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Matt Levine on point again with more information about how this works and why it stopped [0]. I liked the comparison between a bank run and a 'death spiral'. >

Matt Levine on point again with more information about how this works and why it stopped [0]. I liked the comparison between a bank run and a 'death spiral'. > A bank run is a fairly mild event compared to a death spiral. In a bank run, your $1 stablecoin is backed by some pool of assets that are worth something; a rush to withdraw money leads to fire sales that depress their value, but their value derives from something other than confidence in the stablecoin. In a death spiral, the whole system is built on confidence in the stablecoin; if that confidence evaporates then there is no floor at all on the value. [0] https://archive.ph/slgNt https://archive.ph/slgNt

sc68cal 4y ago
> And so the LFG bought a bunch of Bitcoin and promised to use it to defend Terra’s peg to the dollar. If one Terra goes down to $0.90, instead of turning Terra into Luna and selling them in a death spiral, the LFG can buy Terra for $0.90 and pay for it in Bitcoin. If the LFG has enough Bitcoin, and if Bitcoin’s price holds up, then it can defend the peg and keep the price of Terra close to $1. So they've basically re-invented fractional reserve banking, with Terra/Luna taking the place of bank notes in old-timey banking, and BTC being the gold bars sitting in bank vaults, in this metaphor.
Animats 4y ago
Yes. Stablecoins have two stable points: 1 and 0.
- neonnoodle 4y ago
  The space is ripe for a quantum stablecoin
nrp 4y ago
This is truly wild to watch happening in real time. At the time he published his post, UST was around $0.90 and the recovery path to near term stability at $1 seemed plausible. At the time of the HN post, it was at $0.30 and the death spiral to $0 seemed plausible. As I write this comment, it's sitting at just about $0.50, which is obviously not a stable position for a stablecoin. It seems self-evident that a stablecoin ever being closer to $0 than $1 should destroy confidence in it, but apparently not immediately!
xmcqdpt2 4y ago
"As of right now I don’t know what the answer is and wouldn’t want to hazard a guess, though I will say that $0.91 is much closer to $1 than it is to $0." I think what the UST situation is showing is that for a stablecoin 0.90$ is a lot closer to 0$ than to 1.00$. There is probably an interesting mathematical structure to the 0 and 1$ attractors.

sc68cal 4y ago

> And so the LFG bought a bunch of Bitcoin and promised to use it to defend Terra’s peg to the dollar. If one Terra goes down to $0.90, instead of turning Terra into Luna and selling them in a death spiral, the LFG can buy Terra for $0.90 and pay for it in Bitcoin. If the LFG has enough Bitcoin, and if Bitcoin’s price holds up, then it can defend the peg and keep the price of Terra close to $1. So they've basically re-invented fractional reserve banking, with Terra/Luna taking the place of bank notes in old-timey banking, and BTC being the gold bars sitting in bank vaults, in this metaphor.

Animats 4y ago

Yes. Stablecoins have two stable points: 1 and 0.

neonnoodle 4y ago
The space is ripe for a quantum stablecoin

nrp 4y ago

This is truly wild to watch happening in real time. At the time he published his post, UST was around $0.90 and the recovery path to near term stability at $1 seemed plausible. At the time of the HN post, it was at $0.30 and the death spiral to $0 seemed plausible. As I write this comment, it's sitting at just about $0.50, which is obviously not a stable position for a stablecoin. It seems self-evident that a stablecoin ever being closer to $0 than $1 should destroy confidence in it, but apparently not immediately!

xmcqdpt2 4y ago

"As of right now I don’t know what the answer is and wouldn’t want to hazard a guess, though I will say that $0.91 is much closer to $1 than it is to $0." I think what the UST situation is showing is that for a stablecoin 0.90$ is a lot closer to 0$ than to 1.00$. There is probably an interesting mathematical structure to the 0 and 1$ attractors.