# 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?

• 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. Feb 28 '16 at 18:37
• The question is: why it is only $N(d_1)? I want the hedge portfolio to have the same value as the option price at any time. But the option price consits of 2 terms, not just the delta term. Feb 28 '16 at 18:39 • You are trying to hedge the changes in the price of the option as the stock price changes, not match the option value (which we could easily do with a lump of cash equal to the option value, but that is useless). And therefore you have to look at$\frac{\partial C}{\partial S}$not C. And N(d2) does NOT APPEAR in$\frac{\partial C}{\partial S}$. Feb 28 '16 at 20:03 • @AlexC that's an answer, why do you add it in a comment? – SRKX Feb 29 '16 at 4:17 • So i have seen the calculation explaining why$\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)$) it would make sense for me. Do you have an easy argument that show the exercise probability is not relevant ? Is it a question of real world versus risq free ? Apr 21 '16 at 13:24 ## 1 Answer The point is the following: Delta,$\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: $$\frac{\partial C}{\partial S} = N(d_{1}) + \frac{\partial d_{1}}{\partial S} N'(d_{1})S - \frac{\partial d_{2}}{\partial S} N'(d_{2})Ke^{-r\tau} = N(d_{1}) + \frac{\partial d_{1}}{\partial S} N'(d_{1})S - \frac{\partial d_{2}}{\partial S} N'(d_{1})S = N(d_{1}) + \frac{1}{S\sigma\sqrt{\tau}} N'(d_{1})S - \frac{1}{S\sigma\sqrt{\tau}} N'(d_{1})S = N(d_{1}).$$ • 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. Feb 28 '16 at 21:37
• You are right. I added some steps of the derivation. Feb 29 '16 at 6:55