Probability of exactly n cars passing in time T

[Hiero]

Homework Statement

In a short time dt, the probability that a car passes an observer is dt/τ. In a random time interval of length T, what is the probability P(n) that exactly n cars pass the observer?

The Attempt at a Solution

I can only see how to find P(0) the probability that no cars pass the observer.

To do this I treat successive time intervals of length dt as discrete events then take the limit as dt goes to zero:

The probability that no cars pass in dt is (1-dt/τ) so the probability that no cars pass in a time T = n•dt should be (1-dt/τ)ⁿ = (1-dt/τ)^T/dt
The limit as dt→0 involves the famous limit for Euler’s number, and so we get:

P(0) = e^-T/τ

We can say that one minus this is the probability of at least one car passing, i.e.:
1 - P(0) = Σ_{n = 1 to ∞}[P(n)]
But I am at a loss as to how to find the probability for a particular n.

Thanks.

 

[Orodruin]

Take it in steps. What is the probability that exactly one car passes in an interval of length T?

 

[BvU]

Study the properties of the Poisson distribution

 

[Orodruin]

Study the properties of the Poisson distribution

Isn't the entire point of the exercise to derive the probabilities of the Poisson distribution?

 

[BvU]

Maybe - I don't know the context of the exercise - but I figured it doesn't harm it the OP peeks ahead a little ...

 

[Orodruin]

I don't know the context either, it just seems a bit odd to me to pose the question like that if the students are already familiar with the Poisson distribution, but what do I know?

 

[Hiero]

Take it in steps. What is the probability that exactly one car passes in an interval of length T?

I know, I know! It’s P(1)

But in all seriousness I don’t even know how find P(1)

 

[Orodruin]

I know, I know! It’s P(1)

But in all seriousness I don’t even know how find P(1)

What is the probability that a car passes in the first time interval $dt$ and no cars pass after that?
What is the probability that a car passes in the second time interval $dt$ but none before and none after?
...
What is the probability that a car passes in the last time interval $dt$ but none before?
What is the sum of the previous probabilities?

 

[Hiero]

But in all seriousness I don’t even know how find P(1)

Actually I think I’m seeing it. If we break T up into k = T/dt intervals then the chance that exactly one occurs in the first interval is dt/τ(1-dt/τ)^k-1 but we must add the chance of it happening in the 2nd interval (and so on) and so there’s k ways with equal probability so I get:

kdt/τ(1-dt/τ)^k-1 = (T/dt)dt/τ(1-dt/τ)^T/dt-1

Now we take the limit and we get

P(1) = (T/τ) e^-T/τ

[edited to fix it, I took the limit wrong; am in a rush]

(Darn, I tried to type this before your hint haha)

I think I can work it out now, thanks. I got to go though I’ll be back later for you to check my answer.

 

[StoneTemplePython]

I don't know the context either, it just seems a bit odd to me to pose the question like that if the students are already familiar with the Poisson distribution, but what do I know?

I think this happens a lot when people skip 'relevant equations' which is arguably the most important part of the template.

 

[Orodruin]

P(1) = (T/τ) e-T/τ/e

The 1/e should not be there, otherwise fine.

 

[Hiero]

There are no relevant equations. There is no context. I’m not in a course. I don’t know what Poisson distributions are. This is just an exercise in a preliminary chapter in a book on stastitcal mechanics.

...

Following the same idea as we did for P(1) but for general n, we get...
(I’m treating k = T/dt as an integer)

$$\text{P(n)} = \lim_{dt\to 0}\bigg[\binom{k}{n}(dt/\tau)^n(1-dt/\tau)^{k-n}\bigg]$$
$$\text{P(n)} = \lim_{dt\to 0}\bigg[\frac{(T/dt)(T/dt-1)(...)(T/dt-(n-1))}{n!}(dt/\tau)^n(1-dt/\tau)^{T/dt-n}\bigg]$$
The numerator with the ellipsis has exactly n terms, so we can distribute dt^n inside like so:
$$\text{P(n)} = \lim_{dt\to 0}\bigg[\frac{(T)(T-dt)(...)(T-(n-1)dt)}{n!}(1/\tau)^n\frac{(1-dt/\tau)^{T/dt}}{(1-dt/\tau)^n}\bigg]$$

As I said in the OP... $\lim_{dt\to 0}(1-dt/\tau)^{T/dt}=e^{-T/\tau}$
And of course... $\lim_{dt\to 0}(1-dt/\tau)^{n}=1$
And finally... $\lim_{dt\to 0}(T-mdt)=T$ for all finite m

So putting it all together, the final answer comes out as:

$$\text{P(n)} = \frac{(T/\tau)^n}{n!}e^{-T/\tau}$$

 

[Hiero]

There is another part: ‘What is the probability distribution for the time interval between two successive cars?’

Let’s call it p(t).

To solve this, I ask myself ‘what is the probability that it takes “exactly” t seconds between cars?’ (Not sure how to say it properly but of course I mean within dt of t.)

On the one hand, it’s p(t)dt

On the other hand, following the above approach where we take the limit of the discrete case, we get:
$$\text{p(t)}dt= \lim_{dt\to 0}\big[(dt/\tau)(1-dt/\tau)^{t/dt-1}\big]=(dt/\tau)e^{-t/\tau}$$
Thus I think the answer should be:
$$\text{p(t)}=e^{-t/\tau}/\tau$$
Is this correct? Is there an alternate approach? (Or at least a better way to articulate things?)

 

[Ray Vickson]

There is another part: ‘What is the probability distribution for the time interval between two successive cars?’

Let’s call it p(t).

To solve this, I ask myself ‘what is the probability that it takes “exactly” t seconds between cars?’ (Not sure how to say it properly but of course I mean within dt of t.)

On the one hand, it’s p(t)dt

On the other hand, following the above approach where we take the limit of the discrete case, we get:
$$\text{p(t)}dt= \lim_{dt\to 0}\big[(dt/\tau)(1-dt/\tau)^{t/dt-1}\big]=(dt/\tau)e^{-t/\tau}$$
Thus I think the answer should be:
$$\text{p(t)}=e^{-t/\tau}/\tau$$
Is this correct? Is there an alternate approach? (Or at least a better way to articulate things?)

Yes, it is correct.

I applaud your efforts to do it all yourself, and if you have the time, that is certainly the best approach. However, you are re-inventing the wheel, and so if you want results that can be used (rather than having to be derived from scratch), just looking at material on the "exponential distribution" and/or "Poisson process" will get you where you need to be a lot quicker. Of course, you needed to be informed about the appropriate key words to use in an on-line search. (But, being an old-fashioned sort, I still recommend going to the library and taking out some books.)

 

[Hiero]

However, you are re-inventing the wheel, and so if you want results that can be used (rather than having to be derived from scratch), just looking at material on the "exponential distribution" and/or "Poisson process" will get you where you need to be a lot quicker.

Well I have heard of exponential distributions but I didn’t know it would come from this context. (Looking at the Wikipedia page though it does specifically mention “time between events in a Poisson process.”)

I was also confused when I looked up “Poisson distribution” because everything that came up was a discrete distribution, unlike in this problem.
The phrase “Poisson process” comes up with much more relevant information, thanks for mentioning that phrase! I’ll be reading into it.

I prefer to re-invent the wheel at least once, because my memory is not very good. When I am trying to remember the shape of a wheel, I sometimes forget: is it a circle or an octagon?
(That’s a silly analogy haha)

Anyway I am not done with this problem quite yet... there is one final part which has me dumbfounded. (I’ll put it in a separate reply.)

 

[Hiero]

(For context,) between the last part and this part of the problem, they say:

“The mean time between cars is τ. The mean time to the next car should be τ. A little thought should convince you that the mean time since the last car should also be τ. But τ+τ≠τ; how can this be?
The same physical quantity can have different means when averaged in different ensembles! The mean time between cars in part (c) was a gap average: it weighted all gaps between cars equally. The mean time to the next car from part (d) was a time average: the second observer arrives with equal probability at every time, so is twice as likely to arrive during a gap between cars that is twice as long.”

And now the final part reads:

“In the part (c) you found the probability p_gap(t) that a randomly chosen gap was length t.
Find the probability p_time(t) that a second observer, arriving at a random time, will be in a gap between cars of length t.
From p_time(t) calculate the average length of the gaps between cars, using the time-weighted average measured by the second observer.”

I’m not sure how to interpret this p_time distribution (let alone how to find the average gap length from it).

Any clarification is appreciated.

 

[Orodruin]

just looking at material on the "exponential distribution" and/or "Poisson process" will get you where you need to be a lot quicker.

To me this sounds like telling someone who just ran a marathon that they would have gotten there faster if they had used a car. A big problem nowadays is that many people do not take the time to go through things properly. I do not think we should be telling people who do that they should take shortcuts. Sure, he might have arrived at the answer here faster, but he would not have learned the foundations to the same extent and a solid foundation is important.

 

[Ray Vickson]

(For context,) between the last part and this part of the problem, they say:

“The mean time between cars is τ. The mean time to the next car should be τ. A little thought should convince you that the mean time since the last car should also be τ. But τ+τ≠τ; how can this be?
The same physical quantity can have different means when averaged in different ensembles! The mean time between cars in part (c) was a gap average: it weighted all gaps between cars equally. The mean time to the next car from part (d) was a time average: the second observer arrives with equal probability at every time, so is twice as likely to arrive during a gap between cars that is twice as long.”

And now the final part reads:

“In the part (c) you found the probability p_gap(t) that a randomly chosen gap was length t.
Find the probability p_time(t) that a second observer, arriving at a random time, will be in a gap between cars of length t.
From p_time(t) calculate the average length of the gaps between cars, using the time-weighted average measured by the second observer.”

I’m not sure how to interpret this p_time distribution (let alone how to find the average gap length from it).

Any clarification is appreciated.

To me this sounds like telling someone who just ran a marathon that they would have gotten there faster if they had used a car. A big problem nowadays is that many people do not take the time to go through things properly. I do not think we should be telling people who do that they should take shortcuts. Sure, he might have arrived at the answer here faster, but he would not have learned the foundations to the same extent and a solid foundation is important.

I appears that you missed the part where I said "I applaud your efforts to do it all yourself, and if you have the time, that is certainly the best approach." For all I know, the OP may want to know where to read further about these matters, but just does not know the appropriate key words for a search (and, in fact, said as much in his response).