Upper bound on ratio of probability sums

Question

Let us say I have some vector of probabilities $\vec{V}$ such that $\sum_i v_i = 1$. I am trying to find a sharp upper bound on

$\frac{\sum_i v_i^3}{(\sum_i v_i^2)^2}$. I can currently prove via Cauchy-Schwarz a very bad upper bound - this ratio must be less than the number of entries in the vector. But I have numerical evidence that leads me to believe that something like 1.5 should be a sharp upper bound, not dependent on the vector length.

Answer 1

Here is a upper bound:

Assume that $n \ge 3$. Denote $S = \{(v_1, v_2, \cdots, v_n) : ~ v_i \ge 0, \forall i; ~ v_1 + v_2 + \cdots + v_n = 1\}.$

We have $$\max_{(v_1, v_2, \cdots, v_n) \in S} \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2} \le \frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n}. \tag{1}$$

The bound (1) satisfies that $$\lim_{n\to \infty} \left(\frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n} - \max_{(v_1, v_2, \cdots, v_n) \in S} \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2}\right) = 0. \tag{2}$$

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Proof of (1) and (2):

Consider the maximum of $$f(v_1, v_2, \cdots, v_n) = \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2}$$ subject to $v_i \ge 0, \forall i; ~ \sum_{i=1}^n v_i = 1$.

Using Vasc's Equal Variable Theorem (Corollary 1.9, [1]), $f$ is maximal when $0 \le v_1 = v_2 = \cdots = v_{n - 1} \le v_n$.

Letting $v_n = x$ and $v_1 = v_2 = \cdots = v_{n-1} = \frac{1 - x}{n-1}$, we have $$\max_{(v_1, v_2, \cdots, v_n) \in S} \frac{\sum_i v_i^3}{(\sum_i v_i^2)^2} = \max_{x\in [0, 1]} \frac{(n^2 - 2n)x^3 + 3x^2 - 3x + 1}{(nx^2 - 2x + 1)^2}.$$

It suffices to prove that, for all $x\in [0, 1]$ and $n \ge 3$, $$\frac{(n^2 - 2n)x^3 + 3x^2 - 3x + 1}{(nx^2 - 2x + 1)^2} \le \frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n}.$$

Letting $m = \sqrt{3n} - 3\ge 0$. After clearing the denominators, it suffices to prove that, for all $x \in \mathbb{R}$, $$a_1 x^4 + b_1x^3 + c_1x^2 + d_1 + e_1 \ge 0 \tag{3}$$ where \begin{align*} a_1 &= 6\,{m}^{6}+128\,{m}^{5}+1115\,{m}^{4}+5100\,{m}^{3}+12960\,{m}^{2}+ 17388\,m+9639, \\ b_1 &= -32\,{m}^{5}-552\,{m}^{4}-3792\,{m}^{3}-13020\,{m}^{2}-22392\,m-15444, \\ c_1 &= 36\,{m}^{4}+552\,{m}^{3}+3270\,{m}^{2}+8460\,m+8118,\\ d_1 &= -216\,{m}^{2}-1152\,m-1692, \\ e_1 &= 54\,{m}^{2}+216\,m+207. \end{align*}

We need the following auxiliary result.

Fact 1: Let $a_1 > 0, b_1, c_1, d_1, e_1$ be real numbers. Let $f(x) = a_1x^4 + b_1x^3 + c_1x^2 + d_1x + e_1$. Let $\Delta_f$ denote its discriminant. Let $D_f = 64a_1^3e_1-16a_1^2c_1^2+16a_1b_1^2c_1 -16a_1^2b_1d_1-3b_1^4$. If $\Delta_f \ge 0$ and $D_f > 0$, then $f(x)\ge 0$ for any real number $x$.
(See https://en.wikipedia.org/wiki/Quartic_function and the reference [13] therein, i.e., Rees, E. L. (1922). "Graphical Discussion of the Roots of a Quartic Equation". The American Mathematical Monthly. 29 (2): 51–55.)

We have \begin{align*} \Delta_f &= 63700992\, \left( 6\,{m}^{2}+24\,m+23 \right)\\ &\qquad \times \left( 512\,{m}^{5}+ 5886\,{m}^{4}+26604\,{m}^{3}+57887\,{m}^{2}+58242\,m+19326 \right) \\ &\qquad \times \left( {m}^{2}+6\,m+6 \right) ^{4} \left( m+3 \right) ^{6}\\ &> 0 \end{align*} and \begin{align*} D_f &= 12288\, \left( 4\,{m}^{6}+96\,{m}^{5}+916\,{m}^{4}+4460\,{m}^{3}+11389 \,{m}^{2}+13734\,m+5262 \right) \\ &\qquad \times \left( 8\,{m}^{5}+138\,{m}^{4}+900\,{ m}^{3}+2769\,{m}^{2}+4086\,m+2358 \right) \left( m+3 \right) ^{9}\\ &> 0. \end{align*}

By Fact 1, (3) is true.

(2) is true since $$\lim_{n\to \infty} \left(\frac{3\sqrt 3}{16}\sqrt n + \frac58 + \frac{5\sqrt 3}{96\sqrt n} - f\Big(\sqrt{3/n}\Big)\right) = 0$$ where $$f(x) := \frac{(n^2 - 2n)x^3 + 3x^2 - 3x + 1}{(nx^2 - 2x + 1)^2}.$$

We are done.

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Reference:

[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007. Online: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf

Answer 2

Empirically it looks to me as if there is an upper bound which is $O(\sqrt{n})$ and in particular your expression can be above $0.32 \sqrt{n}$ but will never exceed $\frac{\sqrt{n}+2}{3}$ for all vector lengths $n$.

So for example with a vector of length $n=100$, this suggests the maximum possible is somewhere between $3.2$ and $4$. Consider the case where $x_1=0.1585$ and the other ninety-nine $x_i$ are all equal to $0.0085$; your expression would then be about $3.88$.

I suspect it would be possible to prove that the maximal case has

• one value is $x$ and the other $n-1$ values are $\frac{1-x}{n-1}$
• where $x$ is the largest root of $(n^2-2n)x^3+3nx^2-3nx+1=0$

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Added later

Based on River Li and Michael Rozenberg's anaysis, we can tighten this: the maximum possible value is between $$\frac{3\sqrt 3}{16}\sqrt{n} +\frac58 \qquad\text{ and } \qquad\frac{3\sqrt 3}{16}\sqrt{n} +1-\frac{3\sqrt 3}{16}$$ so numerically between $$0.32475\sqrt{n} +0.625\qquad\text{ and } \qquad 0.32476\sqrt{n} +0.67524.$$ My empirical results were consistent with this.

Answer 3

The Henry's suspecting is right!

Indeed, let also $\sum\limits_{i=1}^nv_1^2=constant$ and $n>1$.

Thus, by the Vasc's E-V Method(see here https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf Corollary 1.5)

it's enough to find a maximal value for $v_1=v_2=...=v_{n-1}\leq v_n$.

From here, as Henry says, we have, $v_n=x$, where $x\in[0,1]$, $$v_1=...=v_{n-1}=\frac{1-x}{n-1}$$ and $$\frac{1-x}{n-1}\leq x,$$ which gives $$\frac{1}{n}\leq x\leq1.$$ Thus, we need to find $$\max_{\frac{1}{n}\leq x\leq1}\frac{\frac{(1-x)^3}{(n-1)^2}+x^3}{\left(\frac{(1-x)^2}{n-1}+x^2\right)^2}.$$ Now, $$\left(\frac{\frac{(1-x)^3}{(n-1)^2}+x^3}{\left(\frac{(1-x)^2}{n-1}+x^2\right)^2}\right)'=\frac{(1-nx)((n^2-2n)x^3+3nx^2-3nx+1)}{(nx^2-2x+1)^3}.$$ We can show that for $n>2$ $$x_{max}=\frac{2\sqrt{n-1}\cos\frac{\pi-\arccos\frac{2\sqrt{n-1}}{n}}{3}-1}{n-2}.$$ For $n=2$ $x_{\max}=\frac{1+\sqrt3}{2\sqrt3}.$

I hope it will help.