Deriving Fair Variance from Option-Implied Volatility

Summary

The document examines how to connect two expressions for fair variance: an integral over Black–Scholes implied variance indexed by a transformed variable, and a derivation using log-contract replication and strike derivatives. It identifies a change-of-variables step, setting the transformed coordinate equal to d2 as a function of log-moneyness, and asks whether the resulting reversed integration limits explain a sign difference.

The discussion is a mathematical question rather than a resolved derivation. It highlights that the inverse mapping from d2 back to log-moneyness matters: the transformed coordinate is not generally identical to log-moneyness, and a valid substitution also depends on monotonicity and endpoint behavior. The document gives equations but no proof that the intermediate integrations or boundary terms are valid. It therefore serves as a prompt to check the algebra, signs, and assumptions behind the change of variables rather than as a complete recipe for computing variance.

Key ideas

• The document compares a transformed implied-variance integral with a log-contract replication expression.
• It proposes changing variables from log-moneyness to the Black–Scholes d2 coordinate.
• The proposed substitution reverses the integration limits, which affects the sign.
• The inverse mapping from d2 to log-moneyness is generally not the identity.
• The derivation is left unresolved and requires checking its algebra and assumptions.

Tags

# Recovering expression for the fair value of variance


# Recovering expression for the fair value of variance












In gatheral 2012 page 139, there is an elegant expression for the fair value of variance (Equation 11.5). This is given by \begin{align} z(k) \;=\; d_2 &=\; -\frac{k}{\sigma_{BS}(k)\,\sqrt{T}} \;+\;\frac{\sigma_{BS}(k)\,\sqrt{T}}{2},\\[1ex] \mathbb{E}[W_T] &=\;\int_{-\infty}^{\infty} N'(z)\,\sigma_{BS}^2(z)\,T \,\mathrm{d}z. \end{align}

Next, they propose the following method to brute-force it:

\begin{align*} 2\;\mathbb{E}\bigl[\log(S_T/F)\bigr] &=2\int_{0}^{\infty}dK\;\log\!\bigl(\tfrac{K}{F}\bigr)\, \frac{\partial^{2}C}{\partial K^{2}}\\ &=2\int_{-\infty}^{\infty}dk\;k\,N'(d_{2}) \Bigl\{-\frac{\partial d_{2}}{\partial k}\Bigl(1+d_{2}+\frac{\partial\sqrt{w}}{\partial k}\Bigr) +\frac{\partial^{2}\sqrt{w}}{\partial k^{2}}\Bigr\}\\ &=2\int_{-\infty}^{\infty}dk\;N'(d_{2}) \Bigl\{-\,k\,\frac{\partial d_{2}}{\partial k} \;-\;\frac{\partial\sqrt{w}}{\partial k}\Bigr\}\\ &=\int_{-\infty}^{\infty}dk\;N'(d_{2})\, \frac{\partial d_{2}}{\partial k}\;w. \end{align*}

Where we note the following relationship: $E[W_T] = E[\int^T_0\sigma_{S_t}^2 dt] = -2E[\log S_T/F]$, and $\omega(k) = \sigma^2_{BS}(k)T$.

However, I can't go back to the elegant expression from the brute-force. I get that if $z = d_2(k)$, then $dz = \frac{\partial d_2}{\partial k} dk$. so we get $$ \int_{z=d_{2}(-\infty)}^{z=d_{2}(+\infty)} N'(z)\,w\bigl(k(z)\bigr)\,dz = \int_{\infty}^{-\infty} N'(z)\,w\bigl(k(z)\bigr)\,dz\\ =-\int_{-\infty}^{\infty} N'(z)\,\bigl[\sigma_{\rm BS}(k(z))^2\,T\bigr]\,dz $$ $k(z) \neq z$ right?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.