3 ms·

It's not even needed to introduce weird derivatives, a publicly traded company can announce $1,000,000 dollar prizes for factoring a RSA keys. In fact, you can

It's not even needed to introduce weird derivatives, a publicly traded company can announce $1,000,000 dollar prizes for factoring a RSA keys. In fact, you can embed more powerful oracles like this, e.g. publicly traded company announces a $1,000,000 dollar prize for anyone who submits an inconsistent statement in ZFC. The EMH is a framework. With any scrutiny it's obvious that it can at most be first order accurate since obtaining and processing information is costly. This is not a particularly new insight or one that would be a surprise to any proponent of the EMH.

yowlingcat 6y ago
> The EMH is a framework. With any scrutiny it's obvious that it can at most be first order accurate since obtaining and processing information is costly. This is not a particularly new insight or one that would be a surprise to any proponent of the EMH. Great clarification. I'm always suspicious of interpretations of EMH that are a lot more literal than they should be, drawing effectively false conclusions from double negations. I think the truth is as you said.
whatshisface 6y ago
>at most be first order accurate The EMH is accurate to way better than the tangent line, which is what a first order approximation is.
- murbard2 6y ago
  When someone says a consideration is orthogonal to another one, they are not literally embedded in an Euclidean space. In this context, first order is used loosely to mean that the deviations come from effects which are only secondary in magnitude and relevance, not to indicate that we're looking at the asymptotic behavior of a formula when a parameter is taken to be infinitesimal.
  - whatshisface 6y ago
    The trouble is that EMH is a formula, so the "to first order" thing has a literal meaning that is not what the author meant. It's kind of like saying that "1-x^2/2 is cos(x) to first order," moreso than saying "my business plan is a good idea for every involved party, to first order."
ChrisLomont 6y ago
The complexity of factoring RSA keys is not known, so this method isn’t sufficient. It is known on a quantum computer factoring is polynomially hard.

yowlingcat 6y ago

> The EMH is a framework. With any scrutiny it's obvious that it can at most be first order accurate since obtaining and processing information is costly. This is not a particularly new insight or one that would be a surprise to any proponent of the EMH. Great clarification. I'm always suspicious of interpretations of EMH that are a lot more literal than they should be, drawing effectively false conclusions from double negations. I think the truth is as you said.

whatshisface 6y ago

>at most be first order accurate The EMH is accurate to way better than the tangent line, which is what a first order approximation is.

murbard2 6y ago
When someone says a consideration is orthogonal to another one, they are not literally embedded in an Euclidean space. In this context, first order is used loosely to mean that the deviations come from effects which are only secondary in magnitude and relevance, not to indicate that we're looking at the asymptotic behavior of a formula when a parameter is taken to be infinitesimal.
- whatshisface 6y ago
  The trouble is that EMH is a formula, so the "to first order" thing has a literal meaning that is not what the author meant. It's kind of like saying that "1-x^2/2 is cos(x) to first order," moreso than saying "my business plan is a good idea for every involved party, to first order."

ChrisLomont 6y ago

The complexity of factoring RSA keys is not known, so this method isn’t sufficient. It is known on a quantum computer factoring is polynomially hard.