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Under a constant interest rate, the futures price is given by a deterministic time function times the asset price (I think I understand this). This means that the volatility of the futures price should be the same as that of the underlying asset price.

Not really sure how this is true. Is there any more intuitive explanation as to why this would hold?

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At least from your statement. Provided the coefficient to the asset price is purely a deterministic function of time I.e trend, then the random fluctuations (or volatility) come only from the underlying? Ps i have to admit that I am only a self taught student so I could be completely wrong anyway so please correct if necessary :). – Chinny84 Aug 19 '14 at 12:09
up vote 2 down vote accepted

To avoid confusion (the term futures price could be a type for "future price" or "future's price"), it seems to me that you are talking about the forward price of an asset for which the cost of carry is equal to the interest rate. In that case, indeed, with a fixed interest rate r and an spot S, the forward price F for a time T is given by $F=Se^{r(T-t)}$. If $S$ is governed by the SDE: $\frac{dS}{S} = rdt+\sigma dW$ and by noting that $\frac{\partial F}{\partial t} =-re^{r(T-t)}S$, $\frac{\partial F}{\partial S} = e^{r(T-t)}$ and $\frac{\partial^2 F}{\partial S^2} = 0$, we can apply Ito's lemma. Doing so results after some simple algebra, in the following process for $F$:

$$dF = e^{r(T-t)}S\sigma dW$$

which is equivalent to:

$$\frac{dF}{F} = \sigma dW$$

This is probably what you had in mind?

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How do you get between the last two steps? Do you use a CGM to remove the drift? I am still learning :). – Chinny84 Aug 19 '14 at 14:21
The the "dF=.." to the last equation I'm just substituting the definition of F. – Bram Aug 19 '14 at 15:26
I was talking about the $rdt$ term? I am not saying your wrong, I am just inquiring to see where it goes. – Chinny84 Aug 19 '14 at 15:40
Note that $dF = e^{r(T-t)}dS + S de^{r(T-t)}= e^{r(T-t)}\Big[S(rdt+\sigma dW_t)\Big] -S r e^{r(T-t)}dt = e^{r(T-t)}S \sigma dW_t = F \sigma dW_t$. – Gordon Aug 19 '14 at 19:05
I believe that he made a typo here. That is, it should be $dF = \frac{\partial F}{\partial t} dt + \frac{\partial F}{\partial S} dS$, instead of $dF = \frac{\partial F}{\partial S} dS$. – Gordon Aug 19 '14 at 20:13

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