Change of indices in a double summation

Question

By using a "smart" change of indices $i$ and $j$, I'm trying to show that \begin{equation} \sum_{i=1}^{N}\sum_{j=1}^{N}q_{i}q_{j}a_{i}\left(f_{i}f_{j}^{'}-f_{i}^{'}f_{j}\right) = \sum_{i=1}^{N}\sum_{j=i+1}^{N}q_{i}q_{j}\left(a_{i}-a_{j}\right)\left(f_{i}f_{j}^{'}-f_{i}^{'}f_{j}\right). \end{equation}

Here, $q$'s corresponds to probabilities, so they are all non-negative and less than 1, $f$'s are also non-negative.

I can prove that this holds by induction but I want to understand how to show this equality by using a change of summation indices.

Answer 1

First note that $f_if_j'-f_i'f_j=0$ when $i=j$, so

$$\sum_{i=1}^N\sum_{j=1}^Nq_iq_ja_i\left(f_if_j'-f_i'f_j\right)=\sum_{i=1}^N\sum_{j=1}^{i-1}q_iq_ja_i\left(f_if_j'-f_i'f_j\right)+\sum_{i=1}^N\sum_{j=i+1}^Nq_iq_ja_i\left(f_if_j'-f_i'f_j\right)\;.$$

Then

$$\begin{align*} \sum_{i=1}^N\sum_{j=1}^{i-1}q_iq_ja_i\left(f_if_j'-f_i'f_j\right)&=-\sum_{i=1}^N\sum_{j=1}^{i-1}q_iq_ja_i\left(f_i'f_j-f_if_j'\right)\\\\ &=-\sum_{j=1}^N\sum_{i=j+1}^Nq_iq_ja_i\left(f_i'f_j-f_if_j'\right)\\\\ &=-\sum_{i=1}^N\sum_{j=i+1}^Nq_jq_ia_j\left(f_j'f_i-f_jf_i'\right)\;, \end{align*}$$

where in the last step I simply reversed the names of $i$ and $j$. Thus,

$$\begin{align*} \sum_{i=1}^N\sum_{j=1}^Nq_iq_ja_i\left(f_if_j'-f_i'f_j\right)&=\sum_{i=1}^N\sum_{j=1}^{i-1}q_iq_ja_i\left(f_if_j'-f_i'f_j\right)-\sum_{i=1}^N\sum_{j=i+1}^Nq_jq_ia_j\left(f_j'f_i-f_jf_i'\right)\\\\ &=\sum_{i=1}^N\sum_{j=i+1}^Nq_iq_j(a_i-a_j)\left(f_j'f_i-f_jf_i'\right)\;. \end{align*}$$