_1 | T_1 <a] = \frac{ \int_0^a t e^{-t}dt}{ \int_0^a e^{-t} dt}=\frac{1-e^{-a}(1+a)} {1-e^{-a}}.$$

Putting this back into our formula, we have

$$\begin{align*} E X & = a+ \frac{1-e^{-a}(1+a)} {1-e^{-a}} E (M-1) \\ & = a + \frac{1-e^{-a}(1+a)} {1-e^{-a}} (e^a-1)\\ & = a + e^a-(1+a)=e^a -1. \end{align*}$$