Estimating the AM-GM Inequality

Let I\subset (0,\infty) be an interval, and let f: I\rightarrow (0,\infty) be twice-differentiable. For a parameter \alpha\in\mathbb{R}, consider the function

\displaystyle\varphi(x):=f(x)\cdot x^{-\frac{\alpha\log x}{2}}

We want to figure the values of \alpha for which \varphi is multiplicatively convex (i.e. \varphi(x^{\lambda}y^{1-\lambda})\leq\varphi(x)^{\lambda}\varphi(y)^{1-\lambda}, for all \forall x,y\in I,\forall\lambda\in[0,1]. I leave it as an exercise to the reader to verify that \varphi is multiplicatively convex if and only if \varphi(e^{x}) is convex on \log I. Define \Phi:\log I\rightarrow(0,\infty) by

\displaystyle\Phi(x):=\log\varphi(e^{x})=f(e^{x})+\log(e^{x})^{-\frac{\alpha x}{2}}=f(e^{x})-\dfrac{\alpha x^{2}}{2}

Since \Phi is twice-differentiable, convexity of \Phi is equivalent to \Phi''\geq 0, which implies that \varphi is multiplicatively convex if and only if

\begin{array}{lcl}\displaystyle\dfrac{d^{2}}{dx^{2}}\left[\log f(e^{x})-\dfrac{\alpha x^{2}}{2}\right]&=&\displaystyle\dfrac{d}{dx}\left[\dfrac{f'(e^{x})}{f(e^{x})}e^{x}-\alpha x\right]\\[.7 em]&=&\displaystyle\dfrac{f(e^{x})\left[f''(e^{x})e^{2x}+f'(e^{x})e^{x}\right]-\left(f'(e^{x})\right)^{2}e^{2x}}{\left(f(e^{x})\right)^{2}}-\alpha\geq0\end{array}

Equivalently, \Phi is convex if and only if

\begin{array}{lcl}\displaystyle\alpha(f)&:=&\displaystyle\inf_{x\in\log I}\dfrac{d^{2}}{dx^{2}}\left[\log f(e^{x})\right]\\[.9 em]&=&\displaystyle\inf_{x\in I}\dfrac{x^{2}\left[f(x)f''(x)-(f'(x))^{2}\right]+xf(x)f'(x)}{f(x)^{2}}\\&\geq&\displaystyle\alpha\end{array}

Define \beta(f):=\sup_{x\in I}\dfrac{d^{2}}{dx^{2}}[\log f(e^{x})]. Since for the choice \alpha=\beta(f), we have that \Phi''\leq 0 on \log I and therefore \Phi is concave. We obtain the inequalities

\begin{array}{lcl}\displaystyle\log f\left(\exp\left[\dfrac{1}{n}\sum_{k=1}^{n}\log x_{k}\right]\right)-\dfrac{\alpha(f)}{2n^{2}}\left(\sum_{k=1}^{n}\log x_{k}\right)^{2}&\leq&\displaystyle\dfrac{1}{n}\sum_{k=1}^{n}\log f\left(\exp(\log x_{k})\right)\\&-&\displaystyle\dfrac{\alpha(f)}{2n}\sum_{k=1}^{n}(\log x_{k})^{2},\end{array}

which implies that

\begin{array}{lcl}\displaystyle\dfrac{\left(\prod_{k=1}^{n}f(x_{k})\right)^{\frac{1}{n}}}{f\left(\prod_{k=1}^{n}x_{k}^{\frac{1}{n}}\right)}&\geq&\displaystyle\exp\left(\dfrac{\alpha(f)}{2n^{2}}\left[n\sum_{k=1}^{n}(\log x_{k})^{2}-\left(\sum_{k=1}^{n}\log x_{k}\right)^{2}\right]\right)\\&=&\displaystyle\exp\left(\dfrac{\alpha(f)}{2n^{2}}\sum_{1\leq j<k\leq n}\left(\log x_{j}-\log x_{k}\right)^{2}\right);\end{array}

and

\begin{array}{lcl}\displaystyle\log f\left(\exp\left[\dfrac{1}{n}\sum_{k=1}^{n}\log x_{k}\right]\right)-\dfrac{\beta(f)}{2n^{2}}\left(\sum_{k=1}^{n}\log x_{k}\right)^{2}&\geq&\displaystyle\dfrac{1}{n}\sum_{k=1}^{n}\log f\left(\exp(\log x_{k})\right)\\&-&\displaystyle\dfrac{\beta(f)}{2n}\sum_{k=1}^{n}(\log x_{k})^{2},\end{array}

which implies that

\begin{array}{lcl}\displaystyle\dfrac{\left(\prod_{k=1}^{n}f(x_{k})\right)^{\frac{1}{n}}}{f\left(\prod_{k=1}^{n}x_{k}^{\frac{1}{n}}\right)}&\leq&\displaystyle\exp\left(\dfrac{\beta(f)}{2n^{2}}\left[n\sum_{k=1}^{n}(\log x_{k})^{2}-\left(\log x_{k}\right)^{2}\right]\right)\\&=&\displaystyle\exp\left(\dfrac{\beta(f)}{2n^{2}}\sum_{1\leq j<k\leq n}\left(\log x_{j}-\log x_{k}\right)^{2}\right)\end{array}

In particular, if f(x)=e^{x} restricted to the interval [A,B], where 0<A\leq B<\infty, then \alpha(f)=A, \beta(f)=B.

We can use the preceding estimate of the AM-GM inequality to obtain an inequality for nonnegative L^{\infty} random variables. This inequality sheds light on the probabilistic connections of the AM-GM inequality. Let A and B be as above. If X is the discrete random variable with distribution \mathbb{P}(X=x_{j})=\frac{1}{n} for 1\leq j\leq n, then \mathbb{E}[\log X]=\frac{1}{n}\sum_{j=1}^{n}\log x_{j} and

\displaystyle\text{Var}(\log X)=\dfrac{1}{n}\sum_{j=1}^{n}(\log x_{j})^{2}-\dfrac{1}{n^{2}}\left(\sum_{j=1}^{n}\log x_{j}\right)^{2}=\dfrac{1}{n^{2}}\sum_{1\leq j<k\leq n}\left(\log x_{j}-\log x_{k}\right)^{2}

So we can restate the above inequality in the language of probability theory as

\displaystyle A\text{Var}(\log X)\leq\mathbb{E}[X]-e^{\mathbb{E}[\log X]}\leq B\text{Var}(\log X)

Naturally, we ask if the inequality holds for not necessarily discrete random variables taking values in [A,B]. The answer is yes, as we now show.

Proposition. Let (\Omega,\mathcal{F},\mathbb{P}) be a probability space, and let X:\Omega\rightarrow\mathbb{R} be a random variable taking values in [A,B], where 0<A\leq B. Then

\displaystyle A\leq\dfrac{\mathbb{E}[X]-e^{\mathbb{E}[\log X]}}{\text{Var}(\log X)}\leq B

Proof. Consider the simple function X_{n}=\sum_{k=1}^{{n}}\left[A+(k-1)\frac{B-A}{n}\right]\mathbf{1}_{E_{k}}, where

\displaystyle E_{k}:=\left\{\omega\in\Omega:A+(k-1)\frac{B-A}{n}\leq X(\omega)<A+k\frac{B-A}{n}\right\}

for all 1\leq k\leq n. If we can show that X_{n}\stackrel{L^{1}}\rightarrow X and (\log X_{n})^{2}\stackrel{L^{1}}\rightarrow (\log X)^{2}, then the desired inequality follows from our work above and a limiting argument. But X_{n}\rightarrow X a.s. and by continuity, \log X_{n}\rightarrow \log X a.s. Also, X_{n}\leq X and (\log X_{n})^{2}\leq \max\left\{(\log A)^{2},(\log B)^{2}\right\}. So the desired convergence results follow from the dominated convergence theorem. \Box

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