Change integration order, what happens to the limits?

[infk]

Homework Statement

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?
\int^{\infty}_0\int^{\infty}_xf(t) \mathrm{d}t \mathrm{d}x =
\int^{\infty}_0\int^{t}_0f(t)\mathrm{d}x\mathrm{d}t

Homework Equations

N/A

The Attempt at a Solution

We can change the order of integration but I only come up with:
\int^{\infty}_0\int^{\infty}_xf(t)\mathrm {d}t\mathrm {d}x = \int^{\infty}_x\int^{\infty}_of(t)\mathrm {d}x\mathrm {d}t =

 

[jackmell]

Homework Statement

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?
\int^{\infty}_0\int^{\infty}_xf(t) \mathrm{d}t \mathrm{d}x =
\int^{\infty}_0\int^{t}_0f(t)\mathrm{d}x\mathrm{d}t

Homework Equations

N/A

The Attempt at a Solution

We can change the order of integration but I only come up with:
\int^{\infty}_0\int^{\infty}_xf(t)\mathrm {d}t\mathrm {d}x = \int^{\infty}_x\int^{\infty}_of(t)\mathrm {d}x\mathrm {d}t =

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.

 

[infk]

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.

Yes, I tried doing that but I didn't have an "aha"-moment..
So we have that t ranges from x to \infty and x from 0 to \infty.. How do you draw that?

 

[ehild]

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

 

[DryRun]

Except from the attached figure above, \int^{\infty}_0\int^{\infty}_xf(t) \mathrm{d}t \mathrm{d}x gives the area of required region between the x-axis and the line x=t.

 

[ehild]

Except from the attached figure above, \int^{\infty}_0\int^{\infty}_xf(t) \mathrm{d}t \mathrm{d}x gives the area of required region between the x-axis and the line x=t.

The area is \int^{\infty}_0\int^{\infty}_x \mathrm{d}t \mathrm{d}x

ehild

 

[jackmell]

Yes, I tried doing that but I didn't have an "aha"-moment..
So we have that t ranges from x to \infty and x from 0 to \infty.. How do you draw that?

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:

\int_0^5 \int_x^5 dt dx

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:

\int_0^5 \int_0^t dx dt

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:

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]}]