# Probability distribution of the time interval between two cars

1. Jul 23, 2012

### ENgez

Hello,
I am trying to solve the following problem from Sethna's book on statistical mechanics (not homework).

On a Highway, the probability of a car passing in some interval dt is $\frac{dt}{\tau}$; $\tau=5min$.

what is the probability distribution of time intervals $\Delta$ between two consecutive cars. and what is the mean of this distribution?

My attempt:
In a previous question, I derived the probability distribution for n cars to pass in an interval T

$\rho_{car}(n)=\frac{1}{n!}(\frac{T}{\tau})^ne^{-\frac{T}{\tau}}$

which I believe is correct, as the hint in the question said that I should get a Poisson distribution.

now if I input n=1 i get:

$\rho_{car}(1)=\frac{T}{5}e^{-\frac{T}{5}}$

this is the probability distribution for a single car to pass in a time interval T (at some time t<T).

according to my calculations the mean of this distributions is 50 minutes^2 which really doesn't make sense.

How do you solve this problem?

2. Jul 23, 2012

### mathman

Could you show your calculation in detail?

3. Jul 23, 2012

### haruspex

That is indeed the probability of exactly one car passing in time T, but in what sense is it a pdf (or cdf)? Its integral 0 to infinity is not 1.
Also, it's not clear to me how this is heading towards a solution of the problem. (There is a very easy way to solve it. Hint: think of reversing time.)

4. Jul 24, 2012

### ENgez

I don't get how reversing time can help...

But i tried approaching the problem in the most basic form possible:

suppose car A passes at t=0, we then begin counting time until the next car arrives. in each sliver of time dt there is a probability $\frac{dt}{\tau}$ of a car passing and a probability $1-\frac{dt}{\tau}≈e^{\frac{-dt}{\tau}}$ of a car not passing.
n is the number of intervals of dt, for here we get (1) $Δ=dt\cdot n$.

thus, the chance of a car passing exactly after Δ minutes is (I hope...):

$\frac{dt}{\tau}(e^{\frac{-dt}{\tau}})^{n-1}$

after plugging (1) in (and taking the limit as n→∞) we get:

$\frac{Δ}{\tau\cdot n}e^{\frac{-dt}{\tau}}$

the problem is that when taking n→∞ the n in the denominator will cause the whole expression the go to 0 for all finite Δ, which kinda makes sense, because as dt gets smaller the probability of a car crossing at exactly that interval gets vanishingly small as well.

what am i doing wrong? and this is not a homework question, so feel free to share more than hints.

thank you

5. Jul 24, 2012

### haruspex

Yes, sorry, I was thinking of a slightly different question which usually gets asked. More on that later.
A slightly unconventional approach, but fine so far. We can treat n-1 as n, since n = Δ/δt will tend to infinity as δt shrinks to 0. And you may as well use t, not Δ, for the number of minutes elapsed. Your equation becomes:
$\frac{dt}{\tau}(e^{\frac{-ndt}{\tau}}) = \frac{dt}{\tau}(e^{\frac{-t}{\tau}})$
So that's the pdf of time to next car. The average is obtained by multiplying by t and integrating 0 to ∞.

The more usual method is to put n = 0 in your first equation to get the probability of no cars in time t. Then multiply by the probability of a car in time δt to get the prob of the first car being in interval (t, t+δt).
The problem I was thinking of was: you arrive at the road at a random instant. What is the expected length of time between the last car and the next?

6. Jul 25, 2012

### ENgez

Thank you!
I actually got the answer you provided, but what stumped me is that dt is supposed to be an infinitesimal interval. now i see that the probability of a car passing at exactly time t is zero, and that the pdf only gives out "real" values if integrated over some subset of intervals [t1,t2].

Btw, the question about arriving at a random time is the next one in the book, so thanks for the tip about reversing time.

7. Jul 25, 2012

### ENgez

for an observer arriving at a random time $t_1$, where t=0 is the time when the last car passed, i got the following pdf for $\Delta^{*}$- the time the observe waits until the next car:

$\rho_{\Delta^*}=\frac{1}{\Delta^{*}}\cdot (e^{-\frac{-\Delta^{*}}{\tau}}-e^{-\frac{-2\Delta^{*}}{\tau}})$.

the mean is $\tau$, like the book said and it goes to 0 for $\Delta^{*}→0$ and $\Delta^{*}→∞$, but it still looks kind of weird for a probability distribution.... is this correct?

a short summary of the derivation:

$\Delta^{*}=\Delta-t_1$, where Δ is the time between two consecutive cars (as was found in the previous posts).

$t_1$ has a uniform probability distribution between 0 and Δ, therefore:

$\rho_{\Delta^*}=\int{\rho_{\Delta}(\Delta^*+\zeta)\rho_{t_1}(\zeta) d\zeta}$

for $0<\zeta<\Delta^*$