system
Given the following post, respond as if you're commenting on Reddit:
human
How does the wait time paradox work?

I've read a lot of it on the internet and I found this. If a bus comes every 10 minutes (certain) then you'll on average wait 5 minutes. But if a bus comes on average every 10 minutes (uncertain) then you'll wait on average 10 minutes. How did they calculate the 10 minutes? I tried google scholar but couldnt really find something.
Really interested in this because I'm following a statistics class this semester.
gpt
The waiting time paradox is a good example to show why some naive thinking and calculation can be very wrong. (Also, there is a missing part to this since strictly speaking you do *not* know that the average wait time in the uncertain case is 10 minutes, unless you have additional assumptions.)

For these two problems, let *W* = our waiting time and let *M* = the time between successive bus arrivals. Then *W* and *M* are random variables, and we want to calculate the expectation *E*(*W*).

 1. ***Case 1:*** *M* is constant. In other words, a bus arrives every *M* minutes, on the dot, no matter what. To calculate *W*, we reason that we arrive at the bus stop at some random time between 0 and *M* minutes after the last bus arrived. So *W* is uniformly distributed on the interval [0, M]. Hence its expectation is *E*(*W*) = *M*/2. So if the buses arrive every 10 minutes guaranteed, then our average waiting time is 5 minutes.

 2. ***Case 2:*** *M* is not constant. In other words, *M* is, in principle some random variable supported on the interval [0, ∞). So some buses arrive very soon after the previous bus. Some buses arrive very later after the previous bus. All we know is the value *E*(*M*), i.e., the average time between successive buses. This doesn't really tell you much. If buses alternatingly arrived 5 minutes and 15 minutes after the previous bus, then *E*(*M*) = 10. Our wait time is clearly between 5 and 15 minutes. It is also clearly greater than 5 minutes since there is a positive probability the next bus arrives more than 5 minutes later than the previous one. So already we see that our expected waiting time is not the same as in Case 1. (In this example, *E*(*W*) = 6.25, which is slightly larger than 5.)

The problem in thinking that *E*(*W*) = *E*(*M*)/2 is that the formula *E*(*W*) = *M*/2 for Case 1 was derived under the assumption that *W* was uniform. If *M* is not constant, then we have no reason to believe that *W* is uniform. All we really know is that if the maximum value of *M* is *M**_max_*, then *W* has support on the interval [0, *M**_max_*]. The distribution need not be uniform, c.f., the example in Case 2 above.

Now the missing part... often when people refer to this paradox they give an unqualified statement of "if the bus arrivals are uncertain but 10 minutes on average, then the average waiting time is 10 minutes". That's wrong as stated. Period.

However, there is a specific distribution of *M* that people usually have in mind. Random discrete arrivals with a known mean are often modeled stochastically by a Poisson process. So the arrival time (*M*) between any two buses has an exponential distribution f(m) = Te^(-Tm) where *T* is the known mean arrival time, i.e., *E*(*M*) = *T*. (This assumption is a common approximation.) Since the exponential distribution is memoryless, it follows that any point in time is a point between two successive bus arrivals, the interval between which is again described by the same exponential distribution. So in this case, *E*(*W*) = *E*(*M*) = *T*.

In reality, the distribution of *M* is neither a point mass (as it would be if all the buses arrived every *M* minutes on the dot with *M* constant) nor exponential. For buses that tend to arrive on schedule most often, the distribution of *M* would have a very tall spike at the average *E*(*M*) and decay very quickly beyond that value. (For instance, if the average successive arrival time is 10 minutes, *M* might be approximately normally distributed with mean 10 and standard deviation 1.) In cases where the buses are almost always on schedule, the average wait time is very close to the average successive arrival time. That's why you don't usually feel like you're waiting more than half of the successive arrival time.