delta depreciation

I wrote ATM, not close to ATM. retry.

 

stock = 47

delta = .5173

 

~1.62 (if I did not make miscalculations)

If I am right give people data to test for themselves.

 

credit where credit is due.

You are correct. Pretty cool trick. I see how you do that. Thanks.

 

Thanks for your comments and for keeping the tricks personal. I have been wondering if I should publish such little things (which I have not seen anywhere else) in a book I have been thinking to publish-- On one hand they may provide uniqueness and possibly some demonstrated "expertise" to the book, , but on the other hand readers may not want to see too many people looking from the same angles as themselves.

Cheers!

 

Can you share the trick?

 

Donnap was the first to give the data, and to learn from the solution, so she deserves to know it, as she has earned it.

I think Donnap is doing the right thing to keep it from public, unless the trick is well known elsewhere, which I have not seen. If no one before me has come up with it, I believe I have intellectual rights on it.

 

Okay I think I got it

You derived it like that:

2*spot*(delta-0.5)

ATM, N(d1) and N(d2) are symmetrical around 0.5. That means, N(d1)-N(d2)=2*(N(d1)-0.5)=2*(delta-0.5)
Because there is 0 cost of cary, ATM S=K
And a call is

C=SN(d1)-K(Nd2)
C=SN(d1)-SN(d2)
C=S*2*(N(d1)-N(d2))=S*2*(delta-0.5)

Am I missing something ?

 

stk = 50
delta = .55