2
$\begingroup$

Assume $$X\sim \frac{e^{-\beta Nf(x)}}{Z_{\beta}}$$ where $$ f(x) = -hx -x^2 + \frac{1}{\beta N}(1-x)\ln(1-x) + (1+x)\ln(1+x) $$ and $Z_{\beta}$ is the appropriate normalization factor. The support of $X$ is the lattice $\Gamma_N = \{-1,-1+2/N,\ldots,1-2/N,1\}$ and $N$ is some large positive integer and $\beta>1$.

I would like to know if there's any way to characterize the first three moments of $X$, that is

$$ \mathbb{E}[X], \mathbb{E}[X^2],\mathbb{E}[X^3]$$

as functions of $\beta$ and whatever else comes out, but in particular I'm interested in the overall behaviour when $\beta$ grows large.

If you are familiar with statistical physics this is the Curie Weiss model (or a mean field Ising model).

$\endgroup$
1
$\begingroup$

Since this model constitutes an exponential family, the moments can be derived from the moment generating function of the sufficient statistic: $$\Psi(u)=\mathbb E_\beta[e^{u f(X)}]=\int e^{u f(x)-\beta N f(x)-\log Z_\beta} \text{d} x=e^{\log Z_{\beta-u/N} -\log Z_\beta}$$ Then $$\Bbb E_\beta[f(X)^k] = \frac{\text{d}^k}{\text{d}u^k}\Psi(u) =e^{-\log Z_\beta}\frac{\text{d}^k}{\text{d}u^k} e^{\log Z_{\beta-u/N}}$$ Obviously, this does not answer the question, but I do not believe there is a generic formula for the moments of $X$ itself since this depends on the choice of the representation of $X$.

$\endgroup$
  • $\begingroup$ Notice though that I'm not looking for the expectation of f(X) but rather that of X $\endgroup$ – Three Diag Apr 22 at 19:00
  • $\begingroup$ I'm afraid so. I also wonder what these moment genereting formula are useful for, in the end they will just be a way to write out the definition of the moment $\sum_x g(x) \frac{e^{-\beta f(x)}}{Z_{\beta}}$ where do they ever come in handy? $\endgroup$ – Three Diag Apr 23 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.