I donât understand why you think that even if ETH-FEI were the only place for trading FEI, implementing mint/burn would increase the price for buyers and decrease the price for sellers relative to no mint/burn. (SPOILER: after writing this post, I do understand why you think this; youâre right! )
Ignore mints for a moment. Iâll try to argue that just with burns, the price for buyers is as good as without burns . Mints can then only make the price for buyers even better. I might discover youâre right along the way! In that case Iâll still leave my thought process here for any other skeptics to see.
Iâm going to argue in two scenarios that are easy to think about. Both make a couple of important assumptions: (1) the liquidity pool has $1B FEI in it; (2) the incentivized market is DAI/FEI (to simplify the calculation of the invariant product and prices); (3) the deviation from peg before any trades happen is m_{start} = 0; (4) as claimed in the docs, the burn mechanism is path independent.
The AMM invariant product is k = (10^9)^2 = 10^{18}.
1. The first scenario will assume that thereâs a supply of x FEI that will be sold on the incentivized pool regardless of price. The second scenario will assume that thereâs an unlimited supply of FEI that will be sold until the marketâs price drops to p.
In the first case, the final deviation from the peg will be
m_{end}(x) = 1 - \frac{10^{18}}{(10^{9}+x)^2}
Iâm omitting the derivation of this because itâs just the standard AMM price calculation. (And after making that statement I sure hope itâs right! ) For example, m_{end}(10,000,000) \approx 0.02. The key observation is that m_{end}(x) does not depend on how much is burnt. This is because, as is stated in the docs, âThe entire trade size including the burn is used to calculate the slippage for the end deviation from the peg.â In other words, in this first scenario, the burn only affects the price that sellers get. The final marginal buying price
1 - m_{end}(x)
is the same whether or not there are burns. The conclusion we can draw from this first scenario is limited: if sellers are not responsive to price, then the final marginal buying price is not affected by burns.
2. The second scenario is more complicated. If there are no burns, the sellers will sell x FEI, where x must satisfy
(10^9 + x)(10^9 - px) = 10^{18}
for fixed p. By distributing and rearranging, we can see this holds if and only if
px^2 + (10^9p - 10^9)x = x(px + 10^9p - 10^9) = 0
The solution x = 0 is meaningless. So we get
x(p) = 10^9 \left(\frac{1 - p}{p}\right)
This leads to a marginal buying price of
1 - m_{end}^{unburnt}(p) = \frac{10^9 - px(p)}{10^9 + x(p)} = p^2,
nicely enough. Now what if there are burns? Then if B(x, m_{start}, m_{end}) is the burn function, x must satisfy
[10^9 + (x - B(x, 0, m_{end}^{burnt}(p))](10^9 - px) = 10^{18} \mbox{ (1)}
The burn function in the docs is expressed as an integral, but in the case when m_{start} is 0, the integral can be evaluated to give
B(x, 0, m_{end}^{burnt}(p)) = \frac{100 (m_{end}^{burnt}(p))^2 x}{3}
We can show with tedious algebra that x satisfies (1) if and only if
x(xp\phi(p) + 10^9p - 10^9\phi(p)) = 0
where
\phi(p) = 1 - \frac{100(m_{end}^{burnt}(p))^2}{3}
Again, the solution when x = 0 is meaningless. So we get
x(p) = 10^9\left(\frac{\phi(p) - p}{p\phi(p)}\right)
where the notation x(p) emphasizes that x depends on p. This leads to a marginal buying price of
1 - m_{end}^{burnt}(p) = \frac{10^9 - px(p)}{10^9 + x(p) - B(x(p), 0, 1-p)} = \frac{10^9 - px(p)}{10^9 + \phi(p)x(p)}
What weâre looking for is an expression for m_{end}^{burnt} in terms of p. The best I can do, by first replacing x(p) with itâs expression in terms of p and \phi(p), then replacing \phi(p) by itâs expression in terms of m_{end}^{burnt}(p), and finally doing a bunch of algebra Iâll spare you, is arrive at a fifth degree polynomial:
\frac{-1000}{9} m^5 + \frac{1000}{9}m^4 + \frac{600}{9}m^3 - \frac{600}{9}m^2 - m + 1 - p^2 = 0
where I denoted m_{end}^{burnt}(p) by m to save space.
Back to the original question. Are buyers better or worse off when the protocol burns? I donât know! Hehehe! Letâs find out. I let Wolfram Alpha solve this polynomial for me for different values of p. The polynomial has several solutions for each value of p. I always picked the smallest positive solution. Any of the bigger positive solutions gave nonsensical results. Hereâs a table of what I found:
\begin{array}{ccc}
p & 1 - m_{end}^{unburnt}(p) & 1 - m_{end}^{burnt}(p) \\
0.99 & 0.9801 & 0.9886 \\
0.98 & 0.9604 & 0.9819 \\
0.97 & 0.9409 & 0.9765 \\
0.96 & 0.9216 & 0.9721 \\
0.95 & 0.9025 & 0.9681 \\
0.94 & 0.8836 & 0.9645 \\
0.93 & 0.8649 & 0.9612 \\
0.92 & 0.8464 & 0.9581 \\
0.91 & 0.8281 & 0.9552 \\
0.90 & 0.8100 & 0.9525
\end{array}
The most unintuitive result for me is that none of the formulas in the second scenario depend on the liquidity in the AMM. If I had to make a bet an hour ago, I wouldâve bet that the greater the liquidity, smaller the gap between the burnt and the unburnt marginal buy prices.
Thatâs a very precise way of saying: youâre right! Burns punish the buyer! In restrospect, your intuitive argument is spot on! I think the conclusions we can draw from this are that direct incentives potentially work to return the price to the peg when sellers will sell x FEI independently of the price of FEI. This assumption probably holds true when FEI is within a fraction of a percent of the peg. However, as deviations from the peg become larger, the amount of FEI sellers will sell depends largely on what price they can sell it at. In this case, direct incentives punish buyers as well as sellers, and thus likely create an obstacle for the restoration of the peg. So a solution like the one you (@fei.saver) are proposing, that triggers reweights when the price hasnât deviated too much from the peg for too long, should be able to work nicely with direct incentives that are only triggered when the price is a little bit off peg.
ALSO: I figured someone had to abuse MathJax sooner or later!