Is the average squared distance between two clusters equal to the squared distance of the difference in the two centroids?

Question

Simplify \displaystyle \frac{1}{n_1 n_2}\sum _{\mathbf{x}^{(i)}\in C_1}\sum _{\mathbf{x}^{(j)}\in C_2} \left\| \mathbf{x}^{(i)}-\mathbf{x}^{(j)} \right\| _2^2where C_1 and C_2 are two different clusters.

For simplicity, use the Euclidean distance for \left\| \mathbf{x}^{(i)}-\mathbf{x}^{(j)} \right\|.

Choose all expressions below that are equal to the above expression.

\displaystyle \frac{1}{n_1 n_2}\sum _{\mathbf{x}^{(i)}\in C_1}\sum _{\mathbf{x}^{(j)}\in C_2}\left\| \mathrm{{\boldsymbol \mu }}_1-\mathrm{{\boldsymbol \mu }}_2 \right\| ^2

\displaystyle \frac{1}{n_1 n_2}\sum _{\mathbf{x}^{(i)}\in C_1}\sum _{\mathbf{x}^{(j)}\in C_2} \left\| (\mathbf{x}^{(i)}-\mathrm{{\boldsymbol \mu }}_1)+(\mathrm{{\boldsymbol \mu }}_1-\mathrm{{\boldsymbol \mu }}_2)-(\mathbf{x}^{(j)}-\mathrm{{\boldsymbol \mu }}_2) \right\| ^2

\displaystyle \frac{1}{n_1 n_2}\sum _{\mathbf{x}^{(i)}\in C_1}\sum _{\mathbf{x}^{(j)}\in C_2}\left( \left\| \mathbf{x}^{(i)}-\mathrm{{\boldsymbol \mu }}_1 \right\| ^2+\left\| \mathrm{{\boldsymbol \mu }}_1-\mathrm{{\boldsymbol \mu }}_2 \right\| ^2+\left\| \mathbf{x}^{(j)}-\mathrm{{\boldsymbol \mu }}_2) \right\| ^2\right)
unanswered

Answer 1

The correct expression that is equal to the given expression is:

\displaystyle \frac{1}{n_1 n_2}\sum _{\mathbf{x}^{(i)}\in C_1}\sum _{\mathbf{x}^{(j)}\in C_2}\left\| \mathbf{x}^{(i)}-\mathbf{x}^{(j)} \right\| _2^2

The other expressions provided are not equal to the given expression.