I'm really bad with sigma summation notation and have been puzzling over how my professor in his notes for class got from one step to the next.

It's pretty simple:

$$D_{avg} = \sum_{i=1}^n \sum_{x \in C_i} ||\mathbf x-\mathbf m_i||^2$$ $$D_{IC} = \sum_{i=1}^n \frac{1}{|C_i|} \sum_{x \in C_i} \sum_{x' \in C_i}||\mathbf x-\mathbf x'||^2$$

$$m_i = \frac{1}{|C_i|} \sum_{x \in C_i} \mathbf x$$

therefore $$D_{IC} = 2D_{avg}$$

I tried to simply substitute the expression for $m_i$ into $D_{IC}$, but I think I am getting stuck on manipulating $||\mathbf x - \mathbf x'||$ because they're vectors and not scalars.

Here is what I have thus far:

$$D_{IC} = \sum_{i=1}^n \sum_{x' \in C_i}||\mathbf m_i-\mathbf x'||^2$$

Obviously, I know this is wrong..


\begin{align}D_{IC} &= \sum_{i=1}^n \frac{1}{|C_i|} \sum_{x \in C_i} \sum_{x' \in C_i}||\mathbf x-\mathbf x'||^2 \\ &= \sum_{i=1}^n \frac{1}{|C_i|} \sum_{x \in C_i} \sum_{x' \in C_i}(||\mathbf x||^2+ ||\mathbf x'||^2-2\mathbf x'^T\mathbf x) \\ &= \sum_{i=1}^n \frac{1}{|C_i|} \sum_{x \in C_i}(|C_i|||\mathbf x||^2+\sum_{x' \in C_i} ||\mathbf x'||^2-\sum_{x' \in C_i}2\mathbf x'^T\mathbf x) \\ &= \sum_{i=1}^n \frac{1}{|C_i|} \sum_{x \in C_i}(|C_i|||\mathbf x||^2+\sum_{x' \in C_i} ||\mathbf x'||^2-2|C_i|m_i^T\mathbf x) \\ &= \sum_{i=1}^n \frac{1}{|C_i|} \left( |C_i|\sum_{x \in C_i}||\mathbf x||^2+|C_i|\sum_{x' \in C_i} ||\mathbf x'||^2-2|C_i|^2||m_i||^2\right) \\ &= \sum_{i=1}^n \left( \sum_{x \in C_i}||\mathbf x||^2+\sum_{x' \in C_i} ||\mathbf x'||^2-2|C_i|||m_i||^2\right) \\ &= 2\sum_{i=1}^n \left( \sum_{x \in C_i}||\mathbf x||^2-|C_i|||m_i||^2\right) \\ \end{align}

\begin{align} D_{avg} &= \sum_{i=1}^n \sum_{x \in C_i} ||\mathbf x-\mathbf m_i||^2 \\ &= \sum_{i=1}^n \sum_{x \in C_i} \left(||\mathbf x||^2 + ||m_i||^2-2\mathbf x^T m_i\right) \\ &= \sum_{i=1}^n \left(\sum_{x \in C_i} ||\mathbf x||^2 + |C_i|||m_i||^2-2|C_i||| m_i||^2\right) \\ &= \sum_{i=1}^n \left(\sum_{x \in C_i} ||\mathbf x||^2 -|C_i||| m_i||^2\right) \\ \end{align}

Hence $$D_{IC} =2D_{avg} $$

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