Fixed discouragement paper

papers/casper/discouragement.tex

@@ -82,7 +82,7 @@ $ = 1$
 
 Hence both strategies are unprofitable. For values $r > 2$, the proof would need to be more conditional on specific values of $p$. We can make the claim that, if the griefing factor is bounded by $GF$, i.e. $r \le GF * \frac{\alpha}{1-\alpha}$, then a discouragement attack cannot be profitable if and only if $p \le \frac{1}{GF}$.
 
-We can check this at the boundary $h = 1$ as follows. We want to show that $\frac{1 - hp * \frac{1-\alpha}{\alpha}}{(\alpha + (1-\alpha)(1-h)^{\frac{d}{d+p}})^p} \le 1$, so we show that the numerator is less than or equal to the denominator. At $h = 1$, the numerator simplifies to $1 - \frac{p}{\alpha} + p$ and the denominator to $\alpha^p$. At $\alpha=1$, the two are equal. To show that the numerator is strictly less for $\alpha<1$, we can take the derivative of both with respect to $\alpha$; the numerator becomes $\frac{p}{\alpha^2}$ and the denominator becomes $p * \alpha^{p-1}$, and since $\alpha < 1$ the derivative of the numerator is clearly greater, so for $\alpha < 1$ the original fraction will be less than one. Checking for $0 < h < 1$ is much harder, but analytically it can be verified that it holds.
+We can check this at the boundary $h = 1$ as follows. We want to show that $\frac{1 - hp * \frac{1-\alpha}{\alpha}}{(\alpha + (1-\alpha)(1-h)^{\frac{1}{d+p}})^p} \le 1$, so we show that the numerator is less than or equal to the denominator. At $h = 1$, the numerator simplifies to $1 - \frac{p}{\alpha} + p$ and the denominator to $\alpha^p$. At $\alpha=1$, the two are equal. To show that the numerator is strictly less for $\alpha<1$, we can take the derivative of both with respect to $\alpha$; the numerator becomes $\frac{p}{\alpha^2}$ and the denominator becomes $p * \alpha^{p-1}$, and since $\alpha < 1$ the derivative of the numerator is clearly greater, so for $\alpha < 1$ the original fraction will be less than one. Checking for $0 < h < 1$ is much harder, but analytically it can be verified that it holds.
 
 Hence, if the griefing factor is bounded by 2, we want $p \le \frac{1}{2}$, and similarly for other griefing factors.