# Average wait time arriving at subway randomly

If the subway comes every 10 minutes on average, what is the expected wait time if I arrive at the station randomly? Can someone help me mathematically understand this problem?

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I apologize for the late edit, but a key point that forgot to include is that the trains arrive based on exponential distribution. All answers have been extremely helpful in either of the two cases. Thank you very much. –  Ryan Jul 22 '11 at 14:17
Ryan: Sorry to be blunt but this is quite bad practice to modify your question in extremis (especially when this is to accept an answer literally 3 minutes later). Please avoid this. –  Did Jul 22 '11 at 15:02
Don't mind the bluntness at all. I reversed my changes for the purposes of others. Thanks for the feedback. –  Ryan Jul 22 '11 at 18:00

It depends on the shape of the distribution of train arrivals. To take two easy examples:

• If the train arrives exactly every 10 minutes then the expected waiting time is 5 minutes.
• If the train arrive with an exponential distribution of times between trains (a density of $0.1 e^{-t/10}$) then the expected waiting time is 10 minutes.

And there are many other possibilities.

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Thanks Henry. Could you point me in the right direction to learn about how to calculate the expected wait time in the exponential distribution example? Thanks! –  Ryan Jul 19 '11 at 20:55
Just apply the definition of expected value to the exponential distribution:en.wikipedia.org/wiki/Exponential_distribution –  gary Jul 19 '11 at 21:02
@gary: No (there is a little more to it than that). –  Did Jul 19 '11 at 22:08
@Henry: This is the right answer in these two specific cases but a formula is available for every interarrival (integrable) distribution. –  Did Jul 19 '11 at 22:30
what is the meaning of "the train arrive with an exponential distribution of times between trains"? –  Victor Jul 19 '11 at 22:46

There are two "random" factors interacting here, your time of arrival at the station, and the pattern of "gaps" between arrival times of successive trains.

We will assume two things (i) The arrival times between consecutive trains have an exponential distribution and (ii) Your arrival times at the station and the train arrival times are independent. Assumption (i) is quite unreasonable, but it looks as if you are expected to make it. Given assumption (i), assumption (ii) is quite natural.

I will go into some detail about the exponential distribution and its mean later. But it turns out that you can write down the answer immediately without doing any computation at all!

If the random variable $T$ has exponential distribution, then $T$ has a curious property called memorylessness. Semi-formally, this means that the probability that the next train will arrive within time $\le t$ from right now, given that you have been waiting already for $h$ minutes, does not depend on $h$!

This is not quite as weird as it sounds. Suppose you keep tossing a pair of dice every minute. You are waiting for a double $6$. No matter how long you have been waiting for a double $6$, the probability of getting a double $6$ on the next toss doesn't change. So if you have been waiting in vain for a half-hour, the mean additional waiting time until the next double $6$ is exactly the same as the mean waiting time if you just started tossing. The dice don't remember the past, they are memoryless. In situations such as the dice case, the waiting time has what is called a geometric distribution. The exponential distribution is a continuous analogue of the geometric distribution.

After all this, the answer! We are told the mean time between trains is $10$ minutes. So, by the memorylessness property, if you arrive at the station at a randomly chosen time, your mean waiting time will be $10$ minutes. (We are assuming, among other things, that trains leave instantly after arriving!)

The Exponential Distributions:

A random variable $T$ has exponential distribution if $T$ has density function of the shape $f(t)=e^{-\lambda t}$ for $t \ge 0$, and $f(t)=0$ for $t \lt 0$. Here $\lambda$ is any positive constant.

By the standard formula for expectation (mean), the expectation $E(T)$ of this random variable is given by $$E(T)=\int_0^\infty te^{-\lambda t}\;dt.$$ (The expectation is $\int_{-\infty}^\infty tf(t)\;dt$, but recall that in our case $f(t)=0$ if $t \lt 0$.)

The integral is not very hard to evaluate, it is a standard integration by parts. It turns out that $E(T)=\dfrac{1}{\lambda}$.

Modeling Comment: Real trains don't behave like that, they have schedules that they try to keep. And if you have been waiting for $30$ minutes for a train with mean gap $10$, there probably is a mess up the line, and you may be in for a very long wait. All too often, I have seen questions in the "exponential distribution" segment of a book in which one is clearly being invited to use the exponential distribution, but just as clearly there is good reason to think that an exponential distribution model will not provide a good fit.

However, a surprisingly large number of real "waiting time" phenomena are modeled very well by the exponential distribution.

Take a plutonium atom which has existed for a million years. Its probability of decaying in the next $30$ minutes seems to be exactly the same as the probability that a plutonium atom just created in a nuclear reactor will decay in the next $30$ minutes. Atoms of radioactive substances die, but they do not age.

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Phenomenal answer. Thank you. –  Ryan Jul 22 '11 at 14:09

This is standard renewal theory. If $T$ is the time between successive arrival times and if $T$ has finite mean, the length $\hat T$ of the maximal interval around the present instant without arrivals follows the so called size-biased distribution of $T$ and the present time is uniformly distributed in this interval.

In formulas, the time $W$ until the next arrival is distributed like $U\hat T$ with $U$ uniform on $(0,1)$ and, for every bounded measurable $\varphi$, $$E(\varphi(\hat T))=E(T\varphi(T))/E(T).$$ Hence the distribution of $W$ is characterized by the fact that, for every bounded measurable $\varphi$, $$E(\varphi(W))=E(\varphi(U\hat T))=E(T\varphi(UT))/E(T)=\int_0^1E(T\varphi(uT))\mathrm{d}u/E(T).$$ In particular, for every integrable distribution of $T$, $$E(W)=E(T^2)/(2E(T)),$$ and $W$ has density $$f_W(w)=P(T\ge w)/E(T).$$ Two special cases: if the distribution of $T$ is exponential, then $W$ is distributed like $T$ but this is the only case when this happens; if $T=t_0$ with full probability, the formula above shows that $W$ is uniform on $(0,t_0)$, as it should.

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