write as $$\hat p = \frac{ p \left(\frac{p}{1-p}\right)^{\beta-1} e^\tau}{1+p \left(\frac{p}{1-p}\right)^{\beta-1} e^\tau}.$$ Then $$\nu(p)=\frac{\hat p- p}{p(1-p)} = \frac{p^{\beta-1} e^\tau - (1-p)^{\beta-1}}{p^\beta e^\tau + (1-p)^\beta}.$$ Differentiating this with respect to $p$ we obtain $$\frac{d\nu(p)}{dp}=\frac{ - e^{2\tau} p^{2(\beta-1)}  + e^\tau p^{\beta-2}(1-p)^{\beta-2}(\beta-2 + 2p(1-p)) - (1-p)^{2\beta-2}}
{(p^\beta e^\tau + (1-p)^\beta)^2}.$$ The numerator of $d\nu(p)/dp$ can be written as $$-(e^\tau p^{\beta-1} - (1-p)^{\beta -1})^2 + e^\tau (\beta-1) p^{\beta-2} (1-p)^{\beta-2},$$ which is non-positive for $\beta\le 1$. Hence we have the following proposition.

\[prop:monotonicity\] Under the logistic regression model , for $\beta\le 1$ the betting ratio $\nu_n(p_n)$ is monotone decreasing in $p_n$.

It is natural that $\nu_n$ is monotone decreasing in $p_n$, because if $p_n$ is too high and Skeptic does not believe it, then Skeptic will bet on the non-occurrence $x_n=0$.

For the special case of $\beta=1$, $$\nu_n(p_n)=\frac{e^{\tau_n}-1}{1 + p_n (e^{\tau_n}-1)},$$ which is bounded and monotone in $p_n \in [0,1]$. For $\beta<1$, $\nu_n(p_n)$ is unbounded and it can be easily seen that $$\lim_{p_n \downarrow 0} \frac{\nu_n(p_n)}{1/p_n}=1, \qquad
\lim_{p_n \uparrow 1} \frac{\nu_n(p_n)}{1/(1-p_n)}=-1.$$ We can interpret the first limit as follows. Suppose that $p_n=1/1000$, i.e. the price of a ticket is 1/1000 of a dollar. In this case Skeptic can buy $1000$ tickets with one dollar and has 