# Proving that the $\Delta$ of a call on a futures contract under the B-S model is $N(d_1)$

The author of my textbook says that the $\Delta$ of a call on a futures contract is $N(d_1)$ and not $e^{-rT}N(d_1)$. I wasn't convinced, so I tried to prove this.

Let $F = F_{0, T}(S) = S_0e^{(r - \delta)T}$ and consider a call on a futures contract on a stock $S$ \begin{align}C(F, K, \sigma, r, T, \delta) &= Fe^{-rT}N(d_1) - Ke^{-rT}N(d_2)\\ &= S_0e^{(r - \delta)T}e^{-rT}N(d_1) - Ke^{-rT}N(d_2)\end{align} and

$\frac{\partial C}{\partial S} = e^{-\delta T}N(d_1) + S_0e^{-\delta T}N'(d_1) - Ke^{-rT}N'(d_2).$

Let's relate $\frac{d_1^2}{2}$ and $\frac{d_2^2}{2}$. We know that $d_2 = d_1 - \sigma\sqrt{T}$. So \begin{align*}d_2^2 &= d_1^2 + \sigma^2T - 2d_1\sigma\sqrt{T}\\ &= d_1^2 + \sigma^2T - 2[\ln(F) - ln(K) + (r - r + 0.5\sigma^2)T]\\ &= d_1^2 - 2\ln(F) + 2\ln(K).\end{align*} Then $$\frac{d_2^2}{2} = \frac{d_1^2}{2} - \ln(F) + \ln(K).$$

Now \begin{align*}N'(d_1) &= \frac{e^{-d_1^2/2}}{\sqrt{2\pi}}\\ &= \frac{e^{-d_2^2/2 - \ln(F) + \ln(K)}}{\sqrt{2\pi}}\\ &= \frac{Ke^{-d_2^2/2}}{F\sqrt{2\pi}}\end{align*} and

$$N'(d_2) = \frac{e^{-d_2/2}}{\sqrt{2\pi}}$$.

Hence \begin{align*}\frac{\partial C}{\partial S} &= e^{-\delta T}N(d_1) + \frac{S_0e^{-\delta T}e^{-d_2^2/2}\cdot K}{S_0 e^{(r - \delta)T}\sqrt{2\pi}} - Ke^{-rT}\frac{e^{-d_2^2/2}}{\sqrt{2\pi}}\\ &= e^{-\delta T}N(d_1) + \frac{Ke^{-rT}}{\sqrt{2\pi}} - \frac{Ke^{-rT}}{\sqrt{2\pi}}\\ &= e^{-\delta T} N(d_1).\end{align*}

What am I missing here?

• It's a lot easier to do these B-S derivations by expressing the option price as an integral (via the expectation), and then differentiating under the integral sign. Nov 21, 2016 at 19:13