# Simplifying risk equation for 0/1 loss in machine learning

I'm currently studying machine learning using the textbook Introduction to Machine Learning 3e (Ethem Alpaydin, 2015) and had a question regarding the derivation of a particular equation. For those curious, this specific equation is found on page 53 of 3.3: Losses and Risk.

More specifically, the context of this particular excerpt is where we perform classification and how we can measure loss and risk.

Let us define action $$\alpha_i$$ as the decision to assign the input to class $$C_i$$ and $$\lambda_{ik}$$ as the loss incurred for taking action $$\alpha_i$$ when the input actually belongs to class $$C_k$$. The expected risk for taking action $$\alpha_i$$ is:

$$R(\alpha_i\ |\ \pmb{x}) = \sum_{k = 1}^K \lambda_{ik} P(C_k\ |\pmb{x})$$

and we choose the action with minimum risk:

$$\text{choose}\; \alpha_i\; \text{if}\; R(\alpha_i\ |\ \pmb{x}) = \text{min}_k R$$

Let us define $$K$$ actions $$\alpha_i$$ ($$i = 1, \dots, K$$). In the special case of the $$0/1$$ loss where

$$\lambda_{ik} = \begin{cases}{0\quad \text{if}\; i = k} \\ 1\quad \text{if}\; i \ne k\end{cases}$$

The risk of taking action $$\alpha_i$$ is:

\begin{align} R(\alpha_i\ |\ \pmb{x}) & = \sum_{k = 1}^K \lambda_{ik}P(C_k\ |\ \pmb{x}) \\\ & = \sum_{k \ne i} P(C_k\ |\ \pmb{x}) \\\ & = 1 - P(C_i\ |\ \pmb{x}) \end{align}

because $$\sum_{k}P(C_k\ |\ \pmb{x}) = 1$$.

In the last portion of the excerpt, I'm having trouble understanding how we get from $$\sum_{k \ne 1}P(C_k\ |\ \pmb{x})$$ to $$1 - P(C_i\ |\ \pmb{x})$$.

My understanding of how we got from the first line with $$\lambda$$ to the second line is that we simply removed the cases where $$k = i$$ since $$\lambda = 0$$ in those cases anyway and therefore we only have to consider cases where $$k \ne i$$.

Also, as is stated in the very last line, since

$$\sum_k P(C_k\ |\ \pmb{x}) = \sum_{k = i}P(C_k\ |\ \pmb{x}) + \sum_{k \ne i}P(C_k\ |\ \pmb{x}) = 1$$

we can obtain

$$\sum_{k \ne i}P(C_k\ |\ \pmb{x}) = 1 - \sum_{k = i}P(C_k\ |\ \pmb{x})$$

Assuming that my understanding is correct up until here, I'm having trouble understanding how

$$\sum_{k = i}P(C_k\ |\ \pmb{x}) = P(C_i\ |\ \pmb{x})$$

For any real numbers $$a_1,\dotsc, a_K$$, if $$a_1+\dotsc + a_{i-1} +a_{i}+a_{i+1} +\dotsc+a_{K}=1$$ then, by subtracting $$a_i$$ from both sides, we have $$\sum_{k\neq i, \quad 1\leq k \leq K} a_k = 1-a_i.$$ Apply this to $$a_k=P(C_k|x)$$.