I'm trying to reduce down the RHS of the below equation to be equal to the LHS


$K$ = predefined number of clusters
$C_j$ = is a cluster
$t$ = Assume that an $m\times n$ gray image of pixel intensities has been flattened into a $m\times n =t$ vector
$w_{ij}$ = identity function where 1 if $x_i\in C_j$ and 0 otherwise
$\mu_j$= mean intensity cluster $j$
$\mu_T$= mean intensity of all pixels
$t_j$ = # of elements in cluster $C_j$

I'm having trouble deriving this

  • $\begingroup$ Please undertake a search. This surely has been asked and answered not once already. $\endgroup$
    – ttnphns
    Sep 10, 2020 at 20:57

1 Answer 1


Just expand the brackets:

$$ \sum_{j=1}^K t_j \left( \mu_j^2 + \mu_T^2 - 2 \mu_j \mu_T \right) = \sum_{i=1}^t \left(x_i^2 + \mu_T^2 - 2 x_i \mu_T\right) - \sum_{j=1}^K \sum_{i=1}^t w_{i j} \left( x_i^2 + \mu_j^2 - 2 x_i \mu_j \right) $$

Now use the following manipulations of terms in the sums: $$ \sum_{j=1}^K \sum_{i=1}^t w_{i j} x_i^2 = \sum_{i=1}^t x_i^2, $$

$$ \sum_{j=1}^K \sum_{i=1}^t w_{i j} x_i \mu_j = \sum_{j=1}^K \mu_j \sum_{i=1}^t w_{i j} x_i = \sum_{j=1}^K t_j \mu_j^2, $$

$$ \sum_{i=1}^t x_i \mu_T = t \mu_T^2, $$

$$ \sum_{j=1}^K t_j \mu_j \mu_T = t \mu_T^2. $$

Plug them into the expanded equation.


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