Expectation of exponential distribution (people waiting in line)

Question

Suppose there are people waiting in line for a ride. At $t=0$, the first person enters through the ride's ticket booth, and then people continue to enter at a rate of exactly one per minute (not random). However, the time that the ticket booth remains open is random with an exponential distribution with an average time of $10$ minutes. How many people are expected to enter before the ride closes?

My suspicion is that the desired expectation value is $11$ people. One person at $t=0$, then another for each minute. Since the average time open is $10$ minutes, then $10$ people (at times $> 0$) $+ 1$ person (at time $t=0$) $= 11$ people. However, I'm not sure if this is the right way to think about this problem. Does this being an exponential distribution complicate the problem?

Answer 1

Heuristic way: Your random variable is $Y=1+\lfloor X \rfloor$ where $X\sim \exp\left(\frac{1}{10}\right)$. Assuming that the fractional part of $X$ is $U(0,1)$, on average $Y$ is $1+X-\frac{1}{2}$. So the guess for the expected value is: $10\frac{1}{2}$.
Intro to probability way: Compute the pmf of $Y$, $$ P(Y=k+1)=P(k\le X <k+1)=...\ \ \ k\ge 0 $$ from which, you can easily get, the exact result: $\frac{1}{1-e^{-\frac{1}{10}}}\approx 10.508$. So the first guess was not so bad.

Answer 2

Let $X\sim\mathrm{Exp}(\lambda)$ be the time that the ticket booth remains open and $N = 1+\sum_{n=1}^\infty n\mathsf 1_{[n,n+1)}(X)$ be the number of people that enter before the ride closes. Then by monotone convergence we have \begin{align} \mathbb E[N] &= 1+ \mathbb E\left[\sum_{n=1}^\infty n\mathsf 1_{[n,n+1]}(X)\right]\\ &= 1 + \sum_{n=1}^\infty \mathbb E[n\mathsf 1_{[n,n+1)}(X)]\\ &=1 + \sum_{n=1}^\infty \mathbb n\cdot\mathbb P(X\in[n,n+1))\\ &=1 + \sum_{n=1}^\infty n\int_n^{n+1}\lambda e^{-\lambda t}\ \mathsf dt\\ &=1 + \sum_{n=1}^\infty n \left(e^{\lambda }-1\right) e^{\lambda (-(n+1))}\\ &=1 + \frac1{e^{\lambda}-1}. \end{align} If $\lambda=\frac1{10}$, then $$ \mathbb E[N] = 1 + \frac1{e^{\frac1{10}}-1}\approx 10.50833. $$