Definition#

underlying: base asset of a interest-bearing token interest-bearing token: IOU like cDAI, eUSDC $x$ : Underlying reserve in Pool $y$ : Zero-coupon bond reserve in Pool

Previous Work#

Constant reserve: $V_{rs v} (x, y) = x + y$ , the sum value function. Constant product: $V_{p ro d} (x, y) = x y$ , the product value function.

2-asset#

before and after the swap, the following is invariant.

x^{1 - t} + y^{1 - t} = k (1)

Marginal price and Impliedt rate (IR)#

In general, IR can be expressed:

r = (1/ p)^{1/ t} - 1

$r$ : IR $p$ : Bond price in underlying.

p = (\frac{x}{y})^{t}

r = y / x - 1

Value Function#

V (x, y, t) = (x^{1 - t} + y^{1 - t})^{1/ (1 - t)} (2)

Lemma#

$I f t \to 0, t h e n V (x, y, 1) \to V_{rs v}$ : mStable $I f t \to 1, t h e n V (x, y, 1) \to V_{p ro d}$ : Balancer 2-asset

3-asset#

x^{1 - t} + y^{1 - t} + z^{1 - t} = k (3)

V (x, y, z, t) = (x^{1 - t} + y^{1 - t} + z^{1 - t})^{1/ (1 - t)} (4)

$I f t \to 0, t h e n V (x, y, 1) \to x + y + z$ になるはず。 $I f t \to 1, t h e n V (x, y, z, 1) \to (x yz)^{1/3}$ 3-asset BalancerのValue Funcになるはず。

$P roo f .$

t \to 1 lim (x^{1 - t} + y^{1 - t} + z^{1 - t})^{1/ (1 - t)} = exp ln t \to 1 lim (x^{1 - t} + y^{1 - t} + z^{1 - t})^{1/ (1 - t)} = exp t \to 1 lim \frac{ln ( x ^{1 - t} + y ^{1 - t} + z ^{1 - t} )}{1 - t}

Here, we can apply L’Hopital’s rule. so,

t \to 1 lim \frac{ln ( x ^{1 - t} + y ^{1 - t} + z ^{1 - t} )}{1 - t} = t \to 1 lim \frac{\frac{\partial l n ( x ^{1 - t} + y ^{1 - t} + z ^{1 - t} )}{\partial t}}{\frac{\partial ( 1 - t )}{\partial t}} = \frac{\frac{- l n ( x yz )}{3}}{- 1} = \frac{ln ( x yz )}{3}

as a reference,

Therefore,

t \to 1 lim V (x, y, z, t) = exp \frac{ln ( x yz )}{3} = (x yz)^{1/3}

This is exactly same Value Function of Balancer 3-asset with all weights 1/3.

n-asset#

Let $B_{i}$ denote balance of asset $i$ in a pool and $B$ be vector of balances $[B_{1}, B_{2} ... B_{n}]$ .

Trading Function $f$ :

f (B) = \sum B_{i}^{1 - t} (5)

Value Function $V$ :

V (B, t) = (\sum B_{i}^{1 - t})^{1/ (1 - t)} (6)

Marginal Price $p_{i}$ :

p_{i} = (\frac{B _{i}}{B _{o}})^{t} (7)

Proof#

Let $B_{o}$ denote balance of token $o$ bought by a trader and $B_{i}$ denote balance of token $o$ sold by the trader.

By denition, $p_{i}$ is minus the partial derivative of $B_{i}$ in function of $B_{o}$ :

p_{i} = - \frac{\partial B _{i}}{\partial B _{o}} (8)

From the value function denition we can isolate $B_{i}$ :

B_{i} = ⎩ ⎨ ⎧ V^{1 - t} - k \neq = i \sum B_{k}^{1 - t} - B_{o}^{1 - t} ⎭ ⎬ ⎫^{1/ (1 - t)} (9)

Now Eq8 becomes:

p_{i} = - \frac{\partial B _{i}}{\partial B _{o}} = - \frac{\partial}{\partial B _{o}} ⎩ ⎨ ⎧ V^{1 - t} - k \neq = i \sum B_{k}^{1 - t} - B_{o}^{1 - t} ⎭ ⎬ ⎫^{1/ (1 - t)} = V^{1 - t} - k \neq = i \sum B_{k}^{1 - t} - B_{o}^{1 - t}^{1/ (1 - t) - 1} B_{o}^{- t} = (B_{i}^{1 - t})^{1/ (1 - t) - 1} B_{o}^{- t} = (\frac{B _{i}}{B _{o}})^{t}

References#

Constant Power Root Market Makers

Balancer Whitepaper