Exponential Waiting Times for Bus Arrivals I thought I understood the Memoryless property for Exponential Distributions but I am unable to get the intuition behind the answer to this question:
You are waiting for a bus at a bus station. The buses arrive at the station according to a Poisson process with an average arrival time of 10 mins. If the buses have been running for a long time and you arrive at the bus station at a random time, what is your expected waiting time. On average, how many minutes ago did the last bus leave?
Clearly expected waiting time is 10 mins. However, I dont get this part: "On average, how many minutes ago did the last bus leave". The explanation in the book is that if you look back in time, the memoryless property implies on average the last bus arrived 10 mins ago as well. I'm basically struggling to see how the memoryless property implies this above statement. 
Thanks
 A: I have to say at the start that bus arrivals do not typically follow
an exponential distribution. So it is really hard to get out
of your mind how $actual$ buses work, if someone says interarrival
times are governed by an exponential process.
Maybe it is easier to think about something that really is
exponentially distributed. Suppose you have a very weak radioactive source  and you are capturing  particles it emits
in a counter. Suppose that the detection rate is one per 10
seconds. If you start keeping time at one particular click of
the counter, then the average wait for the next click is 10 seconds.
However, the no-memory property says if we start keeping time
at some arbitrary point in time (click or not), the average
wait until the next click is also 10 seconds. The decaying
particles are not 'keeping track' of each other, and they don't
'know' when you start counting. 
Now suppose you have a paper tape on which clicks are recorded
along a time line. The tape will look pretty much the same whether you
read forwards or backwards: random marks spaced sometimes near 
together, sometimes relatively far apart, but $on\; average$
10 seconds apart. Maybe it is possible to say this is due
to the no-memory property, but in my experience the usual
terminology for this is 'time-reversibility'. 
Both no-memory
and time-reversibility are fundamental properties of exponential
processes, so I suppose it is possible to take a point of view
(for exponential processes) that your statement in bold type is true. But I'm not sure there is a lot of
intuitive value in trying to make this connection between memorylessness and time reversibility when you're just starting
to think about the curious properties of exponential models.
As another example on no-memory, suppose a computer unit in a satellite 
survives an exponentially distributed length of time with
mean lifetime 10 years. Radiation hits are what cause 
such computer units to die. If 8 years have already gone by,
you might think the computer unit is nearing the end of its
life. But if the lifetime really is exponentially distributed,
and it is still alive at 8 years, the expected time of death
from a random radiation hit is still 10 years away. For such
devices, we say "Used is as good as new." This is an appropriate
model for devices that die only because of random radiation
hits. (Of course if you have a census of dead satellite computers
of this type along with their 'death' dates,
you could check back to their 'birth' dates and see that they
were, on average, about 10 years before the death dates, but
that is not much of a profound statement.)
For humans in a certain population we might say that their
average lifetime at birth is 70 years. If such a person is
now 60 years old, it would not be reasonable to say that
he or she has another expected 70 years of life. People
do sometimes die of random accidents, but they also die
by 'wearing out' with age. 
Most things we are familiar with
die from a combination of random accidents and gradual wearing
out: automobiles, light bulbs, pets, T-shirts, and so on.
Other events, like elections, bus arrivals, credit card bills,
and so on tend to happen at rather even intervals--sometimes
without much of a random component.
The reason intuition comes so hard when thinking about
exponentially distributed events is that there are relatively few
events in real life that happen according to an exponential
model. 
In science things get modeled according to exponential distributions
for two reasons: (a) Some things really are exponentially
distributed--at least approximately. Service times at banks,
lives of transistors, radioactive decay, and so on. (b) Because
the no-memory rule makes it unnecessary to take past history
into account, exponential models are mathematically very easy
to handle; that makes it tempting to use exponential models
sometimes when they don't really apply very well. 
A: Let $\{N(t):t\geqslant0\}$ be the arrival process of the busses, with arrival times $\{S_n\}$. What we're concerned with here is the asymptotic distribution of the forward recurrence time, i.e.
$$A(x):=\lim_{t\to\infty} \mathbb P(t-S_{N(t)}\leqslant x).  $$
For intuition, if the current time is $t$, then the time that the last bus arrived is $S_{N(t)}$, since $N(t)$ is the number of arrivals in $(0,t]$. It follows from the key renewal theorem that in general,
$$\lim_{t\to\infty} \mathbb P(t-S_{N(t)}\leqslant x) = \frac1\mu\int_0^x (1-F(y))\ \mathsf d y, $$
where $\mu=\mathbb E[S_n-S_{n-1}]$ is the mean interarrival time and $F$ is the distribution of the arrival times. (Note that this doesn't depend on the interrarival times being exponentially distributed, as they are in a Poisson process.) So for $x>0$ we have
$$A(x)=\frac1{1/\lambda}\int_0^x e^{-\lambda y}\ \mathsf dy = 1-e^{-\lambda x}, $$
and so we find that indeed the time since the last arrival follows $\operatorname{Exp}(\lambda)$ distribution as well. In particular, if $\lambda=\frac1{10}$, then $\mathbb E[A(x)] = 10$.
For an example where this isn't the case, suppose the interarrival times have $U(0,20)$ distribution. Then $\mu = 10$, but $$A(x) = \frac1{10}\int_0^x\left(1-\frac y{20}\right)\ \mathsf d y = \frac1{10}x - \frac1{400}x^2, $$
so
$$\mathbb E[A(x)] = \int_0^{20} (1-A(x))\mathsf dx = \frac{20}3<10. $$
A: The way I understand it is the following:
You are standing on the time point $t_1$, and let us denote the last time that a bus came to be $t_0$. (If $t_1=t_0$, it will be a trivial case, i.e. a bus is coming right now! So we can ignore that in the following discussion and assume there's no bus coming for $t\in (t_0, t_1]$)
Let \Delta t be the time since the left of the last bus, i.e. $\Delta t = t_1 - t_0$. We want to know what is the probability distribution of it. This is equivalent to asking "starting from $t_0$, what is the probability that no bus shows up up to $t=t_1$" or "starting from $t_0$,the first arrival time of a bus would be at least $\Delta t$, i.e. $\tau \geq \Delta t$". So this $\Delta t$ is exponentially distributed with the same parameter.
