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Let the Black-Scholes formula be defined as the function $f(S, X, T, r, v)$.

I'm curious about functions that are computationally simpler than the Black-Scholes that yields results that approximate $f$ for a given set of inputs $S, X, T, r, v$.

I understand that "computationally simpler" is not well-defined. But I mean simpler in terms of number of terms used in the function. Or even more specifically, the number of distinct computational steps that needs to be completed to arrive at the Black-Scholes output.

Obviously Black-Scholes is computationally simple as it is, but I'm ready to trade some accuracy for an even simpler function that would give results that approximate B&S.

Does any such simpler approximations exist?

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up vote 25 down vote accepted

This is just to expand a bit on vonjd's answer.

The approximate formula mentioned by vonjd is due to Brenner and Subrahmanyam ("A simple solution to compute the Implied Standard Deviation", Financial Analysts Journal (1988), pp. 80-83). I do not have a free link to the paper so let me just give a quick and dirty derivation here.

For the at-the-money call option, we have $S=Ke^{r(T-t)}$. Plugging this into the standard Black-Scholes formula $$C(S,t)=N(d_1)S-N(d_2)Ke^{-r(T-t)},$$ we get that $$C(S,t)=\left[N\left(\frac{1}{2}\sigma\sqrt{T-t}\right)-N\left(-\frac{1}{2}\sigma\sqrt{T-t}\right)\right]S.\qquad\qquad(1)$$ Now, Taylor's formula implies for small $x$ that $$N(x)=N(0)+N'(0)x+N''(0)\frac{x^2}{2}+O(x^3).\qquad\qquad\qquad\qquad(2)$$ Combining (1) and (2), we will get with some obvious cancellations that $$C(S,t)=S\left(N'(0)\sigma\sqrt{T-t}+O(\sigma^3\sqrt{(T-t)^3})\right).$$ But $$N'(0)=\frac{1}{\sqrt{2\pi}}=0.39894228...$$ so finally we have, for small $\sigma\sqrt{T-t}$, that $$C(S,t)\approx 0.4S\sigma\sqrt{T-t}.$$ The modified formula $$C(S,t)\approx 0.4Se^{-r(T-t)}\sigma\sqrt{T-t}$$

gives a slightly better approximation.

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Good answer. I wonder if there are any approximations for options that are not at the money? My simple approach would be to assume a delta of 50% for the ATM call option and imply a price for the non ATM option as Option = ATM Price + 0.5*(Strike - Forward). Anyone got anything better? – Robert Feb 7 '12 at 12:08

This one is the best approximation I have ever seen:

If you hate computers and computer languages don't give up it's still hope! What about taking Black-Scholes in your head instead? If the option is about at-the-money-forward and it is a short time to maturity then you can use the following approximation:

call = put = StockPrice * 0.4 * volatility * Sqrt( Time )

Source: http://www.espenhaug.com/black_scholes.html

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In addition to what vonjd already posted I would recommend you to look at the E.G. Haug's article - The Options Genius. Wilmott.com. You can find some aproximations of BS not only for vanilla european call and put but even for some exotics. For example:

  • chooser option: call = put = $0.4F_{0} e^{-\mu T}\sigma(\sqrt{T}-\sqrt{t})$
  • asian option: call = put = $0.23F_{0} e^{-\mu T}\sigma(\sqrt{T}+2\sqrt{t})$
  • floating strike lookback call = $0.8F_{0} e^{-\mu T}\sigma\sqrt{T} - 0.25{\sigma^2}T$
  • floating strike lookback put = $0.8F_{0} e^{-\mu T}\sigma\sqrt{T} + 0.25{\sigma^2}T$
  • forward spread: call = put = $0.4F_{1} e^{-\mu T}\sigma\sqrt{T}$
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The Black-Scholes 'normal-vol' formula leads quickly to a similar approximation to the one described by olaker.

Click here for a paper which contains a formal derivation of the call and put prices based on a normal model (ie a brownian motion rather than a geometric brownian motion).

The formula for the call price is:

$$\text{Call} = (F-K)N(d_1) + \frac{\sigma \sqrt{T-t}}{\sqrt{2 \pi}} e^{-\frac{1}{2} d_1^2},$$


$$d_1 = \frac{F-K}{\sigma\sqrt{T-t}}.$$

BTW, I work in fixed income, so I always tend to write a version that is suitable for swaptions.

Options which are ATM

You can see that for $F=K$ this becomes $\text{ATMCall} = \frac{\sigma \sqrt{T-t}}{\sqrt{2 \pi}} \approx 0.4 \, \sigma \, \sqrt{T-t}$.

Options which are not ATM

I have recently discovered a generalization of this formula which works very well for strikes which are not at the money too. See my blog for a longer discussion, here are the main points.

The standard decomposition for an option is:

$$\text{Option value} = \text{Intrinsic value} + \text{Time value}.$$

In the Black-Scholes normal formula above, if you investigate the term $(F-K)N(d_1)$ in a spreadsheet, you’ll see that for small levels of volatility and maturity (try, for example, $\sigma=0.0025$, Maturity=1) it is actually quite close to $\max(0,F-K)$ – which is the intrinsic value of the call.

Consequently, the BS normal formula is almost:

$$\text{Call Price} = \text{Intrinsic Value} + \text{ATMPrice} \times e^{-\frac{1}{2} d_1^2},$$


$$\text{ATMPrice} = \frac{\sigma \sqrt{T-t}}{\sqrt{2 \pi}}.$$

However, if you compare this approximation to the true BS formula in a spreadsheet, you’ll see that around the strike (especially for larger values of $\sigma$) it gives too much value to the call: basically the term $\text{ATMPrice} \times e^{-\frac{1}{2}d_1^2}$, is too big when $d_1$ is non-zero and small. This is telling us that the difference between $(F-K)N(d_1)$ and $\max(0,F-K)$ gets important near the strike. As I say, have a look in a spreadsheet.

Nonetheless, this simple-but-wrong formula for the Call Price points us in the right direction: it shows that the time value of the option should be written in terms of the price of the ATM option.

Here is my solution.

I call it The Hardy Decomposition:

$$\text{Call Price} = \text{Intrinsic Value} + \text{ATMPrice}\times\text{HardyFactor}$$ where

$$\text{HardyFactor} = e^{-\frac{1}{2} d_1^2} + \frac{d_1}{0.4}N(d_1) – \max(\frac{d_1}{0.4},0).$$

So far this is just a rearrangement of the original Black-Scholes 'normal-vol' formula. The key result is that the $\text{HardyFactor}$ is well approximated by a simple expression:

$$\text{HardyFactor} \approx (1- 0.41 |d_1|) e^{-|d_1|}$$

So you can actually use the following as a pretty-good approximation for call prices:

$$\text{Call Price} = \text{Intrinsic Value} + \text{ATMPrice}\times (1- 0.41 |d_1|) e^{-|d_1|}.$$

A similar result holds for put options.

You can use this Hardy Decomposition to calculate option prices in your head - you only need to remember a few values:

Remember these few values

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As other people have said, you need to approximate the cumulative. The problem is that wherever you look you will find that to approximate it you will need to use exponentials or trigonometric functions which are also very expensive. What you can do is to build yourself a cubic spline with pre-cached values for the cumulative and calculate the value at other points x by (cubic) interpolation. That will make it much faster. Possibly you will also call this method with a series of ordered values so that you can avoid doing the binary search to locate the interval. You will have a cached index for the last interval located and look around that one.

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