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complyorexplain
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### sigma root (T-t)

If we look inside the Black Scholes d1 term, given here https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model#Black%E2%80%93Scholes_formula, we find the denominator term ‘sigma (T-t)’, where 'T-t' signifies the time at expiry.

What do we think this term means? I.e. what quantity, or what sort of quantity, does it signify?

DavidJN
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### Re: sigma root (T-t)

Get thee any introductory text on derivatives. Kindergarten level questions should please be posted in the Student Forum.

complyorexplain
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### Re: sigma root (T-t)

I have read many introductory texts on derivatives and have been pricing options for more than 30 years. My question remains. (Of course it is a rhetorical question, but I am interested in what people answer).

Cuchulainn
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### Re: sigma root (T-t)

What do we think this term means? I.e. what quantity, or what sort of quantity, does it signify?
The question is ambiguous, depends on the observer.
Since no one knows what you are trying to get at it helps to elaborate.

The only thing I can guess at it is: what's the engineering units of $\sqrt{T - t}$. ? $d_1$ should be dimensionless.
Or a number in a formula.
My C++ Boost code gives
262537412640768743.999999999999250072597198185688879353856337336990862707537410378210647910118607313

http://www.datasimfinancial.com
http://www.datasim.nl

complyorexplain
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Joined: November 9th, 2015, 8:59 am

### Re: sigma root (T-t)

What do we think this term means? I.e. what quantity, or what sort of quantity, does it signify?
The question is ambiguous, depends on the observer.
Since no one knows what you are trying to get at it helps to elaborate.

The only thing I can guess at it is: what's the engineering units of $\sqrt{T - t}$. ? $d_1$ should be dimensionless.
Or a number in a formula.
T is the time at expiry, t is the time at which the option is priced. Therefore T-t is the time to expiry. Units would be seconds, hours, or whatever unit of time is chosen. The square root of that is the number which, multiplied by itself, equals T-t.

"Since no one knows what you are trying to get at ...". It should be clear what I am trying to get at, namely what the term means. Time to expiry seems clear. What about sigma? I searched in Hull this afternoon and could find no clear answer. He introduces the term 'volatility' without explaining at any place exactly what he means.

complyorexplain
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### Re: sigma root (T-t)

Samuelson (“A rational theory of warrant pricing” 1965) assumes that the distribution of possible values of the stock *when the warrant matures* is log-normal, which suggests that sigma root t correlates with the standard deviation of the (log of the) distribution *at expiry*. But is that correct?

complyorexplain
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### Re: sigma root (T-t)

Hull (8th edition p.330) says “the standard deviation of ln S(T) is  sigma root T”, where ‘T’ is T-t here, and S(T) is asset price at expiry.

Again, is that everyone's understanding of what sig root t signifies? I.e. the standard deviation of the asset price at expiry?

DavidJN
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### Re: sigma root (T-t)

People have also wondered about the propriety of a similar time scaling technique in the VaR world, namely, scaling VaR calculations made with daily data to longer time intervals (sometimes 10 days is used). This is a very similar problem to the one you posed, and it has been discussed a fair bit in the literature because regulators require banks seeking internal model approvals to "prove" that the scaling is valid.

It is valid given the distributional assumptions, but most people are of course aware that the classic distributional assumptions of the underlying in the Black Scholes world are probably more convenient than true. So, are you perhaps asking whether the scaling is valid given the distributional assumption, or whether the distributional assumption itself is a good one?

DavidJN
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### Re: sigma root (T-t)

https://quant.stackexchange.com/questio ... ot-of-time

They characterize the square root of time rule as a property of geometric Brownian motion. How well it applies to the real world is another story.

Collector
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### Re: sigma root (T-t)

"This is simply to illustrate the importance of understanding the different implications of time, and to get an intuition about why options can both increase and decrease with respect to time to maturity. We could naturally have a different scaling factor in relation to time than the square root of time ‘rule,’ but that is not the topic at this time......But back to uncertainty and time: when it comes to uncertainty, the shorter the time interval, the less uncertainty. This is the case, no matter if it is linked to the square root of time or another time scaling factor. A short time interval will always be inside a longer time interval, so a short time period is always less uncertain than the long time period (that contains that shorter time period). If the time interval is zero, however, then can we even have uncertainty? But let us say that time intervals come in discrete intervals. In modern physics, it is believed...."  Jan W-Mag

To get intuition for T and sigma then also the ATMF approximation  is useful. MC also very useful here!

well I just like to philosophize about things I dont understand, in particular since I am not a philosopher!

complyorexplain
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### Re: sigma root (T-t)

https://quant.stackexchange.com/questio ... ot-of-time

They characterize the square root of time rule as a property of geometric Brownian motion. How well it applies to the real world is another story.
Right, but the question was what people here (who I take to be a representative sample of the quant community) think is the *meaning* of the sigma root t term. I offered on interpretation, namely the s.d. of the asset price at expiry. Can I take it that everyone here agrees with that interpretation?

complyorexplain
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### Re: sigma root (T-t)

"when it comes to uncertainty, the shorter the time interval, the less uncertainty"

I am wondering about that.  Is it 'less' or 'less or equal'? Imagine creatures who live whole lives within a nano second, so that what for us is very short, is for them many generations.

But that's a separate subject.

Alan
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### Re: sigma root (T-t)

https://quant.stackexchange.com/questio ... ot-of-time

They characterize the square root of time rule as a property of geometric Brownian motion. How well it applies to the real world is another story.
Right, but the question was what people here (who I take to be a representative sample of the quant community) think is the *meaning* of the sigma root t term. I offered on interpretation, namely the s.d. of the asset price at expiry. Can I take it that everyone here agrees with that interpretation?

No. Too sloppy.

It is the standard deviation of the probability distribution of the logarithm of the asset price at expiration, under the assumption of GBM.

As was more or less said by Hull.

Collector
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Joined: August 21st, 2001, 12:37 pm

### Re: sigma root (T-t)

https://quant.stackexchange.com/questio ... ot-of-time

They characterize the square root of time rule as a property of geometric Brownian motion. How well it applies to the real world is another story.
Right, but the question was what people here (who I take to be a representative sample of the quant community) think is the *meaning* of the sigma root t term. I offered on interpretation, namely the s.d. of the asset price at expiry. Can I take it that everyone here agrees with that interpretation?

No. Too sloppy.

It is the standard deviation of the probability distribution of the logarithm of the asset price at expiration, under the assumption of GBM.

As was more or less said by Hull.
is this too sloppy? It is the standard deviation of the probability distribution of the logarithm of the asset price at expiration as predicted from now, under the assumption of GBM.

just to be a pain in the a, if the expiration point is exact, then "at expiration" always have zero time interval in theory, so then the std. dev of the probability distribution of the logarithm of the asset price at expiration is sigma*sqrt(0). but then I am not native english. however from now (if now >= Expiry) then T>=0

"namely the s.d. of the asset price at expiry" Bachelier ? Gauss distributed price.

complyorexplain
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Posts: 141
Joined: November 9th, 2015, 8:59 am

### Re: sigma root (T-t)

No. Too sloppy.

It is the standard deviation of the probability distribution of the logarithm of the asset price at expiration, under the assumption of GBM.

As was more or less said by Hull.
You are quite right, and I was being imprecise, but note I quoted Hull further up. But my question, again, is what people think it means. There seems to be broad agreement that it means the standard deviation of the probability distribution of the log of the asset price at expiration.

The 'as predicted from now' is an interesting addition.

Does anyone differ? Can anyone think of any others not present, who might differ? Is there a broad consensus among contemporary financial theorists?