simplify $\log(x^k(1-x)^{n-k}) - \log(x^k(1-x)^{n-k} + (1-x)^k x^{n-k}) - \log(0.5))$ for $x \in (0,1)$

Question

simplify $\log(x^k(1-x)^{n-k}) - \log(x^k(1-x)^{n-k} + (1-x)^k x^{n-k}) - \log(0.5))$ for $x \in (0,1)$ and $k \leq n$

$\textbf{context:}$

I am trying to solve a certain equation in the following form, where $a \in \{-1,1\}$ and $q_a \in [0,1]$:

\begin{align} \sum_{a} p(a)\sum_{k=0}^n {n \choose k} q_{a}^k(1-q_{a})^{n-k} \Bigg[ \log(q_a^k \cdot (1-q_a)^{n-k}) \nonumber - \Bigg(\log\bigg( q_1^{k}(1-q_1)^{n-k}p(a=1)+ q_{-1}^{k}(1-q_{-1})^{n-k}p(a=-1)\bigg)\Bigg)\Bigg] \end{align}

The following properties are given:

1) $p(a = 1) = p(a=-1) = \frac{1}{2}$

2) $q_{1} = 1-q_{-1}$

Using these properties we can deduce that the term in the big brackets can be reduces to:

$\log(x^k(1-x)^{n-k}) - \log(x^k(1-x)^{n-k} + (1-x)^k x^{n-k}) - \log(0.5))$ for $x \in (0,1)$ and $k \leq n$ where $x$ is $q_{1}$ or $q_{-1}$

$\textbf{What i have tried:}$

1) Using taylor expansions, however the right hand term was very complicated thus i eventually did not see any simplifaction in that matter.

2) Trying to find an approximation by using facts as $log(1+x) \approx x$ for small $x$, hower the right hand term still gave me a lot of complications.

3) Plot terms for all possible $n$ and $k$, this methods resulted in nice plots, however i did not seem to find a relation or some sort of fit of the function to k.

$\textbf{My question to you}$

Have i missed out a method that i should try? Is it able to simplify this formula to a simpler form?

With kind regards,

Kees Til

Answer 1

I don't know if it useful to you but you have: \begin{align} \ln\left(x^k (1-x)^{n-k} \right)-\ln\left(x^k (1-x)^{n-k} +x^{n-k} (1-x)^{k} \right)&=-\ln\left(\frac{x^k (1-x)^{n-k} +x^{n-k} (1-x)^{k}}{x^k (1-x)^{n-k}} \right)\\ &=-\ln\left(1+x^{n-k-k} (1-x)^{k-n+k} \right)\\ &=-\ln\left(1+\left(\frac{x}{1-x}\right)^{n-2k} \right)\\ \end{align}