Option delta as a probability

derivstrading's picture
Rank: Neanderthal | 2,546

So ive been meaning to write this for a couple weeks now in response to a thread about delta of options being thought of as the probability the option expires in the money.

The Black Scholes equation is as follows:

c = SN(d1) - PV(K)N(d2)

delta is the sensitivity of the option price to changes in the underlying and by simply differentiation:

dc/dS = N(d1)

Mathematically speaking, the risk neutral probability that an option expires in the money is actually N(d2), not N(d1) as is often believed.

N(d2) = pr ( ln(S) > ln(x) ) at expiry

If we ignore discounting, and take the idea that N(d1) is the risk neutral probability as often believed, therefore we get N(d1) = N(d2)

c = SN(d1) - KN(d1) = N(d1)*(S - X)

but as you can see this can make the call option price negative if S < X. Therefore N(d1) will be strictly greater than N(d2).

The problem is that there are two uncertainties of expiration:
1. If we get anything at all: represented by N(d2)
2. How much we get: represented by N(d1)

The key to take away is that when you are starting out learning about options, it is really not helpful to think of the delta as the probability it expires in the money. It is much more useful and practical to think about the sensitivity to underlying.

Comments (6)

Jan 27, 2011
    • 2
Jan 28, 2011

1)N(d1) = p (S(T)>Xe^-(sigma^2)(T-t)) where as N(d2)= p(S(T)>X) under risk neutral equivalent martingale measure (bond numeraire)
2) N(d1) = p
(S(T)>X) under stock-numeraire equivalent martingale measure.
3) N(d1) does Not equal to N(d2), when you ignore discounting

They could be both probabilities when stock finish in the money depends on which measure you are under.
In general, under risk neutral pricing, s(t)N(d1) is the discounted expected benefit of owning an option under risk neutral measure. I agree w the post for the most part except:
"If we ignore discounting, and take the idea that N(d1) is the risk neutral probability as often believed, therefore we get N(d1) = N(d2)"

Jan 28, 2011
GekkotheGreat:

1)N(d1) = p (S(T)>Xe^-(sigma^2)(T-t)) where as N(d2)= p(S(T)>X) under risk neutral equivalent martingale measure (bond numeraire)
2) N(d1) = p
(S(T)>X) under stock-numeraire equivalent martingale measure.
3) N(d1) does Not equal to N(d2), when you ignore discounting

They could be both probabilities when stock finish in the money depends on which measure you are under.
In general, under risk neutral pricing, s(t)N(d1) is the discounted expected benefit of owning an option under risk neutral measure. I agree w the post for the most part except:
"If we ignore discounting, and take the idea that N(d1) is the risk neutral probability as often believed, therefore we get N(d1) = N(d2)"

I didnt mean that when you ignore discounting N(d1) equals N(d2). It was just an example to show that N(d1) cannot equal N(d2). The ignore discounting is just a simplification statement, not a condition.

Jan 28, 2011

Agreed then. Good post!

Mar 29, 2014

how do derive N(d1)?

Mar 29, 2014
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