# Scaling of moment quotient for discrete probability distribution

## The problem

Let $\{p_i \}_{0 \leq i \leq n}$ be a discrete probability on a set of positive numbers $\{a_i \}_{0 \leq i \leq n}$ (with $\forall i:p_i \geq 0 \text{ and } a_i \geq 0$ and $\sum_{i=0}^{n} p_i =1$).
I am pretty convinced from my application side (see below) that if I scale the largest element $a_k$ of my set in the following way: $$a_k \to \lambda a_k \text{ with } \lambda > 1$$ then the ratio between the second moment and the square of the first one $$\frac{\langle X^2 \rangle}{\langle X \rangle^2}= \frac{\sum_{i=0}^{n}p_i a_i^2}{\left( \sum_{i=0}^{n}p_i a_i \right)^2 }$$ should increase.

## What I have tried

I have written $$\frac{\sum_{i=0}^{n}p_i a_i^2}{\left( \sum_{i=0}^{n}p_i a_i \right)^2 } = \frac{\sum_{i\neq k}p_i a_i^2+p_k a_k^2}{ \sum_{i,j \neq k}p_i p_j a_i a_j + 2 p_k a_k \sum_{i \neq k}p_i a_i + p_k^2 a_k^2}$$ Then I wrote down my assumption that this ratio is increasing when rescaling $a_k$: $$\frac{\sum_{i\neq k}p_i a_i^2+p_k \lambda^2 a_k^2}{ \sum_{i,j \neq k}p_i p_j a_i a_j + 2 p_k \lambda a_k \sum_{i \neq k}p_i a_i + p_k^2 \lambda^2 a_k^2} > \frac{\sum_{i\neq k}p_i a_i^2+p_k a_k^2}{ \sum_{i,j \neq k}p_i p_j a_i a_j + 2 p_k a_k \sum_{i \neq k}p_i a_i + p_k^2 a_k^2}$$ After a bit of algebraic reshuffling, I ended up with: $$\sum_{i\neq k}p_i a_i^2 \left(2 \sum_{i\neq k}p_i a_i + p_k a_k \right) + a_k \left( \sum_{i,j \neq k}p_i p_j a_i a_j \right) (\lambda^2 - 1) + \lambda^2 \left( p_k a_k \sum_{i\neq k}p_i a_i (a_k - a_i) \right) + 2 \lambda \sum_{i\neq k}p_i a_i \left( \sum_{i\neq k}p_i a_k^2-\sum_{i\neq k}p_i a_i^2 \right) + \sum_{i\neq k}p_i a_i a_k^2 \lambda \left( \lambda p_k - 2 \right) > 0$$ I am happy with all the terms but the last one, which is only positive under certain constraints that I do not want.

## The application

In polymer chemistry/physics, $\frac{\langle X^2 \rangle}{\langle X \rangle}$ is called the weight average and $\langle X \rangle$ the number average. It is always stated that having more long chains increase the weight average more than it increases the number average and I wanted to understand that for a simple example (intuitively, it sounds very convincing, but I want to show it formally for a simple case).

• This is my first post here, so I will be happy about any guidance - and also happily remove the question if it does not fit in with the site. Thank you :) – Sanya Sep 8 '16 at 20:35

You can assume without loss of generality that $a_n$ is the largest element. Then, we clearly have $$a_n \sum_{i=0}^{n-1} a_i p_i \ge \sum_{i=0}^{n-1} a_i^2 p_i.$$ Write $$f(\lambda) = \frac{\sum_{i=0}^{n-1} a_i^2 p_i + \lambda^2 a_n^2 p_n}{\left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right)^2},$$ and notice that
\begin{align} f^\prime(\lambda) & = \frac{2 \lambda a_n^2 p_n \left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right)^2 - \left( \sum_{i=0}^{n-1} a_i^2 p_i + \lambda^2 a_n^2 p_n \right) 2 \left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n \right) a_n p_n }{\left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right)^4}\\ & = \frac{2 \lambda a_n^2 p_n \left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right) - \left( \sum_{i=0}^{n-1} a_i^2 p_i + \lambda^2 a_n^2 p_n \right) 2 a_n p_n }{\left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right)^3}\\ & = \frac{2 \lambda a_n^2 p_n \left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right) - \left( \sum_{i=0}^{n-1} a_i^2 p_i + \lambda^2 a_n^2 p_n \right) 2 a_n p_n }{\left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right)^3}\\ & = \frac{2 a_n p_n \left( \lambda a_n \sum_{i=0}^{n-1} a_i p_i - \sum_{i=0}^{n-1} a_i^2 p_i \right)}{\left(\sum_{i=0}^{n-1} a_i p_i + \lambda a_n p_n\right)^3}. \end{align} By the initial observation, and the fact that $\lambda > 1$, we see that $f^\prime(\lambda) > 0$, so $f(\lambda)$ is increasing.