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Change integration order, what happens to the limits?

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Homework about a result in probability theory, but I don't understand one of the steps:

    Let f(x) be the PDF of a continuous R.V which takes only non-negative values.

    Why is the following true?
    [itex]\int^{\infty}_0\int^{\infty}_xf(t) \mathrm{d}t \mathrm{d}x[/itex] =
    [itex]\int^{\infty}_0\int^{t}_0f(t)\mathrm{d}x\mathrm{d}t[/itex]

    2. Relevant equations
    N/A


    3. The attempt at a solution

    We can change the order of integration but I only come up with:
    [itex]\int^{\infty}_0\int^{\infty}_xf(t)\mathrm {d}t\mathrm {d}x[/itex] = [itex]\int^{\infty}_x\int^{\infty}_of(t)\mathrm {d}x\mathrm {d}t[/itex] =
     
  2. jcsd
  3. Sep 29, 2012 #2
    You should first draw the region of integration in the t-x plane, then consider how the double integral sums up small rectangles using dtdx, then consider how the double integral would sum them up using dxdt.
     
  4. Sep 30, 2012 #3
    Yes, I tried doing that but I didn't have an "aha"-moment..
    So we have that t ranges from x to [itex]\infty[/itex] and x from 0 to [itex]\infty[/itex].. How do you draw that? :confused:
     
  5. Sep 30, 2012 #4

    ehild

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    The border line of the integration domain is x=t. How can you cover the shaded area if you integrate with respect to t first and when you integrate with respect to x first?

    ehild
     

    Attached Files:

    Last edited: Sep 30, 2012
  6. Sep 30, 2012 #5

    sharks

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    Except from the attached figure above, [itex]\int^{\infty}_0\int^{\infty}_xf(t) \mathrm{d}t \mathrm{d}x[/itex] gives the area of required region between the x-axis and the line x=t.
     
  7. Sep 30, 2012 #6

    ehild

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    The area is [itex]\int^{\infty}_0\int^{\infty}_x \mathrm{d}t \mathrm{d}x[/itex]:smile:

    ehild
     
  8. Sep 30, 2012 #7
    Mathematica dude. First understand that the integrals are summing up small rectangles in the t-x plane and we can arrange these rectangles going up and down or side to side. Now consider the integral:

    [tex]\int_0^5 \int_x^5 dt dx[/tex]

    and read that as "dt as t goes from t=x to t=5 and then dx as x goes from x=0 to x=5. Now look at my wonderful plot I did in Mathematica, the region we're integrating in light purple, and first the red rectangle illustrating the integration plan above. That rectangle is going from t=x to t=5. Now keep building rectangles above and below it. Then the sum of the rectangles is going from the range of x=0 to x=5 as the integral above.

    Now consider summing rectangles in the vertical direction and consider the blue rectangle and the integral representing this integration plan as:

    [tex]\int_0^5 \int_0^t dx dt[/tex]

    Again, read that as "dx as x goes from x=0 to x=t, and then dt as t goes from t=0 to t=5. Look again at the blue rectangle. It's height is going from x=0 to x=t and if we sum the set along the entire region, that set is going from t=0 to t=5.

    Here's the code in case you want to play with it:

    Code (Text):

    myrec1 = Graphics[{Red, Opacity[0.5],
         Rectangle[{1, 0.8}, {5, 1.2}]}];
    myrec2 = Graphics[{Blue, Opacity[0.5],
        Rectangle[{2.8, 0}, {3.2, 3}]}]
    Show[{Plot[y = x, {x, 0, 5}, Filling -> Bottom],
       myrec1, myrec2}, AxesLabel ->
       {Style["x", 30], Style["t", 30]}]
     
     

    Attached Files:

    Last edited: Sep 30, 2012
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