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Characteristic functions and functionals

  1. Apr 22, 2014 #1

    Matterwave

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    Hello guys, I posted this question in the classical mechanics forum (thinking stochastic mechanics falls into the classical mechanics category) but I haven't gotten an answer there. I was told I'd be better off posting the question here. I don't know how to move a thread, so I'll just copy and paste from the other thread. Hopefully that's OK?

    Hi, I reading a book on functional integration, and to develop its methods, the author takes us first through stochastic methods (which I'm guessing in the later parts of the book will lead to functionals and functional integrals, etc.). I'm just at the beginning parts, and he has defined a characteristic function of a stochastic variable X as:

    $$C(t)\equiv \left<e^{itx}\right>=\int_{-\infty}^\infty e^{itx}d\mu(x)$$

    Where ##\mu(x)## is the probability measure on X. The author gives immediately 4 properties of the characteristic function:

    1. ##C(0)=1##

    2.## |C(t)|\leq 1##

    3. ##\Sigma_{j,k=1}^{K,K}\alpha_j^*\alpha_k C(t_k-t_j) \geq 0,\qquad \alpha_j\in \mathbb{C}##

    4. ##C(t)## is continuous.

    He goes on to say that properties 1-3 are trivial, and proves property 4. I can understand property 1 and 2, but I can't seem to make heads or tails of property 3. What is that property trying to say? Thanks.


    I'll put a second question I found in another chapter of the book (not dealing with characteristic functions) here instead of starting a new thread.

    In this chapter, we are looking at products of probability measures:

    $$d\mu_N(x)=\prod_{n=1}^N\sqrt{\frac{b}{2\pi}}e^{-bx_n^2/2}dx_n$$

    Giving:

    $$d\mu_N(x)=\left(\frac{b}{2\pi}\right)^{N/2}e^{-b\sum_{n=1}^N x_n^2/2}\prod_{n=1}^N dx_n$$

    And we take ##d\mu(x)=\lim_{N\rightarrow\infty}d\mu_N(x)##.

    In the previous section, the author just gave some "bad examples" on how not to take this limit, with explicit examples showing how if we assume that the probability measure is supported on only sequences such that ##\sum_{n=1}^\infty x_n^2<\infty## then our probability measure is only finitely additive. He then goes on to say that we should adopt the view that the measure is supported on a space such that the sum diverges to infinity almost everywhere (?). He does not give the definition of this support space yet.

    Here I have a few questions. First of all, the author seemed to have concentrated mostly on the prefactors in his discussion, with only a passing remark that:

    $$\mathfrak{D}x\equiv \prod_{n=1}^\infty dx_n$$

    Is not a measure because it does not satisfy complete additivity. If this is not a measure, what hope do we have of making ##d\mu## a probability measure? Will the prefactors some how turn it into a measure? Secondly, when the author is talking about "sequences of x_n", we are, in the end, going to integrate over all x_n, for all n, so I don't understand how a "sequence" is coming into play here (unless he was maybe talking about only integrating over some x_n ranges, different for different n, to produce a sequence?).

    Lastly, when he goes on to talk about support properties (which I was hoping to have some of my questions answered in this section) I run into a problem with the first equality he makes. He says, "we note that:

    $$\int x_n^2 d\mu(x)=\int_{-\infty}^\infty y^2\sqrt{\frac{b}{2\pi}}e^{-by^2/2}dy=b^{-1}$$

    independently of n." I don't understand this at all. What happened to the infinite product defined earlier? How did he turn all of those integrals over all the infinitely many x_n into 1 single integral? I have no idea what to make of this. Thanks.
     
    Last edited by a moderator: Apr 23, 2014
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  3. Apr 23, 2014 #2

    jbunniii

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    What this means is that if you choose any set of times (or whatever the ##t## represents), ##t_1,\ldots,t_K##, and form a matrix ##M## whose ##j,k##'th element is ##[M]_{j,k} = C(t_k - t_j)##, then ##M## is positive semidefinite. This means, by definition, that ##C## is a positive-definite function:

    http://en.wikipedia.org/wiki/Positive_definite_function

    It's quite easy to prove: just substitute the integral expression for ##C(t_k-t_j)## and interchange the order of integration and summation.
     
    Last edited: Apr 23, 2014
  4. Apr 23, 2014 #3

    Matterwave

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    Ah, I just needed a more intuitive understanding of what he was trying to say with that property. This was really helpful, thanks!

    Any hints on the latter questions?
     
  5. Apr 23, 2014 #4

    micromass

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    Matterwave, could you perhaps tell us which book you are reading and on which page all of this is happening?
     
  6. Apr 23, 2014 #5

    micromass

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    To explain this, notice that

    [tex]\int_{-\infty}^{+\infty} \sqrt{\frac{b}{2\pi}} e^{-bx^2/2} dx = 1[/tex]

    So take any ##n## and let ##k\leq n##. Thus if we introduce

    [tex]d\nu_n(x_n) = \sqrt{\frac{b}{2\pi}} e^{-bx_n^2/2}dx_n[/tex]

    then we see that

    [tex]\int_{-\infty}^{+\infty} d\nu_n(x_n) = 1[/tex]

    So, if we want to calculate

    [tex]\int_{-\infty}^{+\infty} x_k^2 d\mu_n(x)[/tex]

    then we need to calculate

    [tex]\int_{-\infty}^{+\infty}... \int_{-\infty}^{+\infty} x_k^2 d\nu_1(x_1)....d\nu_n(x_n)[/tex]

    We can reshuffle the integrals so that ##d\nu_k(x_k)## comes last. We get

    [tex]\int_{-\infty}^{+\infty} x_k^2\int_{-\infty}^{+\infty}... \int_{-\infty}^{+\infty} d\nu_1(x_1)....d\nu_n(x_n)d\nu(x_k)[/tex]

    Now we can apply the above result with the ##d\nu_n(x_n)## to obtain that this is equal to

    [tex]\int_{-\infty}^{+\infty} x_k^2 d\nu_k(x_k)[/tex]

    Thus,

    [tex]\int_{-\infty}^{+\infty} x_k^2 \sqrt{\frac{b}{2\pi}} e^{-bx_k^2/2}dx_k[/tex]

    which can then be calculated to yield ##b^{-1}##. So we conclude that

    [tex]\int_{-\infty}^{+\infty} x_k^2 d\mu_n(x) = b^{-1}[/tex]

    Now by taking limits, we see that

    [tex]\int_{-\infty}^{+\infty} x_k^2 d\mu(x) = b^{-1}[/tex]
     
  7. Apr 23, 2014 #6

    Matterwave

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    I'm reading John Klauder's "A Modern approach to Functional Integration". This material is in chapter 2 and 3.

    Thanks for that explanation, that was clear. :)

    I'm still a little confused on what he means when he discusses "sequences" of ##x_n## though since we are going to integrate over all ##x_n##, for every n, from -infinity to +infinity, in the end. This is around section 3.1 and 3.2.
     
  8. Apr 24, 2014 #7

    micromass

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    He wants to define a measure on an infinite-dimensional space. This is a space of sequences. So let ##\mathbb{R}^\mathbb{N}## by definition all the sequences in ##\mathbb{R}##, we want to define integrals of functions ##\mathbb{R}^\mathbb{N}\rightarrow \mathbb{R}##.
     
  9. Apr 24, 2014 #8

    Matterwave

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    Are all the sequences in ##\mathbb{R}## "filling" ##\mathbb{R}^N## so to speak? It seems then that if you preclude certain classes of sequences (because they mess up your measure), then you are leaving "holes" in ##\mathbb{R}^N## or some such?
     
  10. Apr 24, 2014 #9

    micromass

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    Yes, you won't get a measure on entire ##\mathbb{R}^\mathbb{N}##. But I do think that he ends up with a measure on ##\ell^2##, which is also a good thing. The space ##\ell^2## is actually complete, so there are no holes. The space ##\mathbb{R}^\mathbb{N}## is of course a lot larger, maybe too large.
     
  11. Apr 24, 2014 #10

    Matterwave

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    Hmmm, ok, thanks for the help! :)
     
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