3 ms·

He is wrong in that the sentence of his that I quoted contained two statements: "when the volatility of the underlying security increases, [1] arbitrage pressur

He is wrong in that the sentence of his that I quoted contained two statements: "when the volatility of the underlying security increases, [1] arbitrage pressures push the corresponding binary option to trade closer to 50% and [2] become less variable over the remaining time to expiration." His calculation proves [1], but it's [2] that is the basis of his criticism of Silver. And it's just not true, as can be seen even in a simple random-walk model.

mike_ivanov 7y ago
Prove it.
mike_ivanov 7y ago
"[2] become less variable over the remaining time to expiration ... And it's just not true, as can be seen even in a simple random-walk model." Dear @aubreyclayton, I'm genuinely interested in seeing how you arrived at this conclusion. Would you kindly share the proof, or at least explain the logic behind it. Thank you.
- aubreyclayton 7y ago
  Sure. If you've studied stochastic processes like Brownian motion, check out the derivation starting with the last paragraph on p.4 of this note: https://drive.google.com/file/d/1tQj4ZGja6jKADJGiFj-dcQAirWoUFX46/view?usp=sharing https://drive.google.com/file/d/1tQj4ZGja6jKADJGiFj-dcQAirWo... If that's foreign to you, you might be interested in this write-up I did about election-forecasting in which I considered the same example, just in discrete time rather than continuous time: http://nautil.us/issue/70/variables/how-to-improve-political-forecasts http://nautil.us/issue/70/variables/how-to-improve-political... The basic idea is that if you increase the volatility of a random-walk process, say by making the step-sizes larger, that won't actually make the probability of finishing above where you started any less (or more) volatile. The higher volatility means that from any given starting point your final resting place is more dispersed, but you're also more likely to range farther from home as you go. The two effects exactly cancel. Taleb's critique misses the second part of that.
  - mike_ivanov 7y ago
    Thank you.

mike_ivanov 7y ago

Prove it.

mike_ivanov 7y ago

"[2] become less variable over the remaining time to expiration ... And it's just not true, as can be seen even in a simple random-walk model." Dear @aubreyclayton, I'm genuinely interested in seeing how you arrived at this conclusion. Would you kindly share the proof, or at least explain the logic behind it. Thank you.

aubreyclayton 7y ago
Sure. If you've studied stochastic processes like Brownian motion, check out the derivation starting with the last paragraph on p.4 of this note: https://drive.google.com/file/d/1tQj4ZGja6jKADJGiFj-dcQAirWoUFX46/view?usp=sharing https://drive.google.com/file/d/1tQj4ZGja6jKADJGiFj-dcQAirWo... If that's foreign to you, you might be interested in this write-up I did about election-forecasting in which I considered the same example, just in discrete time rather than continuous time: http://nautil.us/issue/70/variables/how-to-improve-political-forecasts http://nautil.us/issue/70/variables/how-to-improve-political... The basic idea is that if you increase the volatility of a random-walk process, say by making the step-sizes larger, that won't actually make the probability of finishing above where you started any less (or more) volatile. The higher volatility means that from any given starting point your final resting place is more dispersed, but you're also more likely to range farther from home as you go. The two effects exactly cancel. Taleb's critique misses the second part of that.
- mike_ivanov 7y ago
  Thank you.