### What are some useful approximations to the Black-Scholes formula?

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?

olaker Correct answer

10 years agoThis 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.

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 I agree with the idea but there seems a sign mistake. Should it be `Option = ATM Price + 0.5*(Forward - Strike)`? So, out-of-the-money option (`Strike>Forward`) should be less valuable than the ATM option.

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Robert 9 years ago

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?