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


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?

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  • $\begingroup$ If I remember correctly from my classes $D_t$ should also be exponentially distributed. $\endgroup$ – Stan Tendijck Feb 28 at 21:06
  • $\begingroup$ $D_t$ and $A_t$ are independent $\mathrm{Exp}(\lambda)$ random variables, so $\mathbb E[D_t+A_t] = 1/\lambda+1/\lambda > 1/\lambda$. $\endgroup$ – Math1000 Feb 28 at 21:22
  • $\begingroup$ I think you're right that the proof is trivial, but perhaps the Question is meant to be thought-provoking (instead of difficult): Since $D_t + A_t$ is exactly one interarrival period, why is it true that $E[D_t + A_t] > 1/\lambda$? BTW, $D_t$ being $Exp(\lambda)$ is not trivial either. $\endgroup$ – antkam Feb 28 at 22:02
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    $\begingroup$ If you (uniformly) randomly pick an INTERVAL between arrivals, the the expected length is 1/lambda. If you randomly (uniformly) pick a POINT IN TIME, the expected length of the interval containing it is 2/lambda as shown here. The latter distribution is called the "size biasing" of the exponential distribution, which comes out to the sum two of independent exponentials, by the forward and backward argument given. This is sometimes called the waiting time paradox. $\endgroup$ – Ned Mar 1 at 1:07
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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.

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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.

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

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    $\begingroup$ Independence is not needed. $E[X+Y] = E[X] + E[Y]$ even if they are dependent. $\endgroup$ – antkam Feb 28 at 21:57
  • $\begingroup$ I don't think so, I would check your reasoning. How do you deduce $Y \neq f(X)$? $\endgroup$ – Jay Land Feb 28 at 22:01
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    $\begingroup$ antkam is right. This is just linearity of expecation $\endgroup$ – Sam T Feb 28 at 22:03
  • $\begingroup$ Congratulations, Jay, you are about to learn one of the best theorems of probability today! :) en.wikipedia.org/wiki/Expected_value#Linearity -- note that it says "arbitrary" random variables, with no mention of independence. (In comparison check out the section on $E[XY]$.) $\endgroup$ – antkam Feb 28 at 22:06
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    $\begingroup$ even if they're dependent, in fact even if $Y=f(X)$, it is still true that $E[X+Y] = E[X] + E[Y]$. see the wiki link for proof. intuitively, if you take the average age of all humans and add it to the average height of all humans (age & height are certainly correlated), it has to equal to the average of (age + height) of all humans, right? $\endgroup$ – antkam Feb 28 at 22:24

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