= Solution
Fix distinct players $i,j$ and write $Ti=T\cup\{i\}$, with analogous notation for $Tj$ and $Tij$. Splitting sampled <coalitions> according to whether they contain the other player gives
$$
\begin{aligned}
2^{n-1}\beta_i(v)&=\sum_{T\subseteq N\setminus\{i,j\}}\bigl[v(Ti)-v(T)+v(Tij)-v(Tj)\bigr],\\
2^{n-1}\beta_j(v)&=\sum_{T\subseteq N\setminus\{i,j\}}\bigl[v(Tj)-v(T)+v(Tij)-v(Ti)\bigr].
\end{aligned}
$$
Subtracting yields the <Banzhaf difference identity>,
$$
\beta_j(v)-\beta_i(v)=\frac1{2^{n-2}}\sum_{T\subseteq N\setminus\{i,j\}}\bigl[v(Tj)-v(Ti)\bigr].
$$
In a <weighted voting game>, $w_j>w_i$ means that $Tj$ has greater total weight than $Ti$. Thus $v(Tj)\geq v(Ti)$ for every $T$: if the lighter <coalition> reaches the quota, the heavier one does too. Every summand is nonnegative, proving \b[$\beta_i(v)\leq\beta_j(v)$]. Equality remains possible even when the weights differ.
Back to article page
