black scholes formula

[lasttrader]

Does anyone out there know if we need to know to this formula by heart.

thanks in advance

[chetan86]

i doubt it very much - the LOS which mentions the BS model is “Explain and evaluate assumptions underlying the BSM model” so i reckon its more to do with the assumptions of the model rather than knowing the formula and being able to input the numbers to calculate anything.

[summerside182]

he is right, you don’t have to know it

[Factor_hedge]

well , if you look at the BSM formula for a call it looks similar to intrinsic value formula for call option = (So - X . e^ -rt) but BSM inculdes 2 more variables the N(D1) and N(D2) wherein the former stands for the delta . I dont think meorizing N(D1) , N(D2) is required .

Similarly , BSM for put is much like the intrinsic value of put = (X . e^ - rt - So) , but the delta of a put is 1-N(D1) .

C = So . N(D1) - X . e^ - rt . N(D2)

p = X . e^ - rt . N(1- D2) - So . (1- n(D1) )

the reason N(D1) , N(D2) are included is to determine the probability the call/put being excersised . N(x) calculates the probability of obtaining a value less than x

Once you get the hang of the variables you dont have to memorize , same goes the case with formulas for N(D1) and N(D2) .

[drakesterling]

@Factor_hedge

well , if you look at the BSM formula for a call it looks similar to intrinsic value formula for call option = (So - X . e^ -rt) but BSM inculdes 2 more variables the N(D1) and N(D2) wherein the former stands for the delta . I dont think meorizing N(D1) , N(D2) is required .

Similarly , BSM for put is much like the intrinsic value of put = (X . e^ - rt - So) , but the delta of a put is 1-N(D1) .

C = So . N(D1) - X . e^ - rt . N(D2)

p = X . e^ - rt . N(1- D2) - So . (1- n(D1) )

the reason N(D1) , N(D2) are included is to determine the probability the call/put being excersised . N(x) calculates the probability of obtaining a value less than x

Once you get the hang of the variables you dont have to memorize , same goes the case with formulas for N(D1) and N(D2) .

Obviously. lol

[makaveli_99]

im hoping we don’t have to memorize it…or the black model or d(1) and d(2). The LoS doesn’t say to calculate or demonstate, just says explain and evaluate the assumptions of the BSM model

[whatsyourgovt]

I think its more prudent to understand some of the dynamics. If i am not mistake, questions like: if the rf rate increases, how will that effect a call price, or given n(d1), how would you hedge a portfolio

[JACT]

@drakesterling

@Factor_hedge

well , if you look at the BSM formula for a call it looks similar to intrinsic value formula for call option = (So - X . e^ -rt) but BSM inculdes 2 more variables the N(D1) and N(D2) wherein the former stands for the delta . I dont think meorizing N(D1) , N(D2) is required .

Similarly , BSM for put is much like the intrinsic value of put = (X . e^ - rt - So) , but the delta of a put is 1-N(D1) .

C = So . N(D1) - X . e^ - rt . N(D2)

p = X . e^ - rt . N(1- D2) - So . (1- n(D1) )

the reason N(D1) , N(D2) are included is to determine the probability the call/put being excersised . N(x) calculates the probability of obtaining a value less than x

Once you get the hang of the variables you dont have to memorize , same goes the case with formulas for N(D1) and N(D2) .

Obviously. lol

seriously. calm down. I can’t compete with this level of geekdom.

[S2000magician]

You do not need to have memorized the Black-Scholes-Merton formula.

[S2000magician]

@whatsyourgovt

I think its more prudent to understand some of the dynamics. If i am not mistake, questions like: if the rf rate increases, how will that a ffect a call price?

That’s a lot easier to see in the put-call parity formula than in Black-Scholes-Merton:

p0 + S0 = c0 + PV(X).

If rf increases, PV(X) decreases (right side is smaller), so either p0 decreases (left side has to get smaller), or c0 increases (right side has to get bigger again):

• When rf increases, call prices increase
• When rf increases, put prices decrease