Discounted Delta for a European Call on a Futures Contract
Summary
The document examines whether the delta of a European call on a futures contract is the normal cumulative probability at the first Black–Scholes term or that probability multiplied by a discount factor. The questioner differentiates a discounted futures option price after relating the futures price to a dividend-paying stock, then attempts to cancel the density terms. The reply confirms that, when delta is expressed with respect to the futures contract, the call delta includes the discount factor.
A zero-strike call is offered as a sanity check: its delta with respect to the futures price is the discount factor. The answer suggests the textbook statement may instead refer to delta with respect to a forward contract. The discussion is brief and does not fully reconcile the questioner’s stock-based differentiation with the futures-price convention, so the underlying variable and contract type must be specified when interpreting delta.
Key ideas
- For a European call on a futures contract, delta with respect to the futures price includes discounting.
- A zero-strike call provides a simple check on the discount factor in the delta.
- A delta stated without a discount factor may use a forward contract convention.
- The underlying variable for the derivative must be identified when comparing formulas.
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Full text
# Proving that the $\Delta$ of a call on a futures contract under the B-S model is $N(d_1)$
# 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?
## Answer by gavbrennan (score 2)
https://quant.stackexchange.com/a/31120
You're correct - the delta in terms of futures contracts is discount factor * N(d1) for a European call-on-future. Consider a zero-strike call as an example, the delta would be 1.0 multiplied by the discount factor. Maybe the author meant to say the delta in terms of forward contracts is N(d1)?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.