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Math Help - Limits when evaluating joint pdfs

  1. #1
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    Critical regions question

    A bus company is about to start a scheduled service between two towns. Before deciding on an appropriate timetable they do ten trial runs to see how long the journey takes. The times, in minutes, are:

    89 92 98 34 98 34 98 98 34 34

    95% interval for mean journey time?
    Last edited by bluebiro; August 31st 2009 at 07:31 AM.
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  2. #2
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    Quote Originally Posted by bluebiro View Post
    I have a joint pdf f(s,t) = 6s, 0<t<s<1

    I know that you integrate the joint pdf with respect to the other variable when finding marginal densities, but I don't know what limits to use.

    So when I integrate 6s w.r.t. t, are the limits 0 and 1? or t and s? or 1 and t? Or s and 0?

    Vice versa, when I integrate 6s w.r.t. s, what are the limits? I integrated it with limits 1 and 0 and got 6 as the marginal density, which is surely wrong.

    explain to me please
    Draw the region defined by 0 < t < s < 1. I put s on the horizontal axis and t on the vertical axis. From what you draw it should be clear that the marginal pdf's are:

    g(s) = \int_{t = 0}^{t = s} f(s, t) \, dt.

    h(t) = \int_{s = t}^{s = 1} f(s, t) \, ds.
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  3. #3
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    Thank you. I sort of see now. And how about the limits when I find the marginal densities, do they change?

    I calculated the marginal density of s given t as 2s/(1-t^2) and t given s as 1/s, how do I know what the range for which this is valid?
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  4. #4
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    Am I on the right track? (Joint prob density question)

    R.v.s s and t have joint p.d.f f(s,t) = 6s, 0<t<s<1. I worked out their marginal densities as
    fs(s) = 6s^2
    ft(t) = 3-3t^2

    and their conditional densities as
    f(t given s) = 1/s
    f(s given t) = 6s/(3-3t^2)

    (a) for what range are the conditional densities valid?
    (b) find (t>0.25 given that s = 0.75)

    I substituted numbers into the f(t given s) = 1/s equation to get
    1/s = 4/3.

    Then I integrated this with respect to t, upper limit 1, lower limit 0.25:
    [4/3t]

    =

    4/3 - 1/3 = 1

    giving the answer as 1. Is this right?

    thanks
    Last edited by bluebiro; August 16th 2009 at 02:10 AM.
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  5. #5
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    Quote Originally Posted by bluebiro View Post
    R.v.s s and t have joint p.d.f f(s,t) = 6s, 0<t<s<1. I worked out their marginal densities as
    fs(s) = 6s^2
    ft(t) = 3-3t^2

    and their conditional densities as
    f(t given s) = 1/s
    f(s given t) = 6s/(3-3t^2)

    (a) for what range are the conditional densities valid?
    (b) find (t>0.25 given that s = 0.75)

    I substituted numbers into the f(t given s) = 1/s equation to get
    1/s = 4/3.

    Then I integrated this with respect to t, upper limit 1, lower limit 0.25:
    [4/3t]

    =

    4/3 - 1/3 = 1

    giving the answer as 1. Is this right?

    thanks
    I've just noticed: The joint pdf you've been given is not valid - the double integral over the region defined by 0 < t < s < 1 is 2 ....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    I've just noticed: The joint pdf you've been given is not valid - the double integral over the region defined by 0 < t < s < 1 is 2 ....
    I get 1 (though I could have made a mistake).

    integral of 6s w.r.t t (limits is 3s^2)
    and the integral of 3s^2 w.r.t s is s^3, which, with limits 1 and 0, gives 1.

    Where have I gone wrong?
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  7. #7
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    Quote Originally Posted by bluebiro View Post
    I get 1 (though I could have made a mistake).

    integral of 6s w.r.t t (limits is 3s^2)
    and the integral of 3s^2 w.r.t s is s^3, which, with limits 1 and 0, gives 1.

    Where have I gone wrong?
    \int_0^s 6s \, dt = \left[6st\right]_0^s = 6s^2 and \int_0^1 6s^2 \, ds = \left[ 2s^3 \right]_0^1 = 2.

    \int_t^1 6s \, ds = \left[3s^2\right]_t^1 = 3 - 3t^2 and \int_0^1 3 - 3t^2 \, dt = \left[ 3t - t^3 \right]_0^1 = 2.

    You have not been given a valid joint pdf.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    \int_0^s 6s \, dt = \left[6st\right]_0^s = 6s^2 and \int_0^1 6s^2 \, ds = \left[ 2s^3 \right]_0^1 = 2.

    \int_t^1 6s \, ds = \left[3s^2\right]_t^1 = 3 - 3t^2 and \int_0^1 3 - 3t^2 \, dt = \left[ 3t - t^3 \right]_0^1 = 2.

    You have not been given a valid joint pdf.
    Well the joint pdf I was originally given was f(s,t)= lambda.s, where lambda is a constant, so to calculate lambda I did

    1=\lambda\int_0^1\int_0^tsdsdt
    1=\frac{\lambda}{2}\int_0^1[s^2]_0^tdt
    1=\frac{\lambda}{2}\int_0^1[t^2]dt
    1=\frac{\lambda}{6}[t^3]_0^1
    1=\frac{\lambda}{6}
    \therefore\lambda=6

    Which is where I got the 6 from.
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  9. #9
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    Quote Originally Posted by bluebiro View Post
    Well the joint pdf I was originally given was f(s,t)= lambda.s, where lambda is a constant, so to calculate lambda I did

    1=\lambda\int_0^1\int_0^tsdsdt
    1=\frac{\lambda}{2}\int_0^1[s^2]_0^tdt
    1=\frac{\lambda}{2}\int_0^1[t^2]dt
    1=\frac{\lambda}{6}[t^3]_0^1
    1=\frac{\lambda}{6}
    \therefore\lambda=6

    Which is where I got the 6 from.
    I assume you're referring to this thread: http://www.mathhelpforum.com/math-he...joint-pdf.html

    In that thread you define the support as 0 < s < t <1. This is NOT the same as the support defined in this thread, namely 0 < t < s < 1. The joint pdfs in each thread are therefore very different from each other.
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