How does this equation hold in the proof of Jensen’s inequality?

Question

While reading the Elements of Information Theory (Cover &Thomas) on the page 27, I cannot understand why this function holds: $p'_{i} = p_i/(1−p_k)$. Please help explain a bit. Thanks.

Answer 1

\begin{align} \sum_{i=1}^k p_if(x_i)&=p_kf(x_k)+\sum_{i=1}^{k-1} p_if(x_i)\\ &=p_kf(x_k)+(1-p_k)\sum_{i=1}^{k-1} \frac{p_i}{1-p_k}f(x_i)\\ &=p_kf(x_k)+(1-p_k)\sum_{i=1}^{k-1} p_i'f(x_i)\\ \end{align}

For convenience, he just defines $$p_i'=\frac{p_i}{1-p_k}.$$

Answer 2

The trick of defining $$p_i'=\frac{p_i}{1-p_k}$$ allows us to have a new probability function (for all the original points except the last one). We divide by ${1-p_k}$ (where $k$ is the last point) in order to normalize the distribution, so that it sums up to one.