Valuing a call option - BSM model

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For computing value of European option using BSM model on a dividend paying stock, in Schweser So is taken as discounted price [So * e-qt] for d1 formula.

Whereas formula doesnot mention anything. There it is spot price only.
Pls comment.


Also if you can give some brief about how to look values of N(d1) and N(d2) for CALL and PUT both in Cumulative Probability Table. And for values like .74644 or similar.
What is use of Alternative table.
If you can post some brief tutorial video for looking the table.

Thanks.


~Anil

 

[David Harper CFA FRM]

Hi Anil

Right, they are correct. There are a couple of ways to show the incorporation of dividends in Black-Scholes. IMO, the easiest is line up behind Hull:

Incorporate dividends by using the Black-Scholes and replacing Stock, S(0) with S(0)*EXP(-qT).

If you replace that into B-S, there are two replacements, and the d1 solves for LN(S0/K) + (r - q + variance/2)T.
It turns out to be the same as using LN(S0*exp(-qT)/K).

The intuition is, dividends aren't enjoyed by option holder, they reduce the option value. Or, i like better: given total return on stock is constant, higher dividend implies lower capital appreciation and option holder only gets the appreciation.

Re N() functions,

* Yes, I understand, I will post that next week (and include in the cram materials)
* It sounds like you know: you don't need to memorize N() values. You should of course know N(-1.645) = 5% and N(-2.33) = 1%, and intuitively that N(0) = 50%, and it would be good to know how to figure out N(1) and N(2) based on the ~68% and ~95.5% thumb rules...
* But I catch your drift, you mean using the lookups, i will add something at least for cram on that

David

 

[NadiaSayyam]

Hi David,
can you please brief that how we can find the N(D1) and N(D2) values for call and put.

Nadia

 

[ShaktiRathore]

Hi,
Let S0=stock selling at =27,X=strike price=30,T=time to maturity=145/365,volatility of stock=sigma=.30,dividend=q=0,continuously compounded risk-free rate =r=4%=0.04
put these values,
d1=(ln(So/X)+(r-q+.5*sigma^2)*T)/(sigma*sqrt(T))=(ln(27/30)+(0.04-0+.5*.30^2)*(145/365))/(.30*sqrt(145/365))=-.37863 =~-.38
Cumulative Standard Normal Probability:

--------------------------------------------------------------------
0.06 | 0.07 | 0.08 | 0.09
-----|-------------------------------------------------------------
0.3 | 0.6406 | 0.6443 | 0.648 | 0.6517
0.4 | 0.6772 | 0.6808 | 0.6844 | 0.6879
0.5 | 0.7123 | 0.7157 | 0.719 | 0.7224
=>for d1=-.38,we look at table above to get value of N(-d1)=N(.38)=0.648,=>N(d1)=N(-.38)=1-N(-d1)=1-N(.38)=1-0.648=0.352
Cumulative Standard Normal Probability:
d2=(ln(So/X)+(r-q-.5*sigma^2)*T]/[sigma*sqrt(T))=(ln(27/30)+(0.04-0-.5*.30^2)*(145/365))/(.30*sqrt(145/365))=-.56771=~-.57
=>for d2=-.57,we look at table above to get value of N(-d2)=N(.57)=0.7157,=>N(d2)=N(-.57)=1-N(-d2)=1-N(.57)=1-0.7157=0.2843
see example here: https://forum.bionicturtle.com/threads/black-scholes-merton-model.7692/#post-30310
thanks

 

[David Harper CFA FRM]

Same answer as @ShaktiRathore, I just wanted to dust off the mathjax, glad to see it still works

Where call and put values are:
\(c={{S}_{0}}N({{d}_{1}})-K{{e}^{-rT}}N({{d}_{2}})\)
\(p=K{{e}^{-rT}}N(-{{d}_{2}})-{{S}_{0}}N(-{{d}_{1}})\)

d1 is given by:
\({{d}_{1}}=\frac{\ln \left( \frac{{{S}_{0}}}{K} \right)+\left( r+\frac{{{\sigma }^{2}}}{2} \right)T}{\sigma \sqrt{T}} \)

d2 is given by:
\({{d}_{2}}=\frac{\ln \left( \frac{{{S}_{0}}}{K} \right)+\left( r-\frac{{{\sigma }^{2}}}{2} \right)T}{\sigma \sqrt{T}}={{d}_{1}}-\sigma \sqrt{T}\)