# First central moment

1. Apr 14, 2007

### gnome

The "first central moment" of a real-valued function

$$\mu_1 \equiv \int_{-\infty}^\infty (x - \mu) f(x)\,dx = 0$$

where

$$\mu \equiv \int_{-\infty}^\infty x\, f(x)\,dx$$

so we have

$$\int_{-\infty}^\infty (x - \left ( \int_{-\infty}^\infty x\, f(x)\,dx \right ) ) f(x)\,dx = 0$$

Intuitively, it seems to make sense, but how do we manipulate those integrals to prove this equality?

2. Apr 14, 2007

### D H

Staff Emeritus
This equality, as you have expressed it, is not true. I can easily come up with a counterexample. There are at least a couple of things you can change to make a true equality.

This looks like homework. You need to show some work before people here will help you.

3. Apr 14, 2007

### gnome

1. It's not homework.

2. Perhaps I should have specified that f(x) is a probability distribution. As for it not being true, I have read it in a number of places such as
http://mathworld.wolfram.com/Moment.html
and
http://en.wikipedia.org/wiki/Moment_(mathematics)
and elsewhere.

The the reason I posted the question is that I realized that I have no idea how to manipulate an integral that contains an integral as part of the integrand, so trying to "show some work" would be pointless. If you know how, I would appreciate any help or examples of how to handle such a function.

4. Apr 14, 2007

### D H

Staff Emeritus
Perhaps you should have. You said $f(x)$ is a real-valued function. The relation is true only if $\int_{-\infty}^{\infty}f(x)dx = 1$.

Use the fact that f(x) has unit area.

5. Apr 14, 2007

### gnome

Sorry, I still don't see it.

I want to show that

$$\int_{-\infty}^\infty x f(x)\,dx = \int_{-\infty}^\infty \left(\int_{-\infty}^\infty x\, f(x)\,dx \right ) f(x)\,dx$$

That would be obvious (given the fact that $\int_{-\infty}^\infty f(x)\,dx = 1$ if I could say

$$\int_{-\infty}^\infty \left(\int_{-\infty}^\infty x\, f(x)\,dx \right ) f(x)\,dx = \int_{-\infty}^\infty f(x)\,dx \cdot \int_{-\infty}^\infty x\, f(x)\,dx[/itex] but what gives me the right to do that? The only thing I've seen that allows that is Fubini's theorem, which I thought is only valid over a rectangle. I can't call this a rectangle, can I? 6. Apr 14, 2007 ### D H Staff Emeritus What gives you the right to do that is that "dx" is a dummy variable. The inner integral is just a number. Look back to your original post: [tex]\mu_1 \equiv \int_{-\infty}^{\infty}(x-\mu)f(x)dx$$

$\mu$ is is just a number. Thus

$$\mu_1 = \int_{-\infty}^{\infty}xf(x)dx - \mu\int_{-\infty}^{\infty}f(x)dx$$

The first term is just $\mu$ by definition. The second term is also $\mu$ since $f(x)$ is a probability distribution function. Thus $\mu_1 = 0$.

7. Apr 14, 2007

### gnome

Heh heh...

I tried to tell myself that before posting the original question, but my self was not convinced. What's not quite clear to me is exactly when is the inner integral "just a number"?

Is the inner integral "just a number", and can I always do this...

$$\int_a^b \left(\int_c^d f(x)\,dx \right ) f(y)\,dy = \int_a^b f(y)\,dy \cdot \int_c^d f(x)\,dx$$

as long as c and d are not functions of y?

(Surely it's not ALWAYS just a number, or there would be no need for Fubini, right?)

Last edited: Apr 14, 2007
8. Apr 14, 2007

### D H

Staff Emeritus
It's not a number when it's a function.

In other words, this is not valid:

$$\int_a^b\int_c^d f(x,y) dy\; g(x) dx = \int_c^d f(x,y) dy \int_a^bg(x) dx$$

because $\int f(x,y) dy$ is a function of x.

9. Apr 14, 2007

### gnome

I think we're saying essentially the same thing with different examples.

Compromising...in

$$\int_a^b\int_{h_1(x)}^{h_2(x)} f(y) dy\; g(x) dx = \int_c^d f(x,y) dy \int_a^bg(x) dx$$

isn't

$$\int_{h_1(x)}^{h_2(x)} f(y) dy$$

also a function of x?