A lower bound for the partition function from Chebyshev's inequality applied to a coin flipping model for the random partition

TL;DR

This paper establishes a lower bound for the growth rate of the partition function p(n) using a coin flipping model and Chebyshev's inequality, providing a new probabilistic approach to partition theory.

Contribution

It introduces a novel probabilistic method employing Chebyshev's inequality to derive a lower bound for the partition function p(n).

Findings

Proves that lim (log p(n))/sqrt(n) ≥ C for an explicit constant C.

Uses a coin flipping model to analyze the random partition.

Provides a new lower bound for the partition function growth rate.

Abstract

We use a coin flipping model for the random partition and Chebyshev's inequality to prove the lower bound $\lim \frac{\log p(n)}{\sqrt{n}} \ge C$ for the number of partitions $p(n)$ of $n$, where $C$ is an explicit constant.

Equations

$$\frac{\log p(n)}{\sqrt{n}}>\frac{\sqrt{8}\log 2}{1+\epsilon }\text{for all n≥N}.$$

$$C_t=\{\begin{array}{ll}{X}_{t-1}& \text{if ωt=T}\\ 0& \text{if ωt=H.}\end{array}$$

$$\mathbf{E}[C_t{X}_u]$$

$$\displaystyle=\frac{1}{2}\bigl{(}\mathbf{E}[X_{t-1}^{2}]+\mathbf{E}[X_{t-1}]\mathbf{E}[W]\bigr{)}$$

$$\displaystyle=\frac{1}{2}\bigl{(}\operatorname{\mathbf{Var}}X_{t-1}+\mathbf{E}[X_{t-1}]^{2}+\mathbf{E}[X_{t-1}]\mathbf{E}[W]\bigr{)}$$

$$\displaystyle=\frac{1}{2}\left(\frac{t-1}{4}+\Bigl{(}\frac{t-1}{2}\Bigr{)}^{2}+\Bigl{(}\frac{t-1}{2}\Bigr{)}\Bigl{(}\frac{u-t}{2}\Bigr{)}\right)$$

$$\displaystyle=\frac{1}{2}\Bigl{(}\frac{t-1}{4}\Bigr{)}\bigl{(}1+t-1+u-t\bigr{)}$$

$$=\mathbf{E}[C_{t-1}]\mathbf{E}[X_u]$$

$$\mathbf{E}[C_t{C}_u]=\frac{1}{2}\mathbf{E}[X_{t-1}{C}_u]=\frac{1}{2}\mathbf{E}[X_{t-1}]\mathbf{E}[C_u]=\mathbf{E}[C_t][C_u]$$

$$\begin{split}\operatorname{\mathbf{Var}}C_{t}=\frac{1}{2}\operatorname{\mathbf{Var}}X_{t-1}+\frac{1}{4}\mathbf{E}[X_{t-1}]^{2}&=\frac{1}{2}\Bigl{(}\frac{t-1}{4}\Bigr{)}+\frac{1}{4}\Bigl{(}\frac{t-1}{2}\Bigr{)}^{2}\\ &\qquad=\Bigl{(}\frac{t-1}{16}\Bigr{)}\bigl{(}2+t-1\bigr{)}=\frac{t^{2}-1}{16}.\end{split}$$

$$\mathrm{Var}N=\frac{m}{4}+\sum _{t=1}^m\frac{t^2-1}{16}=\frac{3m}{16}+\frac{m(m+1)(2m+1)}{96}=\frac{m^3}{48}+\frac{m^2}{32}+\frac{19m}{96}.$$

$$\mathbf{P}\Bigl{[}\,\bigl{|}Z-\mathbf{E}[Z]\bigr{|}\geq d\sqrt{\operatorname{\mathbf{Var}}Z}\,\Bigr{]}\leq\frac{1}{d^{2}}$$

$$\mathbf{P}\Bigl{[}\bigl{|}N-\frac{m(m+3)}{8}\bigr{|}\geq d\frac{m^{3/2}}{4}\Bigr{]}\leq\frac{1}{d^{2}}$$

$$\sum_{n}p(n)>2^{m}\bigl{(}1-\frac{1}{d^{2}}\bigr{)}$$

$$p\Bigl{(}\frac{m(m+3)}{8}+d\frac{m^{3/2}}{4}\Bigr{)}>\frac{2^{m}}{dm^{3/2}}2\bigl{(}1-\frac{1}{d^{2}}\bigr{)},$$

$$p\Bigl{(}\frac{m^{2}}{8}(1+\eta)\Bigr{)}>2^{m}\frac{4}{3\sqrt{3}m^{3/2}}$$

$$\log p(n)\ge \sqrt{\frac{8n}{1+\eta }}\log 2-\frac{3}{2}\log \frac{8n}{1+\eta }+\log \frac{4}{3\sqrt{3}}$$

$$\frac{\log p(n)}{\sqrt{n}}>\frac{\sqrt{8}\log 2}{1+\epsilon }$$

$$p(n)\ge \frac{2^{8\sqrt{n}/3}}{2^{5/2}{n}^{3/4}}\text{for all n≥2.}$$

Full text

A partition of a non-negative integer $n\in\mathbf{N}_{0}$ is a decreasing sequence of natural numbers whose sum is $n$. Let $p(n)$ be the number of partitions of $n$. For example, $p(5)=7$ counts the partitions $(5)$, $(4,1)$, $(3,2)$, $(3,1,1)$, $(2,2,1)$, $(2,1,1,1)$ and $(1,1,1,1,1)$. In this note we use a model for the random partition to prove that for all $\varepsilon>0$ there exists $N\in\mathbf{N}$ such that

[TABLE]

We end with an explicit bound that replaces $\sqrt{8}\log 2$ with the slightly smaller constant ${\textstyle\frac{8}{3}}\log 2$. The proof of ($\star$) is self-contained and intended to be readable by anyone knowing the basics of probability theory.

The asymptotically correct result is $\lim_{n\rightarrow\infty}\frac{\log p(n)}{\sqrt{n}}=2\sqrt{\pi^{2}/6}$. The upper bound $\log p(n)\leq 2\sqrt{\pi^{2}/6}\sqrt{n}$ is relatively easy to prove—see for instance Theorem 15.7 in [5]—but getting a tight lower bound is much more challenging. A fairly lengthy proof using only real analysis was given by Erdős in [2]. Our proof is motivated by the model for the random partition in [1, §4.3], and by the abacus notation for partitions (see [3, page 79]). The latter was used in [4] to prove the uniform lower bound $p(n)\geq\mathrm{e}^{2\sqrt{n}}/14$, and in [6] to prove the upper bound $\log p(n)\leq C(\varepsilon)n^{{\scriptstyle\frac{1}{2}}+\varepsilon}$ for all $\varepsilon>0$. The novel feature here is to combine these motivations to give a simple proof of ($\star$).

The proof begins with a coin flipping model for the random partition. Using linearity of expectation it is easy to show that a partition generated by $m$ flips has expected size about $m^{2}/8$. Critically, the standard is of order $m^{3/2}$. By Chebyshev’s inequality, most of the $2^{m}$ partitions generated by $m$ coin flips have size within a few standard deviations of $m^{2}/8$. This leads easily to the claimed bound.

Coin flipping model

We represent a partition $\lambda$ of length $\ell$ as the set of boxes $\{(i,j):1\leq i\leq\ell,1\leq j\leq\lambda_{i}\}$, forming its Young diagram. We draw Young diagrams in ‘French notation’, so that the box $(i,j)$ is geometrically a unit square with diagonal from $(i-1,j-1)$ to $(i,j)$. For example, the partition $(6,4,2,2)$ of length $4$ is shown in Figure 1 above.

Let $\Omega=\{\mathrm{H},\mathrm{T}\}^{m}$ be the probability space for $m$ flips of an unbiased coin in which each $\omega\in\Omega$ has equal probability $\frac{1}{2^{m}}$. Given $\omega\in\Omega$ with exactly $\ell$ tails, we define the boundary of a corresponding partition $p(\omega)$ of length $\ell$ as follows. Start at $(0,\ell)$ and step right to $(1,\ell)$. Then for each head, step one unit right, and for each tail, step one unit down. For instance if $m=10$ and $\omega=\mathrm{H}\mathrm{T}\mathrm{T}\mathrm{H}\mathrm{H}\mathrm{T}\mathrm{H}\mathrm{H}\mathrm{T}\mathrm{H}$ then $p(\omega)=(6,4,2,2)$; the final head corresponds to a step from $(6,0)$ to $(7,0)$ that is not part of a geometric box.

Let $N$ be the size of $p(\omega)$ and let $X_{t}$ be the number of heads up to and including flip $t$. Let $Y=m-X_{m}$ be the total number of tails; this is the length of $p(\omega)$. A move down at step $t$ adds $X_{t-1}+1$ boxes to the Young diagram. Therefore setting

[TABLE]

we have $N=Y+\sum_{t=1}^{m}C_{t}$.

Expectation and variance

Since $X_{t}$ is distributed binomially as $\mathrm{Bin}(t,{\textstyle\frac{1}{2}})$, we have $\mathbf{E}[X_{t}]=t/2$ and $\operatorname{\mathbf{Var}}X_{t}=t/4$. Hence $\mathbf{E}[Y]=m/2$ and $\operatorname{\mathbf{Var}}Y=m/4$. Observe that $C_{t}=0$ unless flip $t$ is tails. Conditioning on this event shows that $\mathbf{E}[C_{t}]=\frac{1}{2}\mathbf{E}[X_{t-1}]=\frac{t-1}{4}$. Hence, by linearity of expectation, $\mathbf{E}[N]=\mathbf{E}[Y]+\sum_{t=1}^{m}\mathbf{E}[C_{t}]=m/2+\frac{1}{4}\sum_{t=1}^{m}(t-1)=m/2+m(m-1)/8=m(m+3)/8$.

Lemma 1**.**

If $t\leq u$ then the random variables $C_{t}$ and $X_{u}$ are uncorrelated.

Proof.

Again we condition on the event that flip $t$ is tails. In this event, $C_{t}=X_{t-1}$ and $X_{u}=X_{t-1}+W$, where $W$ is the number of heads between flips $t+1$ and $u$, inclusive. Since $W$ is independent of $X_{t-1}$,

[TABLE]

as required. ∎

As a corollary, using a similar conditioning argument, we find that

[TABLE]

whenever $t<u$. Hence $C_{t}$ and $C_{u}$ are uncorrelated for distinct $t$ and $u$. This is perhaps a little surprising, since the inequality $C_{t}\leq C_{u}$ for $t<u$ shows that they are not independent, in general. A final conditioning argument shows that $\operatorname{\mathbf{Var}}C_{t}=\mathbf{E}[C_{t}^{2}]-\mathbf{E}[C_{t}]^{2}=\frac{1}{2}\mathbf{E}[X_{t-1}^{2}]-\frac{1}{4}\mathbf{E}[X_{t-1}]^{2}$, and so

[TABLE]

By Lemma 1, $\mathbf{E}[C_{t}Y]=\mathbf{E}[C_{t}(m-X_{m})]=\mathbf{E}[C_{t}]\mathbf{E}[m-X_{m}]=\mathbf{E}[C_{t}]\mathbf{E}[Y]$ for all $t$. Hence $C_{t}$ and $Y$ are also uncorrelated. If $Z$ and $Z^{\prime}$ are uncorrelated random variables then, by a one-line calculation, $\operatorname{\mathbf{Var}}(Z+Z^{\prime})=\operatorname{\mathbf{Var}}Z+\operatorname{\mathbf{Var}}Z^{\prime}$. We therefore have $\operatorname{\mathbf{Var}}N=\operatorname{\mathbf{Var}}Y+\sum_{t=1}^{m}\operatorname{\mathbf{Var}}C_{t}$ and so

[TABLE]

Critically $\operatorname{\mathbf{Var}}N$ is cubic in $m$, not quartic as one might naively expect. To simplify calculations, we use the crude upper bound $m^{3}/48+m^{2}/32+19m/96\leq 2m^{3}/96+4m^{3}/96=m^{3}/16$ for $m\geq 3$ to get $\operatorname{\mathbf{Var}}N\leq m^{3}/16$.

Lower bound

The concentration of measure estimate in Chebyshev’s inequality

[TABLE]

implies that

[TABLE]

for $m\geq 3$ and any $d>0$.

The probability space $\Omega$ has $2^{m}$ elements. The proportion giving partitions with \bigl{|}N-m(m+3)/8\bigr{|}<dm^{3/2}/4 is more than $1-1/d^{2}$. Since distinct coin flip sequences give distinct partitions, it follows that

[TABLE]

where the sum is over all $n\in\mathbf{N}_{0}$ such that $|n-m(m+3)/8|<dm^{3/2}/4$. Since $p(n)$ is monotonic, we deduce that, for $m\geq 3$,

[TABLE]

where we extend the domain of $p$ to $\mathbf{R}$ by setting $p(x)=p(\lfloor x\rfloor)$. The function d\mapsto\frac{1}{d}\bigl{(}1-\frac{1}{d^{2}}\bigr{)} is maximized when $d=\sqrt{3}$, where it has value ${\textstyle\frac{2}{3\sqrt{3}}}$. Therefore we take $d=\sqrt{3}$. Let $\eta>0$ be given. Provided $m$ is sufficiently large we have $3m/8+\sqrt{3}\hskip 1.0ptm^{3/2}/4<\eta m^{2}/8$. Hence

[TABLE]

for all $m$ sufficiently large. Setting $n=m^{2}(1+\eta)/8$ and taking logs we obtain

[TABLE]

for all $n$ sufficiently large. Since $(\log n)/\sqrt{n}\rightarrow 0$ as $n\rightarrow\infty$ it follows that for all $\varepsilon>0$,

[TABLE]

for all $n$ sufficiently large, as claimed in (A lower bound for the partition function from Chebyshev’s inequality applied to a coin flipping model for the random partition). The constant on the right-hand side is approximately $1.961$, somewhat lower than the asymptotically correct $2\sqrt{\pi^{2}/6}\approx 2.565$. For a concrete lower bound, take $\eta=\frac{1}{8}$ and $m=8\sqrt{n}/3$ in ($\dagger$) to get $p(n)\geq 2^{8\sqrt{n}/3}/2^{5/2}n^{3/4}$ for all $n$ sufficiently large. (One can easily check that $n\geq 10^{6}$ suffices.) Using a computer to check small cases one can show that in fact

[TABLE]