The fear index, VIX and variance swaps

Variance Swap

Before we start the calculations it should be noted that, at maturity, the payoff of a variance swap is $$\begin{array}{c}10000N({\Sigma }^2-K_{var}),\end{array}$$ where

• $N$ = notational amount quoted in $ per volatility point squared (hence the factor of ${100}^2=10000$).
• ${\Sigma }^2$ = realized variance in the underlying asset during the life of the swap.
• $K_{var}$ = strike on variance.

The underlying assumption is that the S&P500, $S_t$, follows a geometric brownian motion

$$\begin{array}{cc}d{S}_t& ={\mu }_t{S}_{t}dt+{\sigma }_t{S}_{t}d{B}_t.\end{array}$$

This can be solved using everyones favourite lemma, Itô’s lemma: Use the substitution $Y_t=\log \left(S_t\right)$ and apply It^o’s Lemma to the function $f(t,x):=\log \left(x\right)$. The partial derivatives are

$$\begin{array}{c}\frac{\partial f}{\partial t}(t,X_t)=0,\frac{\partial f}{\partial x}(t,X_t)=\frac{1}{S_t},\frac{{\partial }^{2}f}{{\partial }^{2}x}(t,X_t)=-\frac{1}{S_t^2}.\end{array}$$ Thus, we obtain $$\begin{array}{cc}d{Y}_t& =(\frac{{\mu }_t{S}_t}{S_t}-\frac{{\sigma }_t^2{S}_t^2}{2S_t^2})dt+\frac{{\sigma }_t{S}_t}{S_t}d{B}_t\text{by Itô's Lemma}\\ & =({\mu }_t-\frac{{\sigma }_t^2}{2})dt+{\sigma }_{t}d{B}_t.\end{array}$$

Dividing the equation for $d{S}_t$ with $S_t$ and subtracting this last equation we have:

$$\begin{array}{cc}\frac{d{S}_t}{S_t}-d\left(\log \right(S_t\left)\right)& ={\mu }_{t}dt+{\sigma }_{t}d{B}_t-(({\mu }_t-\frac{{\sigma }_t^2}{2})dt+{\sigma }_{t}d{B}_t)\\ & =\frac{{\sigma }_t^2}{2}dt.\end{array}$$ Hence, $$\begin{array}{c}Variance=\frac{1}{T}{\int }_0^T{\sigma }_t^{2}dt=\frac{2}{T}({\int }_0^T\frac{d{S}_t}{S_t}-\log (\frac{S_T}{S_0}\left)\right).\end{array}$$

Notice that $$\begin{array}{c}\log \left(\frac{S_T}{S_0}\right)=\log \left(\frac{S_T}{S^{\ast }}\right)+\log \left(\frac{S^{\ast }}{S_0}\right).\end{array}$$

We now make use of the dirac delta function ($\delta \left(x\right)=\infty$ if $x=0$, $0$ otherwise) to write on $-\log \left(\frac{S_T}{S_{\ast }}\right)$ and then integrate by parts twice (this is where the $1/K^2$ comes from):

$$\begin{array}{cc}-\log \left(\frac{S_T}{S_{\ast }}\right)& =\log \left(S^{\ast }\right)-\log \left(S_T\right)\\ & =\log \left(S^{\ast }\right)-{\int }_o^{\infty }log\left(K\right)\delta (S_T-K)dK\text{by propertis of the dirac delta function}\\ & =\log \left(S^{\ast }\right)-{\int }_o^{S^{\ast }}log\left(K\right)\delta (S_T-K)dK-{\int }_{S^{\ast }}^{\infty }log\left(K\right)\delta (S_T-K)dK\text{splitting the integral}\\ & =\log \left(S^{\ast }\right)-{\left[\log \right(K\left)1_{(S_T<K)}\right]}_0^{S^{\ast }}+{\int }_0^{S^{\ast }}\frac{1}{K}{1}_{(S_T<K)}dK\\ & -{\left[\log \right(K\left)1_{(S_T\ge K)}\right]}_{S^{\ast }}^{\infty }+{\int }_{S^{\ast }}^{\infty }\frac{1}{K}{1}_{(S_T\ge K)}dK\text{by integration by parts}\\ & =\log \left(S^{\ast }\right)-\log \left(S^{\ast }\right)1_{(S_T<K)}+{\frac{1}{K}[K-S_T{]}^{+}|}_0^{S^{\ast }}+{\int }_0^{S^{\ast }}\frac{1}{K^2}[K-S_T{]}^{+}dK\\ & -\log \left(S^{\ast }\right)1_{(S_T\ge K)}+{\frac{1}{K}[S_T-K{]}^{+}|}_{S^{\ast }}^{\infty }+{\int }_{S^{\ast }}^{\infty }\frac{1}{K^2}[S_T-K{]}^{+}dK\\ & =\frac{1}{S^{\ast }}({[S^{\ast }-S_T]}^{+}-{[S_T-S^{\ast }]}^{+})+{\int }_0^{S^{\ast }}\frac{1}{K^2}[K-S_T{]}^{+}dK+{\int }_{S^{\ast }}^{\infty }\frac{1}{K^2}[S_T-K{]}^{+}dK\\ & =\frac{1}{S^{\ast }}(S^{\ast }-S_T)+{\int }_0^{S^{\ast }}\frac{1}{K^2}[K-S_T{]}^{+}dK+{\int }_{S^{\ast }}^{\infty }\frac{1}{K^2}[S_T-K{]}^{+}dK.\end{array}$$

This was kinda tedious, but this calculation is the reason for $1/K^2$ part in the final equation.

We now have the final equation for the variance:

$$\begin{array}{c}Variance=\frac{2}{T}({\int }_0^T\frac{d{S}_t}{S_t}-\log (\frac{S^{\ast }}{S_0})+\frac{1}{S^{\ast }}(S^{\ast }-S_T)+{\int }_0^{S^{\ast }}\frac{1}{K^2}[K-S_T{]}^{+}dK+{\int }_{S^{\ast }}^{\infty }\frac{1}{K^2}[S_T-K{]}^{+}dK).\end{array}$$

Say $Variance={\Sigma }^2$ (realized variance in the underlying asset during the life of the swap). To avoid arbitrage in the payoff equation (equation located in the start) we have to set: $$\begin{array}{c}E\left[{\Sigma }^2\right]=K_{var},\end{array}$$ the expectation of the realized variance is equal to the strike on the variance.

For the expecattions, not that $$\begin{array}{cc}E\left[{\int }_0^T\frac{d{S}_t}{S_t}\right]& =E[{\int }_0^T{\mu }_{t}dt+{\int }_0^T{\sigma }_{t}d{B}_t]=\mu T\text{assuming}\mu \text{is constant}\\ E\left[\log \right(\frac{S^{\ast }}{S_0}\left)\right]& =\log \left(\frac{S^{\ast }}{S_0}\right)\text{they're just constants}\\ E\left[S_T\right]& =S_0{e}^{\mu T}.\end{array}$$

Also observe that the price of a put and a call option both with strike $K$ is $$\begin{array}{cc}P\left(K\right)& =e^{-eT}E\left[{(K-S_T)}^{+}\right]\\ C\left(K\right)& =e^{-eT}E\left[{(S_T-K)}^{+}\right].\end{array}$$

Taking expectations of the equation for the realized variance we obtain

$$\begin{array}{c}{K}_{var}=E\left[Variance\right]=\frac{2}{T}(\mu T-(\frac{S_O}{S^{\ast }}{e}^{\mu T}-1)-\log \left(\frac{S^{\ast }}{S_0}\right)+e^{\mu T}{\int }_0^{S^{\ast }}\frac{P\left(K\right)}{K^2}dK+e^{\mu T}{\int }_{S^{\ast }}^{\infty }\frac{C\left(K\right)}{K^2}dK).\end{array}$$ Another way of writing this is: $$\begin{array}{cc}{K}_{var}& =\frac{2}{T}(\log \left(\frac{F_o}{S^{\ast }}\right)-(\frac{F_0}{S^{\ast }}-1)+e^{\mu T}{\int }_0^{S^{\ast }}\frac{P\left(K\right)}{K^2}dK+e^{\mu T}{\int }_{S^{\ast }}^{\infty }\frac{C\left(K\right)}{K^2}dK).\end{array}$$ Now, making the smart choice of setting $S^{\ast }=F_0=S_0{e}^{\mu T}$ a lot cancel out and we have the final equation

$$\begin{array}{c}{K}_{var}=\frac{2e^{\mu T}}{T}[{\int }_0^{F_0}\frac{P\left(K\right)}{K^2}dK+{\int }_{F_0}^{\infty }\frac{C\left(K\right)}{K^2}dK].\end{array}$$

Relationship between variance swap and the VIX

We seek a discretized version (the VIX) of our fair price variance swap formula. Assume that the price of options with strike prices $K_i,(i=1:n)$ are known and that $$\begin{array}{c}{K}_1<K_2<...<K_n.\end{array}$$ Furthermore, choose $S^{\ast }$ equal to the first strike price below $F_0$, define the function

$$\begin{array}{cc}Q\left(K_i\right)& =P\left(K_i\right)1_{(K_i\le S^{\ast })}+C\left(K_i\right)1_{(S^{\ast }<K_i)},\end{array}$$

and then approximate the integrals as following

$$\begin{array}{c}{e}^{\mu T}{\int }_0^{S^{\ast }}\frac{P\left(K\right)}{K^2}dK+e^{\mu T}{\int }_{S^{\ast }}^{\infty }\frac{C\left(K\right)}{K^2}dK=e^{\mu T}\sum _{i=1}^n\frac{\Delta K_i}{K_i^2}Q\left(K_i\right),\end{array}$$

where $$\begin{array}{cc}\Delta K_i& =\frac{K_{i+1}-K_i}{2},i=2:n-1\\ \Delta K_1& =K_2-K_1\\ \Delta K_n& =K_n-K_{n-1}.\end{array}$$

Set this aproximation inside the equation for fair variance swap with and we obtain $$\begin{array}{c}{K}_{var}\approx \frac{2}{T}(\log \left(\frac{F_o}{S^{\ast }}\right)-(\frac{F_0}{S^{\ast }}-1)+e^{\mu T}\sum _{i=1}^n\frac{\Delta K_i}{K_i^2}Q(K_i\left)\right).\end{array}$$

The maclaurin polynomial for $\log \left(F_0/S^{\ast }\right)$ is $$\begin{array}{cc}\log \left(\frac{F_0}{S^{\ast }}\right)& =(\frac{F_0}{S^{\ast }}-1)-\frac{1}{2}{(\frac{F_0}{S^{\ast }}-1)}^2+O\left({(\frac{F_0}{S^{\ast }}-1)}^2\right).\end{array}$$

Thus, $$\begin{array}{c}\log \left(\frac{F_o}{S^{\ast }}\right)-(\frac{F_0}{S^{\ast }}-1)=-\frac{1}{2}{(\frac{F_0}{S^{\ast }}-1)}^2+O\left({(\frac{F_0}{S^{\ast }}-1)}^2\right),\end{array}$$ and we obtain our final approximation

$$\begin{array}{c}{K}_{var}\approx \frac{2}{T}{e}^{\mu T}\sum _{i=1}^n\frac{\Delta K_i}{K_i^2}Q\left(K_i\right)-\frac{1}{2}{(\frac{F_0}{S^{\ast }}-1)}^2,\end{array}$$ which we recognize as the fomula for the VIX.

Remember that the underlying “stock” here is the S&P500 index! I am note sure if there are call and put options that has the S&P as underlying, if there isn’t, you should use underlying stocks that is highly correlated (beta) with the S&P and calculate the VIX for all of them and maybe take the average.