# Integration with absolute value

Gridvvk

## Homework Statement

The problem is in the context of a probability problem; however, my question in regards to a computation regarding a particular integral. All that is needed to know is that the probability density function is 1 in the range 0 < y < 1 , y-1 < x < 1 - y, and 0 otherwise.

I need to find E(XY) and E(|XY|).

## Homework Equations

$$E(XY) = \int_0^{1} \int_{y-1}^{1-y}\ XY \ dx dy$$
$$E(|XY|) = \int_0^{1} \ \int_{y-1}^{1-y} \ |XY| \ dx dy$$

## The Attempt at a Solution

E(XY) = 0. Simple calculation.

For E(|XY|) however, I am getting 1/24, but WolframAlpha computes it out to be 1/12.

I am doing:
$$E(|XY|) = \int_0^{1}\ \int_{y-1}^{1-y}\ |XY| \, dx dy$$ = $$\int_0^{1} \int_0^{1-y} \ XY \, dx dy$$

I am not sure why this is incorrect. Making the lower bound of x 0 worked for me when I was computing E(|X|), but is giving me half the answer for E(|XY|).

Last edited:

Mentor

For definite integrals, the LaTeX markup should look like this:

\int_{lower_bound}^{upper_bound} <integrand> dx dy

If lower_bound or upper_bound is a single character, you don't need the braces, but for expressions such as y - 1, you do need braces, not parentheses as you used.

Also, you had stuff like \1 - y, which was causing problems.

• 1 person
Gridvvk

For definite integrals, the LaTeX markup should look like this:

\int_{lower_bound}^{upper_bound} <integrand> dx dy

If lower_bound or upper_bound is a single character, you don't need the braces, but for expressions such as y - 1, you do need braces, not parentheses as you used.

Also, you had stuff like \1 - y, which was causing problems.

Thanks! I was surprised it magically fixed itself. I'll keep that in mind for future posts.

Mentor
I don't think this is right:
$$\int_0^{1} \int_0^{1-y} \ XY \, dx dy$$
You are changing the shape of the region of integration, removing half of it. The region over which integration should be taking place is the triangle whose vertices are (-1, 0), (0, 1), and (1, 0). The region you are using in the integral above is also a triangle, with vertices (0, 0), (0, 1), and (1, 0), that has half the area of the larger triangle. I believe that's why your answer (1/24) is half of what Wolfram shows (1/12).

Gridvvk
I don't think this is right:
$$\int_0^{1} \int_0^{1-y} \ XY \, dx dy$$
You are changing the shape of the region of integration, removing half of it. The region over which integration should be taking place is the triangle whose vertices are (-1, 0), (0, 1), and (1, 0). The region you are using in the integral above is also a triangle, with vertices (0, 0), (0, 1), and (1, 0), that has half the area of the larger triangle. I believe that's why your answer (1/24) is half of what Wolfram shows (1/12).

That does make sense geometrically, I rechecked and it turns out.

$$2 * \int_0^{1} \int_0^{1-y} \ X \, dx dy = \int_0^{1} \int_{y-1}^{1-y} \ |X| \, dx dy$$

So you would need a factor of 2 for |X| as well. Thanks for the help!