Splitting the summation sign

Question

i am trying to understand the second step in the formula per below - and how the summation sign $\sum_{k=1}^K$ splits into the terms 1-$\sum_{k=1}^{K-1}$ terms. Any help much appreciated

Answer 1

We have that $$ \sum\limits_{k=1}^K x_k=\sum\limits_{k=1}^K \mathbb{I}(x=k)=1 $$ since $x$ takes on exactly one of the values in $\{1,2,\ldots,K\}$. Therefore, $x_K=1-\sum\limits_{k=1}^{K-1} x_k$.

Similarly, $\mu_k=P(x=k)$ and hence $\sum_{k=1}^K \mu_k=1$ by the same argument. Thus, we also have $\mu_K=1-\sum\limits_{k=1}^{K-1}\mu_k$.

Answer 2

First, not 'the summation sign splits' but simply the last term of the sum gets separated: $$ \sum\limits_{k=1}^K x_k \log \mu_k = \left( \sum\limits_{k=1}^{K-1} x_k \log \mu_k \right) + x_K \log \mu_K$$

Then the last term $(x_K \log \mu_k)$ gets replaced with equivalent expression, as @StefanHansen explained.