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First central moment

  1. Apr 14, 2007 #1
    The "first central moment" of a real-valued function

    [tex]\mu_1 \equiv \int_{-\infty}^\infty (x - \mu) f(x)\,dx = 0[/tex]


    [tex]\mu \equiv \int_{-\infty}^\infty x\, f(x)\,dx[/tex]

    so we have

    [tex]\int_{-\infty}^\infty (x - \left ( \int_{-\infty}^\infty x\, f(x)\,dx \right ) ) f(x)\,dx = 0[/tex]

    Intuitively, it seems to make sense, but how do we manipulate those integrals to prove this equality?
  2. jcsd
  3. Apr 14, 2007 #2

    D H

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    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.
  4. Apr 14, 2007 #3
    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
    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.
  5. Apr 14, 2007 #4

    D H

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

    Use the fact that f(x) has unit area.
  6. Apr 14, 2007 #5
    Sorry, I still don't see it.

    I want to show that

    [tex]\int_{-\infty}^\infty x f(x)\,dx = \int_{-\infty}^\infty \left(\int_{-\infty}^\infty x\, f(x)\,dx \right ) f(x)\,dx[/tex]

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

    [tex]\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?
  7. Apr 14, 2007 #6

    D H

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    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[/tex]

    [itex]\mu[/itex] is is just a number. Thus

    [tex]\mu_1 = \int_{-\infty}^{\infty}xf(x)dx - \mu\int_{-\infty}^{\infty}f(x)dx[/tex]

    The first term is just [itex]\mu[/itex] by definition. The second term is also [itex]\mu[/itex] since [itex]f(x)[/itex] is a probability distribution function. Thus [itex]\mu_1 = 0[/itex].
  8. Apr 14, 2007 #7
    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...

    [tex]\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[/tex]

    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
  9. Apr 14, 2007 #8

    D H

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

    In other words, this is not valid:

    [tex]\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[/tex]

    because [itex]\int f(x,y) dy[/itex] is a function of x.
  10. Apr 14, 2007 #9
    I think we're saying essentially the same thing with different examples.


    [tex]\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[/tex]


    [tex]\int_{h_1(x)}^{h_2(x)} f(y) dy[/tex]

    also a function of x?
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