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I have two related questions concerning Black Scholes and delta hedging. I thought about this two questions, but I could not come up with an answer, so maybe you guys & girls can help me:

  1. If an option is at the money, how can the Black Scholes price be calculated in a very fast way (possibly without any big calculations)?

  2. If an option is at the money, how many shares do you have to buy in order to delta-hedge?

Bob Jansen
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user1690846
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    Is this a homework assignment? These are the most commonly presented results in any textbook on options pricing. – chrisaycock Dec 31 '12 at 12:14

2 Answers2

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  1. See this question
  2. You have to buy/sell $\Delta$ shares. $\Delta_{ATM} \approx 0.5$.
Community
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Dmitri Nesteruk
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  1. stock price * volatility * 0.4 * sqt(T), where T denotes time to expiration in years and 0.4 is coming from sqt(1/(2*pi)). The simplifying assumption here is (and that is very important and you will most likely be asked to state the assumptions should such question be asked in the interview): strike price equals underlying asset price AND asset prices are NORMALLY DISTRIBUTED (unlike the assumption in B-S) which assumes the asset price to follow an ARITHMETIC Brownian motion.

  2. As the delta is approximately (stress, not equal) 0.5, you need to hedge with about 1/2 the amount of the underlying asset that the options contract stipulates.

Matt Wolf
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  • ok thanks @Freddy, but what is an arithmetic brownian motion? What is the difference to geometric brownian motion? – user1690846 Dec 31 '12 at 09:54
  • @user1690846, well I kind of pointed to it, its a Brownian motion where the asset price distribution is assumed to be normal and not log-normal. – Matt Wolf Dec 31 '12 at 10:04
  • @chrisaycock, of course, just edited the answer – Matt Wolf Dec 31 '12 at 12:28
  • actually, asset prices do not follow arithmetic brownian motion in .4 S$\sigma * \sqrt{T}$, it is still geometric brownian motion. just do a simple taylor expansion on B-S formula and you will see – Andrew Dec 31 '12 at 21:25
  • @Andrew, I said the assumption is of the asset price to be normally distributed which is the equivalent of an asset price model of arithmetic Brownian motion. – Matt Wolf Jan 01 '13 at 05:54
  • @Andrew, and by the way, B-S assumes log normally distributed prices that is why the driving BM is of geometric nature. – Matt Wolf Jan 01 '13 at 08:02
  • @Freddy, I am saying that for the .4S$\sigma * \sqrt{T}$, this derivation comes from B-S value of a call/put. if you assume 0 interest rates, you can taylor expand it and you will see this result and as you noted, B-S follows geometric brownian motion. so it should be the asset price follows geometric brownian motion – Andrew Jan 02 '13 at 15:53
  • @Freddy, assume interest rates 0, dividend 0, then $value = SN(D1) - KN(D2) $ in geometric brownian motion. and S=K $value \approx S(N(0) + N'(0) * d1) - K(N(0) + N'(0) * d2)$ by 1st order taylor expansion $value \approx S(1/2 + \frac{\sigma\sqrt{T}}{2\sqrt{2\pi}})- K(1/2 - \frac{\sigma\sqrt{T}}{2\sqrt{2\pi}})$, since S=K $value \approx \frac{S\sigma\sqrt{T}}{\sqrt{2\pi}}$, since $1/\sqrt{2\pi} \approx 0.4$, $value \approx .4*S\sigma\sqrt{T}$. again, still gbm – Andrew Jan 02 '13 at 16:24
  • @Andrew, my friend this approximation was derived a long time before Black and Scholes were ever born by Bachelier. The assumption of his approximation is an arithmetic browning motion no matter how much you argue about it. – Matt Wolf Jan 02 '13 at 16:32
  • Here is a link which makes the difference between BS and Bachelier clearer. For short time to expiration options with strikes at the money this comes out to be very close to the Taylor expansion you gave, showing that indeed it is still a valid and good enough approximation even with the arithmetic BM assumption. http://www.mat.univie.ac.at/~schachermayer/preprnts/prpr0121.pdf – Matt Wolf Jan 02 '13 at 16:44
  • interesting. did not know that even under stochastic vol bachelier and bs are that close – Andrew Jan 04 '13 at 18:44