# Price Equilibrium and Arbitrage

Given a price range

$[F,C]$

within which the price of the specific cryptocurrency varies, where $F$

is the strike price for call and $C$

is the strike price for put. Let $c\in [F,C]$

be the market price of the cryptocurrency on the platform. Thus $c-F$

is the intrinsic value of a call token, while $C-c$

is the intrinsic value of a put token. Let $x,y$

be the amount of call and put token respectively, and $z$

be the total reserved value. The total reserved value is all the asset collected that are used to generate Antimatter tokens. If call and put tokens increase at the same ratio, the price should not change. We aim to build a model that $z$

is a function of $x$

and $y$

, $z=f(x,y)$

, such that $kf(x,y)=f(kx,ky)$

With this equation, we are able to define the price of each token: $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}$

When the volumes of both tokens are equal, it is expected that the prices of both tokens are equal(In reality, the price of call tokens might be higher, because it has more upside potential). When the volume of call tokens far exceeds the volume of put tokens, it is expected that the price of put token approaches $0$

. When the volume of put tokens far exceeds the volume of call tokens, it is expected that the price of put token approaches $C-F$

. The behavior of call token is wilder but in the same way. One note is that the price of both tokens depends on the ratio of volume of both tokens, rather than the difference between them.For example, we work with ETH and USDT. The price of ETH in terms of USDT varies within the interval

$[1000,4000]$

. Here $1000$

is the strike price of call, and $4000$

is the strike price of put. If the ratio between call token and put token generated is $6:4$, it is expected that the market price is $3500$

In this case, the cost to generate a call token is about $6000$

and the cost to generate a put token is about $240$

. When there is a difference between the market price and market price of ETH, one can buy and sell call or put token in two market to make profit.The actual model goes as follows. Let

$z=\frac{(C-F)y^2}{\sqrt{x^2+y^2}}+e\cdot\frac{C-F}{C}\frac{x^2}{\sqrt{x^2+y^2}}$

where $e$

is the price of ETH. This expression ensures that $\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}\geq0$

. This model works well with prices of tokens because of positive definiteness. Because the lowest possible price of ETH is $0$

and the highest price of ETH is infinity, the price of call token will be appreciated to a greater extend, if the price moves in the right direction. The prices of both tokens remain in the interval $[0,C-F]$

for current version.Starting from

$z=f(x,y)$

, such that $kf(x,y)=f(kx,ky)$

, we have the equivalent following $xf_x+yf_y=f$

. Let $r=\sqrt{x^2+y^2},\theta=\tan^{-1}(\frac{y}{x})\in (0,\frac{\pi}{2})$

. Also, $x=r\cos\theta,y=r\sin\theta$

. We further transform:
$xf_x+yf_y=f\equiv\frac{x}{r}f_x+\frac{y}{r}f_y=\frac{f}{r}$

. By chain rule, we have $\frac{\partial f}{\partial r}=\frac{f}{r}$

Therefore $f=A(\theta)r$

for some function $A$

(This is true for any function of $\theta$

). Now we put these aside and consider the second question. The idea is that because $(C-F)f=U+Ee$

, we may find $U,E$

first and then sum them up because of homogeneity. We need $\frac{\partial U}{\partial x}<0,\frac{\partial U}{\partial y}>0,\frac{\partial E}{\partial x}>0,\frac{\partial E}{\partial y}<0$

. It worth checking that $U=y\cdot(\frac{y}{\sqrt{x^2+y^2}})^i,E=x\cdot(\frac{x}{\sqrt{x^2+y^2}})^i$

satisfy the above inequalities. By homogeneity conditions, it turns out summing through index $i$

does no harm anything (although it turns out that the summation is redundant). We relate $U$

and $E$

with $\theta$

by $\sin\theta=\frac{y}{\sqrt{x^2+y^2}},\cos\theta\frac{x}{\sqrt{x^2+y^2}}$

. Then by choice of $A$

, we can obtain that $z=f(x,y)=x\cdot e\cdot\sum_{i=1}^\infty e_i(\frac{x}{\sqrt{x^2+y^2}})^i+y\cdot\sum_{i=1}^\infty u_i(\frac{y}{\sqrt{x^2+y^2}})^i$

Next, we consider why $I$

must be equal to $1$

. For simplicity, we let those constants be $1$

Suppose $I=2$

, then $E=\frac{x^3}{x^2+y^2},E_x=\frac{x^2(x^2+3y^3)}{(x^2+y^2)^2}$

. Similarly, $U_x=\frac{-2y^3x}{(x^2+y^2)^2}$

. If we consider $z_x=U_x+eE_x=\frac{ex^4+y^2x(3ex-2y)}{(x^2+y^2)^2}$

and $z_y=U_y+eU_y=\frac{y^4+3x^2y^2-2ex^3y}{(x^2+y^2)^2}$

That their numerators are different means that we need more relations on $x$

and $y$

, which is implausible since $x$

and $y$

are the token volume. Thus $k=1$

. The final thing is to take some constants that have good numerical performance and it is $E=\frac{C-F}{C}\cdot\frac{x^2}{\sqrt{x^2+y^2}},U=(C-F)\cdot\frac{y^2}{\sqrt{x^2+y^2}}$

. $z$

follows directly.

Last modified 1yr ago