From Gaussian fluctuations to large deviations

The Gaussian variable \(V_N=m_h+G/\sqrt N\), where \(G\sim\mathcal N(0,v_h)\), has continuous density \(g_{N,h}\). Consequently, the probability of an interval is an integral:

In the central region \(|x_k-m_h|=O(N^{-1/2})\), the density changes only a little over one lattice spacing \(2/N\). Thus the integral is approximately \(2g_{N,h}(x_k)/N\), motivating

The ordinary CLT alone controls probabilities of intervals on the \(N^{-1/2}\) scale; relative-error accuracy for individual lattice masses is the local CLT proved from Stirling's formula in the next extra. Importantly, \(\int_{\mathbb R}g_{N,h}(x)\,dx=1\), whereas \(2g_{N,h}/N\) is not itself a probability density. It is evaluated at the lattice points to approximate the mass at each point.

Finally, the Gaussian laws converge to a point mass: for every bounded continuous function \(\psi\),

This means \(\operatorname{Law}(V_N)\Rightarrow\delta_{m_h}\). By the law of large numbers, \(\operatorname{Law}(\widehat m_N)\Rightarrow\delta_{m_h}\) as well. Their agreement near the mean is what the interactive plot illustrates, on the scale of their individual lattice probabilities.

The ordinary CLT gives convergence of probabilities of intervals after rescaling. A probability at one lattice point needs more information. Here it can be computed from the binomial law in Proposition 3.1.

Let \(x_N=-1+2k_N/N\) be a lattice point and \(q_N=k_N/N\). If \(q_N\) stays in a compact subinterval of \((0,1)\), the multiplicative form of Stirling gives

Multiplying by \(p_h^{k_N}(1-p_h)^{N-k_N}\) therefore gives

To examine the CLT region, put \(y_N=\sqrt N(x_N-m_h)/s_h\) and assume \((y_N)\) is bounded. Then \(x_N\to m_h\), \(q_N\to p_h\), and \(p_h(1-p_h)=v_h/4\). The derivatives computed in the derivatives following Proposition 3.1 imply, uniformly for bounded \(y_N\),

The prefactor satisfies \(\sqrt{2\pi Nq_N(1-q_N)}\sim s_h\sqrt{2\pi N}/2\). Combining these estimates yields

This proves the local approximation in the central window. For lattice points \(x_N\to a\in(-1,1)\) with \(a\ne m_h\), the binomial formula retains \(I_h(x_N)\), and \(N^{-1}\log\mu_{N;0,h}(\widehat m_N=x_N)\to-I_h(a)\). Replacing this rate by its quadratic expansion is generally incorrect at the exponential scale.

See the factorial substitutions behind Stirling’s formula.

For every integer \(n\ge1\), Stirling’s formula can be written as

Taking logarithms gives

For \(1\le k\le N-1\), substitute this formula into each factorial:

The linear terms cancel because \(-N+k+(N-k)=0\). If \(q=k/N\), the terms involving \(\log N\) cancel as well:

If \(q\) stays away from zero and one, the logarithmic prefactor is \(O(\log N)\) and the remainder \(r_N-r_k-r_{N-k}\) is \(O(N^{-1})\). This gives both the logarithmic estimate in the main proof and the multiplicative estimate used for the local Gaussian approximation.

Why the estimate is uniform, even at the endpoints. For a stronger estimate uniform in every lattice index, \(k\) may be close to zero or \(N\). Define

Since \(0!=1\) and \(0\log0=0\), the identity \(\log(n!)=n\log n-n+R(n)\) now holds for all integers \(n\ge0\). Stirling implies that \(|R(n)|\le C_0\log(n+1)\) for a constant \(C_0\), uniformly in \(n\). Therefore

It follows that

Finally, add the two log weights:

This is a uniform estimate over all \(N+1\) masses; the fixed-threshold proof above needs only the interior-index case. At \(k=0\) or \(k=N\), the original probability is simply \((1-p_h)^N\) or \(p_h^N\), respectively, in agreement with the rate’s endpoint values. See also Friedli–Velenik, Lemma B.3.

For \(q=(1+a)/2\), the condition \(\widehat m_N\ge a\) is exactly \(K_N\ge Nq\). Since \(K_N\) is integer-valued, the smallest allowed value is \(k_N=\lceil Nq\rceil\). Thus

The ratio of consecutive binomial probabilities is obtained by cancellation:

If \(k\ge k_N\ge Nq\), then \(N-k\le N(1-q)\) and \(k+1\ge Nq\). When \(q>p_h\),

because the last inequality is equivalent to \(p_h(1-q)<q(1-p_h)\), or \(p_h<q\). Therefore \(\mu_{N;0,h}(K_N=k_N)\) is the largest summand in the tail:

The factor \(N+1\) disappears after taking \(N^{-1}\log\), since \(\log(N+1)/N\to0\). This is the precise sense in which one largest term determines the exponential scale of this sum.

For one spin with law \(\rho_{0,h}\),

If \(-1<x<1\), the derivative of \(t\mapsto tx-\Lambda_h(t)\) is \(x-\tanh(h+t)\). It vanishes at the unique point

The second derivative is \(-1/\cosh^2(h+t)<0\), so this critical point is the global maximum. Since \(\cosh(\operatorname{arctanh}x)=1/\sqrt{1-x^2}\), substitution gives

Use \(\log(2p_h)=h-\log\cosh h\) and \(\log(2(1-p_h))=-h-\log\cosh h\). Expanding the logarithms in Proposition 3.1 now shows that this expression is exactly \(I_h(x)\).

At \(x=1\), the function \(t-\Lambda_h(t)\) is increasing, and its supremum is the limit as \(t\to+\infty\):

At \(x=-1\), the derivative of \(t\mapsto-t-\Lambda_h(t)\) is \(-1-\tanh(h+t)<0\). Hence this function is decreasing, and its supremum is its limit as \(t\to-\infty\), equal to \(-\log(1-p_h)\). If \(x>1\), then \(tx-\Lambda_h(t)\to+\infty\) as \(t\to+\infty\); if \(x<-1\), let \(t\to-\infty\). Thus \(\Lambda_h^*=I_h\) on all of \(\mathbb R\), including the endpoints and the infinite values outside the spin range.

For a function \(J:\mathbb R\to[0,+\infty]\), lower semicontinuity means \(J(x)\le\liminf_nJ(x_n)\) whenever \(x_n\to x\). Equivalently, every finite sublevel set \(\{x:J(x)\le c\}\) is closed.

To prove the forward implication, take \(x_n\) in that sublevel set with \(x_n\to x\). Then

so \(x\) remains in the set. Conversely, suppose the sublevel sets are closed but lower semicontinuity fails along \(x_n\to x\). Choose a finite \(c\) strictly between \(\liminf_nJ(x_n)\) and \(J(x)\). Infinitely many \(x_n\) then lie in \(\{J\le c\}\); a subsequence of those points still converges to \(x\). Closedness gives \(J(x)\le c\), a contradiction.

For a Legendre transform, there is also a direct set formula:

Each set on the right is closed because \(x\mapsto tx-\Lambda(t)\) is continuous, and an arbitrary intersection of closed sets is closed.

Let \(\mathbb P(Z_N=-1)=\mathbb P(Z_N=1)=1/2\) for every \(N\). For any Borel set \(B\), its probability is zero if \(B\) misses \(\{-1,1\}\), is \(1/2\) if it contains exactly one of these points, and is one if it contains both. Therefore

This equals \(-\inf_{x\in B}J(x)\) for the function in Remark 3.6. This verifies the claimed exponential rate directly for every fixed Borel set. Every finite nonempty sublevel set is \(\{-1,1\}\), so \(J\) is lower semicontinuous and good. Its failure of convexity follows from the midpoint inequality already displayed in the remark.

Now compute the scaled log moment-generating function:

The inequalities \(\tfrac12e^{N|t|}\le\cosh(Nt)\le e^{N|t|}\) give

so \(\Lambda(t)=|t|\). Its Legendre–Fenchel transform is

Indeed, if \(|x|\le1\), then \(tx-|t|\le|t||x|-|t|\le0\), and \(t=0\) attains zero. If \(x>1\), let \(t\to+\infty\); if \(x<-1\), let \(t\to-\infty\). In both exterior cases the supremum is infinite.

In particular, \(\Lambda^*(0)=0\) although \(J(0)=+\infty\). The transform assigns zero cost to the entire interval between the two support points, whereas the actual laws never visit any interior point. The limiting \(\Lambda\) has left derivative \(-1\) and right derivative \(+1\) at zero, so the differentiability hypothesis in Remark 3.4 fails. Existence of the scaled log-mgf limit alone does not identify a general LDP rate with its Legendre transform.