AM-GM-HM Inequality: A Statistical Point of View

In this post we shall give another proof of the famous AM-GM-HM inequality:

If x_1,\cdots,x_n are positive real numbers, then AM \geq GM \geq HM, precisely

\displaystyle \frac{1}{n} \sum_{i=1}^n x_i\geq \sqrt[n]{\prod_{i=1}^n x_i} \geq \frac{n}{\sum_{i=1}^n x_i^{-1}}.

I remember in high school the AM-GM inequality is proved using direct induction or calculus. Later on I knew proofs using backward induction, convexity, Jensen’s inequality etc (See the above Wikipedia link for details). The following proof, which I learnt in a statistics class, comes from Stefanski (1996). The only prerequisite is 100-200 level calculus.

We first make the proof mathematics-like. After that, we describe its meaning in statistics.

Proof. Fix x_1,\cdots,x_n>0. Consider a map L_1: \mathbb{R}_{++}^n \rightarrow \mathbb{R} given by

\displaystyle L_1 (\lambda_1,\cdots,\lambda_n) = \left(\prod_{i=1}^n \lambda_i \right) \exp \left( - \sum_{i=1}^n \lambda_i x_i\right),

where \mathbb{R}_{++} \triangleq \{\lambda\in \mathbb{R}: \lambda > 0\}.

By standard differentiation techniques we learnt in calculus,

\displaystyle \max_{(\lambda_1,\cdots,\lambda_n) \in \mathbb{R}_{++}^n} L_1(\lambda_1,\cdots,\lambda_n) = \left(\prod_{i=1}^n x_i \right)^{-1} e^{-n}

and

\displaystyle \max_{\lambda>0} L_1(\lambda,\cdots,\lambda) = \left(\frac{1}{n} \sum_{i=1}^n x_i \right)^{-n} e^{-n}.

Using the fact \displaystyle \max_{\lambda>0} L_1(\lambda,\cdots,\lambda)\leq \max_{(\lambda_1,\cdots,\lambda_n) \in \mathbb{R}_{++}^n} L_1(\lambda_1,\cdots,\lambda_n) we get the AM-GM inequality.

Next, we consider L_2: \mathbb{R}_{++}^n \rightarrow \mathbb{R} given by

\displaystyle L_2 (\lambda_1,\cdots,\lambda_n) = \frac{\prod_{i=1}^n \lambda_i}{\left(\prod_{i=1}^n x_i\right)^2}\exp\left( -\sum_{i=1}^n \frac{\lambda_i}{x_i}\right).

Then

\displaystyle \max_{(\lambda_1,\cdots,\lambda_n) \in \mathbb{R}_{++}^n} L_2(\lambda_1,\cdots,\lambda_n)= \left(\prod_{i=1}^n x_i\right)^{-1}e^{-n}

and

\displaystyle \max_{\lambda>0} L_2(\lambda,\cdots,\lambda)=\frac{n^n}{\left(\sum_{i=1}^n x_i^{-1}\right)^n \left( \prod_{i=1}^n x_i\right)^2}e^{-n}.

The fact \displaystyle \max_{\lambda>0} L_2(\lambda,\cdots,\lambda)\leq \max_{(\lambda_1,\cdots,\lambda_n) \in \mathbb{R}_{++}^n} L_2(\lambda_1,\cdots,\lambda_n) implies the GM-HM inequality. The proof is completed.          \square

The function L_1 is indeed the joint probability density function (also called likelihood) of independent random variables X_i\sim Exponential(\lambda_i) with density x_i \mapsto \lambda_i e^{-\lambda_i x_i}\delta_{(0,+\infty)}(x_i) (\delta is the indicator function). Then for a fixed (x_1,\cdots,x_n),

\displaystyle \Lambda_1 \triangleq \frac{\max_{\lambda>0} L_1(\lambda,\cdots,\lambda)}{\max_{(\lambda_1,\cdots,\lambda_n) \in \mathbb{R}_{++}^n} L_1(\lambda_1,\cdots,\lambda_n)}

is the Likelihood Ratio Test statistic of testing the null hypothesis H_0:\lambda_1=\cdots=\lambda_n against the alternative hypothesis H_1: \lambda_i are not all equal. Obviously 0\leq \Lambda_1\leq 1. The test rejects the null hypothesis if and only if \Lambda_1 is too small. This is because, if the numerator of \Lambda_1, which is the likelihood when H_0 is true, is small then H_0 is less likely to occur.

The function L_2 is the likelihood of independent random variables X_i^{-1} where X_i\sim Exponential(\lambda_i). Then given (x_1,\cdots,x_n),

\displaystyle \Lambda_2 \triangleq \frac{\max_{\lambda>0} L_2(\lambda,\cdots,\lambda)}{\max_{(\lambda_1,\cdots,\lambda_n) \in \mathbb{R}_{++}^n} L_2(\lambda_1,\cdots,\lambda_n)}

is the Likelihood Ratio Test statistic of testing the null hypothesis H_0:\lambda_1=\cdots=\lambda_n against the alternative hypothesis H_1: \lambda_i are not all equal. The test rejects the null hypothesis if and only if \Lambda_2 is too small.

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This entry was posted in Applied mathematics, Calculus, Optimization, Statistics. Bookmark the permalink.

One Response to AM-GM-HM Inequality: A Statistical Point of View

  1. NOOR UN NISA says:

    what is the basic reason of am >gm ???????
    why am >gm?????

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