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Probability distribution of the time interval between two cars

  1. Jul 23, 2012 #1
    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 [itex]\frac{dt}{\tau}[/itex]; [itex]\tau=5min[/itex].

    what is the probability distribution of time intervals [itex]\Delta[/itex] 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

    [itex]\rho_{car}(n)=\frac{1}{n!}(\frac{T}{\tau})^ne^{-\frac{T}{\tau}}[/itex]

    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:

    [itex]\rho_{car}(1)=\frac{T}{5}e^{-\frac{T}{5}}[/itex]

    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?
     
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  3. Jul 23, 2012 #2

    mathman

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    Could you show your calculation in detail?
     
  4. Jul 23, 2012 #3

    haruspex

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    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.)
     
  5. Jul 24, 2012 #4
    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 [itex]\frac{dt}{\tau}[/itex] of a car passing and a probability [itex]1-\frac{dt}{\tau}≈e^{\frac{-dt}{\tau}}[/itex] of a car not passing.
    n is the number of intervals of dt, for here we get (1) [itex]Δ=dt\cdot n[/itex].

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

    [itex]\frac{dt}{\tau}(e^{\frac{-dt}{\tau}})^{n-1}[/itex]

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

    [itex]\frac{Δ}{\tau\cdot n}e^{\frac{-dt}{\tau}}[/itex]

    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
     
  6. Jul 24, 2012 #5

    haruspex

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    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:
    [itex]\frac{dt}{\tau}(e^{\frac{-ndt}{\tau}}) = \frac{dt}{\tau}(e^{\frac{-t}{\tau}})[/itex]
    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?
     
  7. Jul 25, 2012 #6
    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.
     
  8. Jul 25, 2012 #7
    for an observer arriving at a random time [itex]t_1[/itex], where t=0 is the time when the last car passed, i got the following pdf for [itex]\Delta^{*}[/itex]- the time the observe waits until the next car:

    [itex]\rho_{\Delta^*}=\frac{1}{\Delta^{*}}\cdot (e^{-\frac{-\Delta^{*}}{\tau}}-e^{-\frac{-2\Delta^{*}}{\tau}})[/itex].

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



    a short summary of the derivation:

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

    [itex]t_1[/itex] has a uniform probability distribution between 0 and Δ, therefore:

    [itex]\rho_{\Delta^*}=\int{\rho_{\Delta}(\Delta^*+\zeta)\rho_{t_1}(\zeta) d\zeta}[/itex]

    for [itex]0<\zeta<\Delta^*[/itex]
     
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