I'm trying to understand delta hedging. If I sell a plain vanilla call option, in order to delta hedge it, I have to buy delta amount of stocks.

What I don't understand is that the BS price of the call is:

$$C = SN(d_1) - e^{-rT}XN(d_2)$$

I want to construct the hedge portfolio which has the same value as the option price at any time. But the option price consists of 2 terms, not just the delta term.

What about the second term? Why don't I need it for hedging?

  • $\begingroup$ Because formula for delta is $N(d_1)$. That is why we do not need second term for hedging. Please elaborate more, what specific you want to know, otherwise I am voting for close this question. $\endgroup$
    – Neeraj
    Feb 28, 2016 at 18:37
  • $\begingroup$ The question is: why it is only $N(d_1)? I want the hedge portfolio to have the same value as the option price at any time. But the option price consits of 2 terms, not just the delta term. $\endgroup$ Feb 28, 2016 at 18:39
  • 4
    $\begingroup$ You are trying to hedge the changes in the price of the option as the stock price changes, not match the option value (which we could easily do with a lump of cash equal to the option value, but that is useless). And therefore you have to look at $\frac{\partial C}{\partial S}$ not C. And N(d2) does NOT APPEAR in $\frac{\partial C}{\partial S}$. $\endgroup$
    – Alex C
    Feb 28, 2016 at 20:03
  • 1
    $\begingroup$ @AlexC that's an answer, why do you add it in a comment? $\endgroup$
    – SRKX
    Feb 29, 2016 at 4:17
  • $\begingroup$ So i have seen the calculation explaining why $\Delta = N(d_1)$ and I know the calculation that show we need to use $\Delta$ for hedging, but I can't wrap my mind around the fact we don't use the probability of exercice (i.e. $N(d_2)$) it would make sense for me. Do you have an easy argument that show the exercise probability is not relevant ? Is it a question of real world versus risq free ? $\endgroup$ Apr 21, 2016 at 13:24

1 Answer 1


The point is the following:

Delta, $\Delta$, is defined as $\frac{\partial C}{\partial S}$, where $C$ is the value of the call option, and $S$ is the price of the underlying asset.

So, given that the value of a call option for a non-dividend-paying underlying stock in terms of the Black–Scholes parameters is

$$C = N(d_{1})S - N(d_{2})Ke^{-rT},$$

$$\Delta = \frac{\partial C}{\partial S} = N(d_{1}).$$

Basically, Delta is just the first partial derivative of $C$ with respect to $S$.

How to derive $\Delta$

  • $N(x)$ is the cumulative probability that a variable with a standardized normal distribution will be less than x;
  • $N'(x)$ is the probability density function for a standardized normal distribution:

$$N'(X) = \frac{1}{\sqrt{2\pi}}e^{\frac{x^2}{2}}.$$

Then, defining $\tau = T - t$, we have $$ d_{1} = \frac{\ln(\frac{S}{K}) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}$$


$$ d_{2} = \frac{\ln(\frac{S}{K}) + (r - \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}$$

It follows that

$$ N'(d_{1}) = N'(d_{2} + \sigma\sqrt{\tau}) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(d_{2} + \sigma\sqrt{\tau})^2}{2}} = N'(d_{2})e^{-d_{2}\sigma\sqrt{\tau} - \frac{\sigma^2\tau}{2}} = N'(d_{2})\frac{Ke^{-r\tau}}{S}$$


$$N'(d_{1})S = N'(d_{2})Ke^{-r\tau}.$$


$$ \frac{\partial d_{1}}{\partial S} = \frac{\partial d_{2}}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}}$$

Since there is an $S$ in $N(d_{1})$ and $N(d_{2})$, we use the chain-rule:

$$ \frac{\partial C}{\partial S} = N(d_{1}) + \frac{\partial d_{1}}{\partial S} N'(d_{1})S - \frac{\partial d_{2}}{\partial S} N'(d_{2})Ke^{-r\tau} = N(d_{1}) + \frac{\partial d_{1}}{\partial S} N'(d_{1})S - \frac{\partial d_{2}}{\partial S} N'(d_{1})S = N(d_{1}) + \frac{1}{S\sigma\sqrt{\tau}} N'(d_{1})S - \frac{1}{S\sigma\sqrt{\tau}} N'(d_{1})S = N(d_{1}).$$

  • $\begingroup$ I find that this can cause confusion, as $S$ appears also in $d_{1,2}$. One might think that the partial derivative is straightforward, but it's not. OP might be asking exactly that. $\endgroup$
    – Kiwiakos
    Feb 28, 2016 at 21:37
  • $\begingroup$ You are right. I added some steps of the derivation. $\endgroup$ Feb 29, 2016 at 6:55

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