Why Black–Scholes Call Delta Is N(d₁), Not Both Price Terms

# Why is $N(d_2)$ not needed for hedging?


# Why is $N(d_2)$ not needed for hedging?












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?

## Answer by stochazesthai (score 8, accepted)

https://quant.stackexchange.com/a/24607

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 $$ d_{1} = \frac{\ln(\frac{S}{K}) + (r + \frac{\sigma^2}{2})\tau}{\sigma\sqrt{\tau}}$$

and

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

It follows that

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

Thus,

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

Then

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

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

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