BSM option pricing model

[k_gopala]

dear david,
one of the assumptions of the model is the (continuous ) risk free rate is constant and known.

But in the calculation of d1, the denominator ( sigma * sqrt(time) is common for both the two parts in the numerator. The volatility is represented for continuously compounded returns on the stock(in totality). If so, how the assumption on risk free return is taken care? Can you please enlighten with your explanation and correct me if i am missing a point. Thanks.

 

[David Harper CFA FRM]

Hi risklearner,

"The volatility is represented for continuously compounded returns on the stock (in totality)"
That's correct in a way, as the stock's return does not enter Black-Scholes, but rather the stochastic property of the stock is "represented" entirely by volatility.

The riskless rate enters three times. The rate (r) enters both d1 and d2. N(d2) is the probability the option will be struck if the stock grows at the riskless rate (r). N(d2) is effectively a Merton model (de Servigny credit Ch 2). N(d2) with riskless rate assumes the stock grows at the riskless rate and gives back the probability the option will be struck under this assumption that stock grows at riskless rate. Black-Scholes can use the riskless rate due to the risk-neutral valuation idea (namely, the option is riskier but that higher future gain is offset by a higher discount rate). It is a common misunderstanding to believe that N(d2) in B-S is the probability the option will be struck. This is why the riskless rate is important to the risk neutral idea: N(d2) understates the probability the option will be struck in the real world.

But (r) also enters in the 'parent' B-S: c = Stock*N(d1) - K*EXP(-rT)*N(d2). Here it is discounting the future strike price at the riskless rate. This owes to the premise of the B-S: we can replicate (no arbitrage) the option by purchasing the stock and borrowing the exercise price; i.e., long share + short cash. As the strike is in the future, we only need to borrow the present value (K*EXP[-rT]).

In summary, I'd say the riskless rate enters primarily by assuming the option buyer can borrow the future strike price at the riskless rate, in the theoretical risk-less world and, by the strange and wonderful risk-neutral idea, we are lucky that the option value (but not the N(d2)!) can be transferred "back through the looking glass" into the real world.

If you are looking to find the intuition (which is not totally necessary for the exam), i recommend you study N(2) and compare it to N(distance to default). Here is related screencast from last week.

The Merton Model for credit risk, which uses N(d2) except with growth rate instead of riskless rate, is an good place to start because N(d2) is not option pricing per se; it is growth math plus a distribution. We go from growth rate in Merton model to riskless rate (r) in the Black-Scholes and this becomes un-natural to the extent we grapple with the risk-neutral idea (i.e., that using riskless rate is still valid in the real world, not for N(d2), but rather for the option price).

Hope that helps!

David

 

[ahnnecabiles]

Hi David,

Jorion and Hull have different formula for computing d1. The Handbook uses ln (S/Ke-rt) while Hull has ln (S/K) --face value of K. Likewise, Jorion uses sigma while Hull uses sigma squared in the numerator. Are they the same?

Thanks!

 

[David Harper CFA FRM]

Hi Chinquee,

Yes, exactly the same. Jorion's is an unusual way but it's good to see why they are they same (this is why finance uses continuous compounding, it is easier to manipulate logs and exponents):

Jorion d1 = ln(S/(K*exp[-rt]))/(sigma*SQRT[T]) + sigma*SQRT[T]/2

multiply SECOND TERM above only by 1: numerator and denominator by (1/2)(sigma)*SQRT[T]. So terms can be added

d1 = ln(S/(K*exp[-rt]))/(sigma*SQRT[T]) + (1/2)sigma^2*T/(sigma*SQRT[T])
d1 = (ln(S/(K*exp[-rt]) + (1/2)sigma^2*T)/(sigma*SQRT[T])

and unpacking (ln(S/(K*exp[-rt]) + (1/2)sigma^2*T):
ln(S/(K*exp[-rt]) + (1/2)sigma^2*T
= ln(S) - ln(K*exp[-rt]) + (1/2)sigma^2*T
= ln(S) - [ln(K)+ ln(exp[-rt])] + (1/2)sigma^2*T
= ln(S) - ln(K) - [-rt] + (1/2)sigma^2*T

now replace back into d1:
from: d1 = (ln(S/(K*exp[-rt]) + (1/2)sigma^2*T)/(sigma*SQRT[T])
to: d1 = (ln(S) - ln(K) - [-rt] + (1/2)sigma^2*T)/(sigma*SQRT[T])
= (ln(S/K) + rt + (1/2)sigma^2*T)/(sigma*SQRT[T])
= (ln(S/K) + t*(r+ (1/2)sigma^2)/(sigma*SQRT[T])

David