Harashima Senju
Ton up Member
I know how to derive the formula but I have no idea what it means or how it is useful for this course. Q&A doesn't seem to test it as well. Can anyone kindly explain to me the importance of this PDE
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Meaning and purpose of Black-Scholes PDE
- Thread starter Harashima Senju
- Start date
Hi Harashima,
It's used to sign off derivative formulae, subject to the relevant boundary conditions...
eg
1. I propose that f(St,t) = cos(St) - 4tK is the fair price of a call option. Check if this satisfies the PDE. It doesn't, it's nonsense, juts made it up. Chuck it in the bin.
2. I propose that f(St,t) = St - Kexp(-r(T-t)) is the fair price of a call option. Check if this satisfies the PDE. It does. You could check this, homework task ;-) However, in my formula, we have f(ST,T) = ST - K, which is not the correct BC for a call. Chuck it in the bin.
I give up :-(
3. Black and Scholes propose that f(St,t) = see page 47 with q=0 is the fair price of a call option. Check if this satisfies the PDE. It does. Check it gives the right BC, it does, f(ST,T) = max{ST - K,0}. Wahoo! We sign it off, "well done Black and Scholes, your proposal is correct. Here, have a Nobel prize (but only Scholes can have it because Black is dead)."
Good luck!
John
It's used to sign off derivative formulae, subject to the relevant boundary conditions...
eg
1. I propose that f(St,t) = cos(St) - 4tK is the fair price of a call option. Check if this satisfies the PDE. It doesn't, it's nonsense, juts made it up. Chuck it in the bin.
2. I propose that f(St,t) = St - Kexp(-r(T-t)) is the fair price of a call option. Check if this satisfies the PDE. It does. You could check this, homework task ;-) However, in my formula, we have f(ST,T) = ST - K, which is not the correct BC for a call. Chuck it in the bin.
I give up :-(
3. Black and Scholes propose that f(St,t) = see page 47 with q=0 is the fair price of a call option. Check if this satisfies the PDE. It does. Check it gives the right BC, it does, f(ST,T) = max{ST - K,0}. Wahoo! We sign it off, "well done Black and Scholes, your proposal is correct. Here, have a Nobel prize (but only Scholes can have it because Black is dead)."
Good luck!
John
Harashima Senju
Ton up Member
It took me a while to figure out page 47 (Gold Tables Booklet), doh!
Wow! Great explanation and the homework made it crystal clear
Writing the BC as ST-K is false in the exam, I would have made this mistake instead of checking if BC (max{ST-K, 0}) is satisfied.
Wow! Great explanation and the homework made it crystal clear
Writing the BC as ST-K is false in the exam, I would have made this mistake instead of checking if BC (max{ST-K, 0}) is satisfied.
It's actually very clever how you get the maximum, it uses the fact that ln(S/K) > 0 when S > K and < 0 when S < K. I'd give you that as another homework task were it not for it being extremely unlikely to be asked in a CT8 exam,
Good luck!
John
Good luck!
John
Harashima Senju
Ton up Member
I think I see what you mean. Both phi(d1) and phi(d2) approach zero when S = 0?
No, as t tends to T...
when S < K, Phi (- infinity) = 0, so you get 0 - 0 = 0
when S > K, Phi ( infinity) = 1, so you get S - K
when S=K Phi(0) = 0.5 but 0.5(S-K) = 0
Don't spend any time on this though, it's interesting and great if you understand it quickly, but it has never been asked in the CT8 exam. So, just ignore this now if it's not clear
when S < K, Phi (- infinity) = 0, so you get 0 - 0 = 0
when S > K, Phi ( infinity) = 1, so you get S - K
when S=K Phi(0) = 0.5 but 0.5(S-K) = 0
Don't spend any time on this though, it's interesting and great if you understand it quickly, but it has never been asked in the CT8 exam. So, just ignore this now if it's not clear