When Three Options Cannot Provide Gamma and Vega Neutrality

Summary

The document considers whether two call options can hedge a short call so that the resulting portfolio is both gamma-neutral and vega-neutral. It expresses the hedge as a two-equation linear system, with the other options’ vegas and gammas forming a two-by-two coefficient matrix. A hedge solution exists only when that matrix has a nonzero determinant; if the determinant is zero, the two hedge instruments do not provide independent exposures for solving both constraints.

The accepted response begins with a Black–Scholes relationship linking an option’s vega to its gamma, volatility, time to expiry, and the squared underlying price. This provides a way to examine when the hedge exposures may be dependent. The excerpt does not include the rest of the derivation, specific conditions for singularity, or a numerical example, so it does not establish how often the problem occurs in practice. It is a compact illustration of checking hedge-system solvability before assuming a neutralizing portfolio can be formed.

Key ideas

• Gamma and vega neutrality with two hedge options requires solving a two-equation linear system.
• A solution requires the matrix of hedge vegas and gammas to be invertible.
• A zero determinant means the hedge instruments’ exposures cannot independently satisfy both constraints.
• The response invokes a Black–Scholes relationship between vega and gamma.
• The excerpt does not show the full singularity conditions or quantify their practical frequency.

Tags

# Gamma-Vega Neutral Portfolio Not Possible with Only 3 Options


# Gamma-Vega Neutral Portfolio Not Possible with Only 3 Options












Let's say we have sold a call option, x, on a share and we have 2 other call options, y & z, with different strikes and maturities to try and achieve a portfolio that is both Gamma and Vega neutral. We just need to solve the following system of equations:

$$\begin{bmatrix} V_x \\ \Gamma_x \\ \end{bmatrix} = \begin{bmatrix} V_y & V_z \\ \Gamma_y & \Gamma_z \end{bmatrix} % \begin{bmatrix} y \\ z \end{bmatrix} $$

However, a solution only exists if the 2x2 matrix above is invertible, i.e. $V_y \Gamma_z-V_z\Gamma_y\ne0$.

Is there any reason why the solution wouldn't exist or a name given to the situation when it happens? Or does this happen just by chance, that the determinant is 0?

## Answer by ir7 (score 1, accepted)

https://quant.stackexchange.com/a/51700

In Black-Scholes world, we have:

$$V_y= \sigma_y \tau_y S^2 \Gamma_y $$

and similarly for $z$.

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.