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Statistics with pdfs

  1. Jul 8, 2006 #1
    A car rental company receives cars at n = 1000 cars/year for the first 5 years and none thereafter.

    The pdf for retiring a car is,

    Code (Text):
    p(t) = (1/5)(1 - exp(-t/3))             , 0<t<5
         = (1/5)(exp[-(t-5)/3] - exp(-t/3)) , t > 5
    Show that after 6 years cars are being retired from the scheme at approxiamtely 11 per week.

    To answer that I did,

    N = int[t1 to t2] t*p(t) dt

    t1 = 6 (years)
    t2 = 6 1/52 (years)
    N would be the number of retired cars during that week.

    I thought that was how I was supposed to do it. Is that correct ?

    When I worked out that integral, I got N = 0.0134.

    How do I get 11 as the answer ?
    Where does the n = 1000 come into it, if at all ?
  2. jcsd
  3. Jul 8, 2006 #2


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    Ok, I think that what you're saying is that p(t) is the probability density function that a given car will be retired after having been in use for t years. So at time three years, some cars will have a probability density of p(3) for being retired, while some will be at p(2.55) and some will be of p(.0003). Also since you say that these are PDFs the 1000 cars/year I will also assume to be a continuous rate of car acquisition instead of getting 1000 cars in a lump at the start of each year. Correct me if this is not what you mean.

    The thing you can find from this information is the rate at which cars are being retired. Let's say that a car is acquired at time t, and we are now at time t + 2.1. Then the probability density for this car being retired is p(2.1)--if this car were divisible and retired part by part, then this would be the expected rate at which the car is being retired. At a time T, if we integrate up all those rates of retirement for all cars that could possibly exist at time T, we should get the expected total rate at which the cars are being retired. Let that rate be R(T). To aid notation, let c(t) be a function determining the rate at which cars are acquired, so c(t) = 1000 for 0 <= t <= 5 and c(t) = 0 for t > 5. So,
    [tex]R(T) = \int_0^T p(T-t)c(t) dt[/tex]

    so you just have to find R(6) and then convert from cars per year to cars per week.
    Last edited: Jul 8, 2006
  4. Jul 8, 2006 #3
    Sorry, the pdf I gave is for retiring a car, t years after the start of the program (t=0), where the program is the supply of cars at 1000 cars/year for 5 years. This supply is to be taken as a constant rate.

    The pdf that a given car will be retired after t' years of service/use is,

    pser(t') = (1/3)exp(-t'/3), t' > 0

    That was in an earlier part of this question. I don't think I need it for this part.
  5. Jul 8, 2006 #4
    Here's the full question.

    A car rental company signs a contract to take delivery of a new model starting in January 2006, at the rate of n = 1000 cars per year, until the contract ends, after T = 5 years, in January 2011. The cars are removed from service when they are damaged or start to look old: the probability density for retiring a car after t years of service is ρser(t) = exp(-t/τ)/τ, with τ = 3 years.

    (a) What is the mean value <t> of the time from delivery of a car until it is retired?

    (b) Assuming that the cars are supplied at a constant rate, what is the probablity denasity ρsup(t) that a given car will be supplied t years after the start of the program ?

    (c) Assuming that the service llife of a car is independent of the time when it is supplied, show that the probability density ρret(t) that a given car being retired from service at a time t after the start of the program is,

    Code (Text):
    ρ(t) = (1/T)(1 - exp(-t/τ))             , 0<t<5
         = (1/T)(exp[-(t-T)/τ] - exp(-t/τ)) , t > 5
    (d) Show that, after 6 years, cars are being retired from the scheme at a rate of approximately 11 per week.
  6. Jul 8, 2006 #5


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    Well, if you have the pdf for retiring a car over the lifetime of the car, you can use it.

    Edit: ah, perhaps I have misunderstood. Your p(t) is the probability density function that a given car _out of any of the 5000 supplied_ ends at a given time. Then it is a bit easier, since you can simply find p(6) and multiply it by 5000, then convert it into cars per week.
    Last edited: Jul 8, 2006
  7. Jul 8, 2006 #6
    Thanks. That gives me the right answer.
    p(6) works out at 0.11624 which gives 11.18 pro woche.

    I've got the answer now, but I still don't understand this stats stuff. I'm pretty good at other maths disciplines, but I just can't seem to follow the "logic" of statistical reasoning.

    Anyway, thanks one more for the help.
  8. Jul 8, 2006 #7


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    Basically in this problem you just have to realize the connection between the probability density function of any one of a group of cars being retired and the rate at which the whole group decreases in number. If the PDF indicates that the "instantaneous" probability of a car being retired is .2 per year, then you make the connection that if you have 1000 of these cars, they are declining at an instantaneous rate of 200 per year.
    Last edited: Jul 8, 2006
  9. Jul 9, 2006 #8


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    ok forget about that last post it *was* over compicating it. In fact it was so much nonsense that I've deleted it.
    Last edited: Jul 10, 2006
  10. Jul 9, 2006 #9


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    BTW, you could do it as an ODE if you really want. The correct ODE in this case would be :

    dN/dt = 1000 - N/3 for 0 <= t < 5


    dN/dt = -N/3 for t>=5

    This give the same answer as above. dN/dt = -581.2 cars/year (approx -11 cars per week) at t=6
  11. Jul 9, 2006 #10
    Is this correct ?

    When doing part (b), I got (with some help from elsewhere) that the probability of any given car being suppplied (t<5) is P = t/1000.
    I believe that the probability is defined as,

    P(t1,t2) = int[t1 to t2] psup(t) dt

    where P(t1,t2) is the probability that a car will be supplied between t1 and t2.


    psup(t) = dP/dt
    psup(t) = 1/1000

    It doesn't feel right that it should be a constant function for this pdf.
    I've made a mistake, yes ?
  12. Jul 10, 2006 #11


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    No that's not right. psup(t) is a constant but it's 1/5, not 1/1000. Take the case for example of t2-t1 = one year and use this in your P(t1,t2) = int[t1 to t2] psup(t) dt formula. If you evaluate this with psup(t) = 1/5 then you get the correct answer that the probabilty of a particular car being supplied in a given one year period is 1/5. That makes sense right, it's just a uniform probability densisty function spread of 5 years, the probabilty of receiving a particular car in the 5 year period is unity.

    BTW. The full answer they are looking for in part (b) is the uniform pdf :
    psup(t) = 1/5 for 0<=t<=5, zero otherwise.

    Part c) is the most important of the four parts for your understanding of this. What method did you use to get the stated asnswer to part c). Did you use joint and/or marginal probabilty density functions or did it just come from some formula that perhaps you dont understand? If your not fully sure of how the result in part c) came about then that's the area I'd recommend concentrating on.
  13. Jul 10, 2006 #12
    Thanks. Working backwards from that statement makes sense of the previous parts, viz psup(t) = 1/5, 0<=t<=5, zero otherwise.

    The mark scheme for this question is,

    (a) [5]
    (b) [4]
    (c) [10]
    (d) [6]

    For part (c) I'm to use a convolution. I should do,

    pret = pser⊕psup

    Where I'm using ⊕ as the convolution operator.

    But thats why I said, in my last post, I thought having psup(t) = constant didn't feel right.
    If psup(t) is a constant then the convolution becomes trivial, has no effect ???

    Convolutions I can do :smile:
    Statistics I suck at :cry:

    Edit: in post #4, in the code section, I have
    Code (Text):
    ρ(t) = ...
    . That should have been
    Code (Text):
    ρ[sub]ret[/sub](t) = ...
    Last edited: Jul 10, 2006
  14. Jul 10, 2006 #13


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    That's good, convolution is a good way to do it. Convolution works in this case because of a result that says "if you construct a new random variable as the sum of two other random variables (that are independant of each other) then the pdf of the new random variable is the convolution of the pdf's of the two original random variables". In this case if you let say "x" be the time from delivery of a car until it is retired, and "y" be the time of delivery, then if you let "t" denote the total time from the start of the program to retirement then clearly t = x + y. So that's why convolution works in this case.

    You can also solve it by marginal and joint pdf's if you've covered that material. This method is kind of lower level, more like first principles and might give a better insight.

    You can start by looking at the conditional pdf for retirement at time "t" given that the car is delivered at time "y". That is,

    p(t|y) = 1/3 exp(-(t-y)/3) : for t>y, and zero otherwise.

    You then construct the joint pdf "p(t,y)" from the Bayes formula, p(t,y) = p(t|y) p(y). (Sorry I'm being lazy and not subscripting these properly).

    In this case,
    [tex]p(t,y) = 1/3 \ e^{-(t-y)/3} \times 1/5[/tex] : for t>y and 0<=y<=5, and zero otherwise.

    Properly specifying the region "t>y and 0<=y<=5, and zero otherwise" here is very important. It's a region in the (t,y) plane that you need to take into account when you integrate out the "y" in the next step.

    The final step is to integrate out the "y" to leave the desired pdf for "t" alone. That is

    p_ret(t) = int[y=-infinty to +infinity], p(t,y) dy

    If you evaluate the above integral paying carful attention to the boundarys where the p(t,y) is zero then you'll get exactly the same result as the convoution integral. Doing it this way might give more insight into the problem.
    Last edited: Jul 10, 2006
  15. Jul 10, 2006 #14
    Ive never heard of marginal and joint pdf's :frown:

    I did my convolution though and got the right answer :smile:

    Many thanks for the help.
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