So, I get of course that I can plot the entropy $- x \log_2(x)$ just as is which gives us:
This is because we say that $p = (p_1, \ldots, p_n)$ and we have
\begin{align*} H(p) = H(p_1, \ldots, p_n) &= - \sum_{k=1}^n p_k \log_2 p_k \end{align*}
Now I have here an example that defines $p_0 \mapsto H(p_0, 1-p_0)$ (continuous on $[0,1]$) for $p_0 = \frac{1}{2} = 1 - p_0$ and right below I have this plot:
So in this example $p$ is actually not really defined this is why I don't understand what $H(p, 1-p)$ actually is and also how it comes that I can actually plot it since the only thing I have given is $p_0 = \frac{1}{2}$ and its complement.


\begin{align} H(p_0, p_1) &= -\sum_{k=1}^2 p_k \log_2 p_k = - p_0\log_2 p_0 - p_1 \log_2 p_1 \ &= - \frac{1}{2} \log_2 \frac{1}{2} - (1- \frac{1}{2}) \log_2 (1 - \frac{1}{2}) \ &= 1 \end{align}
and to plot it what they actually do is
\begin{align} H(p, (1-p)) &= - p\log_2 p - (1-p)\log_2 (1-p)\ \end{align}
This just feels strange as $p$ got previously defined as a probability distribution $p(\cdot)$.
– Stefan Falk Nov 03 '16 at 15:16