# Change integration order, what happens to the limits?

1. Sep 29, 2012

### infk

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

2. Relevant equations
N/A

3. The attempt at a solution

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

2. Sep 29, 2012

### jackmell

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.

3. Sep 30, 2012

### infk

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?

4. Sep 30, 2012

### 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

#### Attached Files:

• ###### double int.JPG
File size:
4.8 KB
Views:
39
Last edited: Sep 30, 2012
5. Sep 30, 2012

### sharks

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.

6. Sep 30, 2012

### ehild

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

ehild

7. Sep 30, 2012

### jackmell

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:

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:

• ###### myarea.jpg
File size:
6.5 KB
Views:
33
Last edited: Sep 30, 2012