# Change integration order, what happens to the limits?

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

N/A

## The Attempt at a Solution

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

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

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.

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
Homework Helper
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

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DryRun
Gold Member
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
Homework Helper
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

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:

Code:
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]}]

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