inequality between entropy measure and Herfindahl index

Question

I am considering the entropy measure and the Herfindahl-Hirschmann- (or Hirschmann-Herfindahl)-Index to measure diversification of firms.

A firm is perfectly focused if it generates all sales in one industry, i.e. the percentage of sales generated from each industry $i$ is denoted as $p_i$, and the perfectly focused firm has $$p_1 = 1; p_k = 0 \text{ for } k\neq1.$$

There are at most $N$ industries, thus, the perfectly diversified firm has $p_i = 1/N\ \forall \ 1\dots i\dots n$.

When a firm is not active in an industry at all, that $p_i$ is equal to zero. All $p_i$ sum to 1, and they are all between 0 and 1.

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The Herfindahl index is defined as $$HHI :=\sum_{i=1}^n p_i^2$$ and I take the "true diversity"-measure, a special case of the entropy measure as explained and defined in the intro to this wikipedia page: $$ ^1 D := \prod_i p_i^{p_i} = \exp\left(\sum_{i=1}^n \ln(p_i)p_i\right)$$ I leave out the negative sign to make the measures comparable. This way, both of the measures are now technically focus measures, i.e. as diversification increases, they decrease.

Now, I believe that one can find an inequality along the lines of $HHI\ \geq\ ^1D$. Both measures are equal to 1 in case of perfect focus, and decrease from there. In case of perfect diversification into $N$ industries, they also both equal $1/N$. In between, however, the entropy measure seems to be smaller. Examples:

My goal is to prove (or disprove) $$\exp\left( \sum_i\ln(p_i)p_i\right) \leq \sum_i p_i^2$$

Note that the summands on the left side are undefined if $p_i=0$, but they converge to 0 as $p_i$ goes to zero, so we just take them as zero. Alternatively, we can assume for the proof that $p_i$ cannot be zero.

I cannot prove the inequality, although it appears to me as something along the lines of "taking a logarithm of numbers, then adding these, and then taking the exponent of the sum again leaves us with something smaller than just adding the numbers in the first place", perhaps due to the slope of the curve $f(x) = \ln x$ being more than 1 when $x$ is below 1 (convexity of logarithm and exponential functions?).

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I have come up with: $$\ln p_i < p_i-1 \ \text{ (equality if $p_i = 1$)}\Rightarrow \\ \exp \left(\sum_i \ln (p_i) p_i \right) < \exp \sum_i (p_i-1)p_i = \exp \sum_i (p_i^2 - p_i )\\ = \exp (HHI-1) \stackrel{!}{<} HHI $$ But this is incorrect, as, when the HHI is $0.9$, the expression on the left is $ \exp(-0.1) \approx 0.905$. In fact, when diversification increases s.t. HHI goes to zero (i.e., large $N$ and perfect diversification), then $e^{-1}>0$.

In particular, the opposite is true, namely $$\exp(HHI-1)>HHI$$ if $0<HHI<1$, with equality at 1.

Answer 1

My idea is to use the Weighted AM–GM inequality. Firstly we can transform $^{1} D$:

$$\exp\left( \sum_i\ln(p_i)p_i\right)=e^{\ln(p_1)\cdot p_1}\cdot e^{\ln(p_2)\cdot p_2}\cdot \ldots \cdot e^{\ln(p_n)\cdot p_n} $$

$$=\left( e^{\ln(p_1)}\right)^{p_1} \cdot \left( e^{\ln(p_2)}\right)^{p_2} \cdot \ldots \cdots \left( e^{\ln(p_n)}\right)^{p_n} =p_1^{p_1}\cdot p_2^{p_2}\cdot \ldots \cdot p_n^{p_n} =\prod_{i=1}^n p_i^{p_i}$$

For the moment we can replace the exponents $p_i$ by $w_i$ (weights) and we obtain $\prod_{i=1}^n p_i^{w_i}$, where $\sum_{i=1}^n w_i=1$. The same can be done for the $HHI: \sum_{i=1}^n w_i\cdot p_i$

Then the Weighted AM–GM inequality (page 4) states

If $x_1, . . . , x_n > 0$ and $w_1, . . . , w_n ≥ 0$ and $w_1 + · · · + w_n = 1$, then

$$w_1x_1+w_2x_2+w_3x_3+\ldots + w_nx_n\geq x_1^{w_1}\cdot x_2^{w_2}\cdot x_3^{w_3}\cdot \ldots \cdot x_1^{w_n}$$

That means $ HHI\geq ^{1}D$

Answer 2

I have come up with using a generalized form of the concavity of the logarithmic function:

$$\ln [a+t(b-a)] > \ln a + t [ \ln b- \ln a ] \\ \Rightarrow \ln [xa + (1-x)b] > x\ln a + (1-x) \ln b $$

But I'm not sure if the desired results follows already clearly from the fact that $\sum_i p_i = 1 $, much like the $x$ and $1-x$ above, or whether that generalization requires further proof (or is even wrong).