# Integral of the Square of density probability function

1. Jan 16, 2010

### BoMa

Hi,
I'm looking for the value of the integral of the square of a density probability function
on a bounded interval.

Tks

2. Jan 16, 2010

### mathman

It could be anything, depending on the density function itself.

3. Jan 17, 2010

### BoMa

Is there a way to characterise this "anything" , you're talking about.

$$\int \int f^{2}(x,y) dx\,dy$$

Some norm on the probability space ????

4. Jan 17, 2010

### winterfors

You wouldn't use it as a norm (the L^2 norm), having a probability space implies that you are using the L^1 norm (and that this norm equals one).

The integral of the squared PDF can however be interpreted as a measure of the average probability density of the PDF, i.e. how concentrated the probability is. Further more, you know that it will be larger or equal to the average probability density of a constant PDF:

$$\int \int f^{2}(x,y) dx\,dy \geq 1 / \int \int dx\,dy$$

5. Jan 17, 2010

Just a question: you seem to be automatically assuming that the density is a joint density of two variables, hence you write

$$\iint f^2(x,y) \, dxdy$$

If, however, you consider a univariate density, the proper expression for the integral of the density squared is

$$\int f^2(x) \, dx$$

6. Jan 18, 2010

### BoMa

Yes I'm talking about the bivariate bounded probability density function (pdf) $$f(x,y).$$

Sorry I can't understand the difference between L1 or L2 norm on the probability space.

About not using the L2 norm , I thought that the pdf could be written as
$$\int^{b}_{a} \int^{b}_{a} f^{2}(x,y)dxdy =\int^{b}_{a} \int^{b}_{a} |f^{2}(x,y)|dxdy=||f||^2$$
which looks like the L2 norm , more than the L1 norm.

Could you please explain in more details why this case can't be seen as L2 norm ?

7. Jan 18, 2010

### frustr8photon

what are L1 and L2?

8. Jan 18, 2010

### mathman

To get the L2 norm, take a square root.

I really don't understand what you are trying to achieve. To give a simple example. Assume that f is constant over a square with area 1/A, then f=A over this area and the integral of f2 is A. Since A is completely arbitrary, the integral can have any value.

9. Jan 18, 2010

### BoMa

I understood that the value will be any constant depending on the choice of f which is arbitry chosen here. So I wanted to say that it is some L2 norm. But someone on the list said that It should be L1 norm (not L2 norm !!), because the problem here is on a probabity space. So I wonder why, he said L1 norm instead of L2 norm ?

10. Jan 19, 2010

### mathman

Sorry, I am not a mind reader. Just make sure you take the square root to get the norm. In the simple example the L2 norm is √A.