Poisson Process: Time until next arrival

Question

Question:

Suppose that busses arrive at a bus stop as a Poisson process with rate $\lambda$ starting from time $t=0$ (that is, the interarrival time between busses is exponentially distributed with parameter $\lambda$).

You arrive at the bus stop at a particular (given) time $t$. Let $D_t$ be the amount of time since the last bus has departed, and $A_t$ be the amount of time until the next bus arrives.

What is the distribution of $D_t$ and $A_t$?

Show that $\Bbb E[D_t+A_t]>\dfrac 1\lambda$.

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Attempt:

I think that we should have $A_t \sim \exp (\lambda)$, because the Poisson process has the memoryless property. So at any given time $t$, the next arrival disregards whatever happened before $t$ and arrives in $\exp(\lambda)$ time starting from $t$.

However, I feel that this is not quite right, or else the last part would be rather trivial:

$$\Bbb E[D_t + A_t] = \Bbb E[D_t] + \Bbb E[A_t] \geq \Bbb E[A_t] = \frac 1\lambda$$

As for $D_t$, I have no idea.

Any hints?

Answer 1

We know that $A_t$ is independent of $D_t$: this is simply the memoryless property for the exponential interarrival times of the Poisson process (ie buses).

Now, suppose that $P = (P_t)_{t\ge0}$ is a standard, rate 1 Poisson point process (which I'll call a $\text{PPP}$)---ie arrivals have exponential interarrival times with rate $1$ (so mean $1$ too). Write $N_t$ for the number of arrivals by time $t$. So $N_t \sim \text{Poisson}(t)$. It is standard that $$ \text{given $N_t = n$}\quad\text{the arrival times $T_1, ..., T_n$ have the distribution of an ordered uniform sample};$$ that is, draw $U_1, ..., U_n$ iid uniformly from $[0,t]$, then order them into $T_1 \le \cdots \le T_n$. It is clear that "looking from $t$ back to $0$" (ie backwards in time), rather than "from $0$ up to $t$", this ordered sample has the same distribution. That is, if I define $T'_1 = t - T_n$, $T'_2 = t - T_{n-1}$, ..., $T'_n = t - T_1$, then $(T_1, ..., T_n)$ and $(T'_1, ..., T'_n)$ have the same distribution.

Noting that $A_t + D_t$ is the interarrival time, this implies that $D_t$ is also distributed as $\text{Exponential}(1)$. (That is, if we reverse a Poisson process, then we still get a Poisson process.

In general you can read more about this stuff in these Applied Probability lecture notes.

Answer 2

Intuitively, $\mathop{\mathbb{E}}[D_t] > 0$. This is because the bus must have left before you arrived at time $t$. It would not be a surprise to me if the distribution of $D_t$ was also exponential, although with statistics these kind of non-rigorous deductions often result in errors. Regardless, we can say $\mathop{\mathbb{E}}[D_t] > 0$.

We do not need to show that the random variables $A_t$ and $D_t$ are independent because the linearity of expectations holds for dependent variables. This proof can be done without the use of the memoryless property.

The expectation of an exponential random variable is one over the rate parameter ($\frac{1}{\lambda}$).

We now have all the parts we need to answer the question.

$$ \mathop{\mathbb{E}}[D_t + A_t] \: (given)$$ $$= \mathop{\mathbb{E}}[D_t] + \mathop{\mathbb{E}}[A_t] \: $$ $$\leq \mathop{\mathbb{E}}[A_t] \: \: \: \: \: \: \: \: \: (\mathop{\mathbb{E}}[D_t] > 0)$$ $$= \frac{1}{\lambda} \: \: \: \: \: \: \: \: \: \square $$