Can someone help me interpret this definition of delta?

Delta is a conditional probability of terminal value (St) being greater than the Strike (X) given that St > X for a call option.

Is the second part of the statement not completely redundant? This may be completely obvious, but I don't see why this is conditional.

The source of the definition is:

Initially found here, where three definitions are given:


Also in:

Farid (2015) - An Option Greeks Primer: Building Intuition with Delta Hedging and Monte Carlo Simulation in Excel.

  • $\begingroup$ where is this definition take from? $\endgroup$
    – Richi W
    May 3, 2016 at 11:27

1 Answer 1


IMHO the 'definition' you mention is not a mathematical definition per se, but rather an approximation used by some practitioners.

Mathematically, it is $N(d_2)$ in the BS formula which figures the conditional probability that the terminal asset price $S_T$ will finish above the strike level $X$ given the information we possess today (represented by $S_t$), and that, under a risk-neutral measure which uses the risk-free money market account as numéraire. Omitting discounting, $N(d_2)$ is indeed the price of a binary or digital option assuming GBM.

The delta (omitting dividend payments) is on the other hand given by $N(d_1)$ not $N(d_2)$. Only for extremely short time to maturity or low volatility can $d_1$ be approximated by $d_2$ and hence $\Delta$ be approximated by $N(d_2)$ (again assuming no divs).

Still, we could also say that $\Delta$ corresponds to a probability of finishing in the money, but in an equivalent martingale measure which uses the asset itself as numéraire.

It therefore depends on the probability measure in which you are working/interpreting your results. In the sell-side we like to think in terms of risk-neutral probabilities (or binary options).

See the definition of $d_1$ and $d_2$ here for instance.

  • $\begingroup$ So what you say seems to be right, look at this. The prob ending in the money is $N(d_2)$ and $N(d_1)$ is more complicated. $\endgroup$
    – Richi W
    May 3, 2016 at 14:33
  • 1
    $\begingroup$ @Richard But you are right too: $N(d_1)$ is indeed the probability of finishing in the money under an EMM which uses the asset $S_t$ itself as numéraire instead of the risk-free money market account $B_t$. It is worth noting that both these conditional probabilities of finishing ITM will be different from the 'true' one attached to the physical measure $\mathbb{P}$ so it is merely a question of interpretation and terminology. $\endgroup$
    – Quantuple
    May 3, 2016 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.