Sunday, May 22, 2011

Black Scholes Conundrum Concised

I have never been enthused by this model after I left college. Not only large parts of this model’s derivation is taken from the seminal yet unnoticed paper of its time by Louis Bachelier but also the assumption of normality reduces it to be worth for the books of economic and financial history than otherwise.  What’s even worse is that this started a trend that tried to bring this subject of financial economics more apt as a social science into the realms of physical sciences. Anyways for those inclined to use this model or still revere it with some awe let me make your life simple. In one of my insomniac nights I worked upon this derivation, I would cut the entire crap used in the derivation of Black Scholes band bring out the same model in a very simple form. So here we go.
PS: The content would be a little technical but can be easily understood with a few minutes of patience as the mathematics is pretty simple (as was the aim of this whole exercise)
Basically if I buy a call and a put, the price that I should pay is the expected movement of the stock in that period. So in essence

C+P = E[S]  - Eq 1
After time t, where t is the maturity of the option. Now in the realm of Black Scholes the distribution of the returns from the stock is Normal. So instead of using the expected return E[S] I have to be using the Standard deviation or √E[S^2] and the relationship between √E[S^2] and E[S]in the world of Normality is pretty simple:
E[S]/ √E[S^2] = √(2/π) – Eq 2
We can also represent √E[S^2]  by σ so in essence Eq 2 now becomes
E[S] = √(2/π) * σ
Hence Eq 1 now becomes
(C+P)/S =  √(2/π) * σ
For At the money options (the starting point from where deviations are computed can be taken as 0) and assuming interest rates as negligible C≈P. Hence
2C/S = √(2/π) * σ or C/S = √(1/2π) * σ
That’s it……….. the price of the option is the standard deviation of the underlying asset divided by the square root of 2π

Now let me get to this equation by reducing the Black Scholes model. Later some other day I shall give you the proof by using geometric interpretation.
The call price (assuming interest rates as negligible) c = SN(d1) – KN(d2), the put formula is also accordingly KN(-d2) – SN(-d1), We are taking the case of At the money option as it is convenient to show. So for At the money option.
c= SN(σ/2*√t) – KN(-σ/2*√t), Since We are taking volatility for the entire duration of the option so I am going to replace σ*√t with σ and K by S (as the option is At the money). Hence
c= SN(σ/2) – SN(-σ/2), N(-σ/2) = 1- N(σ/2),
c= S(2 N(σ/2) -1), Similary p = S(2 N(σ/2) -1), Hence
c+p = 2S(2 N(σ/2) -1), as c=p Hence,
c/S = (2 N(σ/2) -1) – Eq 1, Now
N(σ/2) =[∫-∞0exp^- (σ/2)^2  + ∫0 σ/2exp^- (σ/2)^2]/sqrt(2 π)
The first part is equal to 0.5, to integrate the second part of the expression let me expand it using the Taylor series. The Taylor series expansion of the exponential term is
ex = 1 + x + x^2/2! + x^3/3! ……
replacing x by σ/2 and then integrating we get
N(σ/2) = 0.5+ [0 σ/20n (-1) n σ2n+1/2nn!(2n+1)] /sqrt(2 π)
As SD is small number (in decimals) we can ignore the higher power of σ,
Please note that if σ is high the whole notion of Normality falls apart and as does the Black Scholes model.
So expanding this expression and putting it in Eq 1 we get
c/S = [σ]/ sqrt(2 π) as we got it earlier without this whole and dance.
This expression is for At the Money options at negligible interest rates. We can do it for out of money/in the money options as well and introducing a little bit of interest rates also. So in anycase, next time if you feel bound to calculate volatility or option price using the assumptions of Normality for At the money options your formula is
C/S = √(1/2π) * σ
I wasn’t able to write anything last month so the next one would hopefully follow soon…….