Moments of Partition Functions of 2d Gaussian Polymers in the Weak Disorder Regime-I

Abstract

Let $W_N(\beta ) = {\mathrm E}_0\left[ e^{ \sum _{n=1}^N \beta \omega (n,S_n) - N\beta ^2/2}\right]$ be the partition function of a two-dimensional directed polymer in a random environment, where $\omega (i,x), i\in \mathbb {N}, x\in \mathbb {Z}^2$ are i.i.d. standard normal and $\{S_n\}$ is the path of a random walk. With $\beta =\beta _N={\hat{\beta }} \sqrt{\pi /\log N}$ and ${\hat{\beta }}\in (0,1)$ (the subcritical window), $\log W_N(\beta _N)$ is known to converge in distribution to a Gaussian law of mean $-\lambda ^2/2$ and variance $\lambda ^2$, with $\lambda ^2=\log (1/(1-{\hat{\beta }}^2))$ (Caravenna et al. in Ann Appl Probab 27(5):3050–3112, 2017). We study in this paper the moments ${{\mathbb {E}}}[W_N( \beta _N)^q]$ in the subcritical window, for $q=O(\sqrt{\log N})$. The analysis is based on ruling out triple intersections.

Appendices

Appendix A: Proof of (19)

First note that $p_{2n}^\star \le p_{2n}(0)$ since, by the Cauchy-Schwarz inequality,

$$\begin{aligned}p_{2n}(x)=\sum _{y} p_n(x-y) p_n(y) \le \sum _y p_n(y)^2 = p_{2n}(0).\end{aligned}$$

Let $p^{(d)}_{2n}$ be the return probability of d-dimensional SRW to 0. A direct computation gives that $p^{(2)}_{2n}=(p^{(1)}_{2n})^2$ (see e.g. [20, Page 184]). We will show that $a_n=\sqrt{2n} p^{(1)}_{2n}$ is increasing. We have,

$$\begin{aligned} a_n =\sqrt{2n} 2^{-2n} \begin{pmatrix} 2n\\ n \end{pmatrix}.\end{aligned}$$

Hence,

$$\begin{aligned} \frac{a_{n+1}}{a_n}= & {} \frac{1}{4} \sqrt{\frac{n+1}{n}} \frac{ (2n+2)(2n+1)}{(n+1)^2}\\= & {} \sqrt{1/(n(n+1))} (n+ (n+1))/2. \end{aligned}$$

Since $(a+b)/2\ge \sqrt{ab}$, we conclude (using $a=n$ and $b=n+1$) that $a_{n+1}/a_n\ge 1$.

Let $p_{2n+1}^{(1)}$ be the probability of the 1-dimensional SRW to come back to 1 in $2n+1$ steps. By the random walk representation [20, Remark in Pg. 185], we have that $p_{2n+1}^\star \le (p^{(1)}_{2n+1})^2$. A similar line of argument to the above shows that $b_n=\sqrt{2n+1}p_{2n+1}^{(1)}$ is increasing in n. Indeed,

$$\begin{aligned} \frac{b_{n+1}}{b_n}&= \frac{1}{4} \frac{\sqrt{2n+3}}{\sqrt{2n+1}} \frac{(2n+3)(2n+2)}{(n+2)(n+1)}\\&= \frac{2n+3}{2\sqrt{(n+1)(n+2)}} \frac{\sqrt{(2n+3)(n+1)}}{\sqrt{(2n+1)(n+2)}}, \end{aligned}$$

where the first fraction is bigger than 1 by the formula $(a+b)/2\ge \sqrt{ab}$, as well as the second fraction by expanding the products.

Now, we know from the local limit theorem that $a_n$ and $b_n$ converge to $2/\sqrt{2\pi }$, thus they are always smaller than this limit. This leads to (19). $\square$

Appendix B: Improved Estimates on $U_N$

When n is taken large enough, the estimate (59) can be improved as follows.

Proposition B.1

There exists $\varepsilon _{n}=\varepsilon (n,{\hat{\beta }})\rightarrow 0$ such that as $n\rightarrow \infty$, uniformly for $N\ge n$,

$$\begin{aligned} U_N(n) = (1+\varepsilon _{n})\frac{{\hat{\beta }}^2}{\left( 1-\hat{\beta }^2 \frac{\log n}{\log N}\right) ^2} \frac{1}{ n \log N}. \end{aligned}$$

(82)

Proof

Since $(S_n^1-S_n^2){\mathop {=}\limits ^{(d)}}(S_{2n})$, we can write

$$\begin{aligned} U_N(n+1) = \sigma _N^2 {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{n} \textbf{1}_{S_{2i} = 0}} {\textbf{1}}_{S^1_{2n}= 0} \right] . \end{aligned}$$

(83)

Consider $\ell =\ell _n = n^{1-\varepsilon _n}$ with $\varepsilon _n = \frac{1}{\log \log n}$, so that $\ell _n=o(n)$ and $\varepsilon _n\rightarrow 0$.

First step: As $n\rightarrow \infty$ with $n\le N$,

$$\begin{aligned} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{n} {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2n}= 0} \right] \sim {\mathrm E}_{0}\left[ e^{ \beta _N^2(\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}+ \sum _{i=n-\ell }^{n} {\textbf{1}}_{S_{2i} = 0})} {\textbf{1}}_{S_{2n}= 0 } \right] . \end{aligned}$$

(84)

We compute the norm of the difference which, using that $|e^{-x}-1|\le |x|$ for $x\ge 0$, is less than

$$\begin{aligned}&{\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{n} {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2n}= 0} \times \beta _N^2 \sum _{j=\ell }^{n-\ell } {\textbf{1}}_{S_{2j} = 0} \right] \\&= \beta _N^2 \sum _{j=\ell }^{n-\ell } {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{j} {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2j} = 0}\right] {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{n-j} \textbf{1}_{S_{2i} = 0}} {\textbf{1}}_{S_{2(n-j)} = 0} \right] . \end{aligned}$$

where we have used Markov’s property in the second line. By (83) and (59), the last sum is smaller than

$$\begin{aligned} C\beta _N^2 \sum _{j=\ell }^{n-\ell } \frac{1}{j} \frac{1}{n-j} \le 2C\beta _N^2 \sum _{j=\ell }^{n/2} \frac{1}{j} \frac{1}{n/2}&\le \frac{1}{n}C'\beta _N^2 \log \left( \frac{n}{\ell _n}\right) \le \frac{1}{n}C''\varepsilon _n =o(n^{-1}). \end{aligned}$$

Since the left hand side of (84) is bigger than $c n^{-1}$ for some constant $c>0$, this shows (84).

Second step: As $n\rightarrow \infty$ with $n\le N$,

$$\begin{aligned} \begin{aligned}&{\mathrm E}_{0}\left[ e^{ \beta _N^2(\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}+ \sum _{i=n-\ell }^{n} {\textbf{1}}_{S_{2i} = 0})} {\textbf{1}}_{S_{2n}= 0 } \right] \\&\sim {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i}=0}} \right] {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=n-\ell }^{n} {\textbf{1}}_{S_{2i} = 0}}{\textbf{1}}_{S_{2n} = 0} \right] . \end{aligned} \end{aligned}$$

(85)

By Markov’s property, we can write the LHS of (85) as

$$\begin{aligned}&\sum _{x\in {\mathbb {Z}}^2} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2\ell } = x} {\mathrm E}_x \left[ e^{\beta _N^2 \sum _{i=n-2\ell }^{n-\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2{n-\ell }}= 0 }\right] \right] \\&= \sum _{x\in {\mathbb {Z}}^2} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2\ell } = x} {\mathrm E}_0 \left[ e^{\beta _N^2 \sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2n-\ell }=x} \right] \right] . \end{aligned}$$

Therefore the difference in (85) writes $\sum _{x\in {\mathbb {Z}}^2} \Delta _x$ with

$$\begin{aligned}&\Delta _x := {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } \textbf{1}_{S_{2i} = 0}} {\textbf{1}}_{S_{2\ell } = x}{\mathrm E}_0 \left[ e^{\beta _N^2 \sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \left( {\textbf{1}}_{S_{2{n-\ell }}= 0 } - {\textbf{1}}_{S_{2{n-\ell }}= x }\right) \right] \right] . \end{aligned}$$

Since ${\mathrm E}_0 \left[ e^{\beta _N^2 \sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \right] \le C({\hat{\beta }})$ by (54), we have

$$\begin{aligned} \sum _{|x|> \sqrt{\ell } n^{\varepsilon /4}} |\Delta _x| \le C\sum _{|x| > \sqrt{\ell } n^{\varepsilon /4}} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2\ell } = x}\right] . \end{aligned}$$

By Hölder’s inequality with $p^{-1}+q^{-1}=1$, and p small enough so that $\sqrt{p} \hat{\beta } <1$,

$$\begin{aligned} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2\ell } = x}\right]&\le {\mathrm E}_{0}\left[ e^{ p\beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}}\right] ^{\frac{1}{p}} p_{2\ell }(x)^{\frac{1}{q}} \le C({\hat{\beta }}) \ell _n^{-1}e^{-\frac{1}{2q} \frac{|x|^2}{\ell _n}}, \end{aligned}$$

for n large enough. Therefore,

$$\begin{aligned} \sum _{|x|> \sqrt{\ell } n^{\varepsilon /4}} |\Delta _x|&\le C \sum _{|x| > \sqrt{\ell } n^{\varepsilon /4}} \ell _n^{-1}e^{-\frac{1}{2q} \frac{|x|^2}{\ell _n}}, \le C e^{-\frac{1}{2q} n^{\varepsilon /2}} = o(n^{-1}). \end{aligned}$$

We now estimate the sum on $\Delta _x$ for $|x| \le \sqrt{\ell } n^{\varepsilon /4}$. We start by bounding the expectation inside the definition of $\Delta _x$:

$$\begin{aligned} \begin{aligned}&{\mathrm E}_0 \left[ e^{\beta _N^2 \sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \left( {\textbf{1}}_{S_{2{n-\ell }}= 0 } - {\textbf{1}}_{S_{2{n-\ell }}= x }\right) \right] \\&= \sum _{y\in {\mathbb {Z}}^2} {\mathrm E}_0 \left[ e^{\beta _N^2 \sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{\ell }= y} \right] \left( p_{2n-2\ell }(y)-p_{2n-2\ell }(y-x)\right) . \end{aligned} \end{aligned}$$

(86)

By the same argument as above, we can prove that the above sum restricted to $|y|\ge \sqrt{\ell }n^{\varepsilon /4}$ is negligible with respect to $n^{-1}$, uniformly for $|x| \le \sqrt{\ell } n^{\varepsilon /4}$. On the other hand, by the local limit theorem we have

$$\begin{aligned} \sup _{|x|\le \sqrt{\ell }n^{\varepsilon /4},|y|\le \sqrt{\ell }n^{\varepsilon /4}} \left| p_{2n-2\ell }(y)-p_{2n-2\ell }(y-x)\right| = o(n^{-1}), \end{aligned}$$

since $\ell _n n^{\varepsilon /2} = n^{1-\varepsilon _n/2}=o(n)$. Thus, the quantity in (86) is bounded uniformly for $|x|\le \sqrt{\ell }n^{\varepsilon /4}$ by

$$\begin{aligned} {\mathrm E}_0 \left[ e^{\beta _N^2 \sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \right] \times o(n^{-1})=o(n^{-1}).\end{aligned}$$

This completes the proof of (85).

Third step: As $n\rightarrow \infty$ with $n\le N$,

$$\begin{aligned} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=n-\ell }^{n} {\textbf{1}}_{S_{2i} = 0}}{\textbf{1}}_{S_{2n} = 0} \right] \sim {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \right] p_{2n}(0). \end{aligned}$$

(87)

Equivalence (87) can be proven by following the same line of arguments as used to prove (85), hence we omit its proof.

Now, combining the three steps leads to the equivalence

$$\begin{aligned} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{n} {\textbf{1}}_{S_{2i} = 0}} {\textbf{1}}_{S_{2n}= 0} \right] \sim {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \right] ^2 p_{2n}(0). \end{aligned}$$

By (55), as $\log \ell \sim \log n$, we have

$$\begin{aligned} {\mathrm E}_{0}\left[ e^{ \beta _N^2\sum _{i=1}^{\ell } {\textbf{1}}_{S_{2i} = 0}} \right] \sim \frac{1}{1-{\hat{\beta }}^2 \frac{\log n}{\log N}}, \end{aligned}$$

and so (82) follows from (83) and the last two displays. $\square$

Appendix C: Khas’minskii’s Lemma for Discrete Markov Chains

The following theorem is another discrete analogue of Khas’minskii’s lemma, compare with Lemma 2.2.

Theorem C.1

Let $(Y_n)_n$ be any markov chain on a discrete state-space E and let $f:E \rightarrow {\mathbb {R}}_+$. Then for all $k\in {\mathbb {N}}$, if

$$\begin{aligned} \eta _0:=\sup _{x\in E} {\mathrm E}_{x} \left[ \sum _{n=1}^k (e^{f(Y_n)}-1)\right] < 1, \end{aligned}$$

(88)

one has

$$\begin{aligned} \sup _{x\in E} {\mathrm E}_{x} \left[ e^{\sum _{n=1}^k f(Y_n)}\right] \le \frac{1}{1-\eta _0}. \end{aligned}$$

(89)

Proof

Denote by $D_n = e^{f(Y_n)}-1$. We have,

$$\begin{aligned}&{\mathrm E}_{x} \left[ e^{\sum _{n=1}^N f(Y_n)}\right] = {\mathrm E}_{x} \left[ \prod _{n=1}^N (1+D_n)\right] = \sum _{p=0}^\infty \sum _{1\le n_1< \dots< n_p \le k}{\mathrm E}_{x} \left[ \prod _{i=1}^p D_{n_i}\right] \\&= \sum _{p=0}^\infty \sum _{1\le n_1< \dots< n_{p-1} \le k} {\mathrm E}_{x} \left[ \prod _{i=1}^{p-1} D_{n_i} {\mathrm E}_{Y_{n_{p-1}}} \left[ \sum _{n=1}^{k-n_{p-1}} D_n \right] \right] \\&{\mathop {\le }\limits ^{(88)}}\sum _{p=0}^\infty \eta _0 \sum _{1\le n_1< \dots < n_{p-1} \le k} {\mathrm E}_{x} \left[ \prod _{i=1}^{p-1} D_{n_i} \right] \le \dots \le \sum _{p=0}^\infty \eta _0^p = \frac{1}{1-\eta _0}. \end{aligned}$$

$\square$

Corollary C.2

Let $(Y_n)_n$ be any markov chain on a discrete state-space E and let $f:E \rightarrow [0,1]$. Then for all $k\in {\mathbb {N}}$, if

$$\begin{aligned} \eta _1:=\sup _{x\in E} {\mathrm E}_{x} \left[ \sum _{n=1}^k f(Y_n)\right] < 1, \end{aligned}$$

(90)

one has

$$\begin{aligned} \sup _{x\in E} {\mathrm E}_{x} \left[ e^{\sum _{n=1}^k f(Y_n)}\right] \le \frac{1}{1-\eta _1}. \end{aligned}$$

(91)

Proof

Simply observe that $e^{f(x)}-1\le e^c f(x)$ and apply Theorem C.1. $\square$