If IV is 100%

[Herkfsu]

Is it correct to say a 2 standard deviation move to the downside is impossible(on a yearly basis)? ie something can't go down 200%.

Same question for a stock with a 50% IV and a 3 standard deviation move. Not possible? Am I thinking about this correctly?

EDIT: I thought I remember watching a TastyTrades video saying 30 delta is ~1 SD, which makes sense if 1 SD covers 68% of possibilities. If IV is 100%, where would the 30 delta line up(on the surface it seems like it would be at 0, but I know that makes no sense).

Thanks!

 

[oldmonk]

The log price changes are normally distributed, not the price itself.

If IV is 100%, then the implied annual 1SD log price change would be 1. If the current price is 100, then a 1SD down move would be 100*exp(-1) = 37. A 2SD down move would be 100*exp(-2) = 13.5, and so on. It will go to zero in the limit of -Inf.

 

[Herkfsu]

Ok, so if I see a SPY IV of 13, that is a log 13%, not a linear(Or geometric? Whats the "opposite" of log?)? So the actual 1SD change over the next year would be 13.88%?

 

[guru]

https://www.tastytrade.com/tt/learn/standard-deviation

 

[Herkfsu]

https://www.tastytrade.com/tt/learn/standard-deviation

Is this right though?? From the link...

For example, if a $100 stock is trading with a 20% implied volatility, the standard deviation ranges are:
- Between $80 and $120 for 1 standard deviation
- Between $60 and $140 for 2 standard deviations
- Between $40 and $160 for 3 standard deviations

From this, we can conclude that market participants are pricing in a:
- 68% probability of the stock closing between $80 and $120 a year from now
- 95% probability of the stock closing between $60 and $140 a year from now
- 99.7% probability of the stock closing between $40 and $160 a year from now

Normal distribution of returns = log normal distribution of prices. This would indicate otherwise.

Using the example above, a 3 SD move, would be 100*e^-.6 on the downside and 100*e^.6 on the upside, right?

 

[oldmonk]

Is this right though?? From the link...

For example, if a $100 stock is trading with a 20% implied volatility, the standard deviation ranges are:
- Between $80 and $120 for 1 standard deviation
- Between $60 and $140 for 2 standard deviations
- Between $40 and $160 for 3 standard deviations

From this, we can conclude that market participants are pricing in a:
- 68% probability of the stock closing between $80 and $120 a year from now
- 95% probability of the stock closing between $60 and $140 a year from now
- 99.7% probability of the stock closing between $40 and $160 a year from now

Normal distribution of returns = log normal distribution of prices. This would indicate otherwise.

Using the example above, a 3 SD move, would be 100*e^-.6 on the downside and 100*e^.6 on the upside, right?

The difference between normal and log normal only becomes apparent for larger IVs. For 20% IV the actual 1 SD implied move would be 100*exp(-0.2) = 81 and 100*exp(0.2) = 122. That's fairly close to what you have. The approximation works because the first order Taylor expansion for exp(x) is 1 + x. For larger values, like what you had in your OP (and for the 3 SD move), the same approximation would not work as well.

 

[Herkfsu]

The difference between normal and log normal only becomes apparent for larger IVs. For 20% IV the actual 1 SD implied move would be 100*exp(-0.2) = 81 and 100*exp(0.2) = 122. That's fairly close to what you have. The approximation works because the first order Taylor expansion for exp(x) is 1 + x. For larger values, like what you had in your OP (and for the 3 SD move), the same approximation would not work as well.

Thanks. Is my example correct though? The link to Tastytrade has a wrong example, right?

 

[oldmonk]

Thanks. Is my example correct though? The link to Tastytrade has a wrong example, right?

Yes, a 3SD move would be 100*exp(+/- 0.6), which is 54 and 182. The tastytrade numbers are clearly off here. That's because they're assuming the prices are normally distributed. But otherwise their logic seems right.