\(b > (1-\tau _g)^2 \frac{\delta \phi - r(1-\phi )}{(x + n + \delta )r(1-\tau _p)(\phi (1-\tau _g) + (1-\phi )(1-\tau _d) )}\) implies that \(\frac{\partial \tilde{\hat{k}}}{\partial \tau _g} < 0\). Note that \(d\dot{\hat{\nu }}\) is...