In More Than You Ever Wanted to Know About Volatility Swaps the fair value of a future variance swap can be replicated from market prices for calls and puts. The fair put and call strike is shown to be

$K_{var} = \frac{2}{T}\left( rT - \left(\frac{S_0}{S_*}e^{rT} - 1 \right) - \log \frac{S_*}{S_0} + e^{rT}\int_0^{S_*} \frac{1}{K^2} P(K)dK + e^{rT}\int^\infty_{S_*} \frac{1}{K^2} C(K)dK \right)$

where $P(K)$ is current market put prices and $C(K)$ is market call prices.

Is there an intuition to the scaling factor $\frac{1}{T\cdot K^2}$? It also shows up in the discretized version of variance estimates, or the VIX calculation.

Derman, et al. also states it provides a "direct connection between the market cost of options and the strategy for capturing future realized volatility, even there is an implied volatility skew and the simple Black-Scholes formula is invalid". So I think there is a risk-neutral takeaway that doesn't regard the model.


I believe you want to know why the VIX is a weighted portfolio of calls and puts with weights proportional to $\frac{1}{K^2}$ (NB: obviously the T is there to adjust for time to maturity, hence is not very enlightening).

Let's start with the basics. As shown by Breeden and Litzenberger (1978) the second derivative of an option price with respect to the strike is proportional to the risk-neutral probability, i.e. $\frac{\partial^2C}{\partial K^2}=\frac{\partial^2P}{\partial K^2} \propto q(K)$ where $q(K)$ is the risk-neutral pdf at $K$. The intuition is that an infinitely tight butterfly spread around $K$ gives us the risk-neutral likelihood of $K$.

Now that we have $q(K)$ we can construct in principle any desired payoff $f(S_T)$, given that $$price(f(S_T))=e^{-rT}E^Q[f(S_T)]=e^{-rT}\int_0^\infty f(K)q(K)dK=\int_0^\infty f(K) \frac{\partial^2C}{\partial K^2}dK$$ If you integrate by parts the previous integral and assume that the function is well behaved at the boundaries as shown here you obtain that: $$price(f(S_T))=e^{-rT}f(F)+\int_0^F\frac{\partial^2 f}{\partial K^2}\left(K\right)P(K)dK+\int_F^\infty\frac{\partial^2 f}{\partial K^2}\left(K\right)C(K)dK$$ where $F$ is the current Forward price, $C$ and $P$ are call and put prices.

As shown by Neuberger, if prices follow a geometric Brownian motion, i.e. $dS=S\mu dt+S\sigma dW$, then $$log E^Q[S_T]-E^Q[log S_T]\propto\sigma^2$$ (NB: just write down the cumulant-generating function of a Normal distribution). This shows that under lognormal stock prices if we set $f(S_T)$ we can recover the implied volatility (NB: the implied risk-neutral variance, but the physical and risk-neutral variances are equal because of Girsanov's theorem in a continuous time diffusion.) Hence, you see immediately that to recover the price of the log contract you have to compute $\frac{\partial^2}{\partial K^2}log(K)=-\frac{1}{K^2}$. This is the reason why in the VIX$^2$ computation the weights are inversely proportional to $K^2$.

By the way, as a side note, if you are interested in risk-neutral variances outside of the log-normal paradigm it is enough to notice that $Var^Q(S_T)=E^Q[S_T^2]-E^Q[S_T]^2$, hence if you set $f(S_T)=S_T^2$ then you will notice that $\frac{\partial^2}{\partial K^2}K^2=1$ and therefore: $$Var^Q(S_T)=...\int_0^F P(K)dK+\int_F^\infty C(K)dK$$

  • 1
    $\begingroup$ Can you clarify the portion "Hence, you see immediately that to recover the price of the log contract you have to compute..."? Is it because hedging the log contract is how I gain exposure to variance (or volatility), and with the log contract the payoff $f(S_T) = \log (S_T)$? $\endgroup$
    – Jared
    Jul 9 '17 at 20:23
  • $\begingroup$ Yes, Neuberger showed that to gain exposure to variance you hedge a log contract. Which seemed a clever but useless result because "log contracts" do not exist in any real market. Then someone figured out you can essentially simulate the log-contract (among other payoffs of the form $f(S_T)$ for some $f$) with options using the other formula with the two integrals. $\endgroup$
    – Alex C
    Jul 10 '17 at 0:06
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    $\begingroup$ @Jared, start from the second equation and rewrite it as $$e^{-rT}\left(f\left( F\right)-E^Q[ f(S_T)]\right)=e^{-rT}\left(log E^Q[S_T]-E^Q[log S_T]\right)=-\left(\int_0^Ff''(K)put(K)dK + \int_F^\infty f''(K)call(K)dK \right)=\int_0^F K^{-2}put(K)dK+\int_F^\infty K^{-2}call(K)dK\propto \sigma^2$$ $\endgroup$
    – fni
    Jul 10 '17 at 9:19

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