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I believe that having a negative forward variance on a ATMF implied volatility curve of a volatility surface could imply the existence of a static arbitrage (for example, a calendar arbitrage). Although I tried to look at this from multiple perspectives, I could not come up with an answer. Therefore, I thought it might be appropriate to ask the community, in case someone has more insight into this than I do.

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    $\begingroup$ Yes, if the forward moneyness are equal. Hint: Look at the answer in the following post and adjust the strikes in the answer in such a way that your desired result flows from it. quant.stackexchange.com/questions/75799/… $\endgroup$
    – Frido
    Jun 9, 2023 at 19:14
  • $\begingroup$ @Frido I added the missing details below. $\endgroup$
    – fwd_T
    Jun 9, 2023 at 21:15

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Let $$ V_t^{T_1,T_2}=\frac{(T_2-t)V_t^{T_2}-(T_1-t)V_t^{T_1}}{T_2-T_1} $$ be our forward variance where $t<T_1<T_2$, $V_t^{T_1}$ is the ATMF implied vol as seen at time $t$ for slice at maturity $T_1$ and $V_t^{T_2}$ is the AMTF implied vol seen at time $t$ for tenor $T_2$. Let $P_1$ be the PV at time $t$ of an ATMF long call position of maturity $T_1$, and $P_2$ be the PV at time $t$ of an ATMF long call position of maturity $T_2$. We know that $$ \frac{\partial P_1}{\partial V_t^{T_1}}>0; \frac{\partial P_1}{\partial V_t^{T_1}}>0 $$ Let $\widetilde{V}_t^{T_1}=(T_1-t)V_t^{T_1}$ and $\widetilde{V}_t^{T_2}=(T_2-t)V_t^{T_2}$. Then, $$ \frac{\partial P_1}{\partial \widetilde{V}_t^{T_1}}=\frac{\partial P_1}{\partial V_t^{T_1}}\frac{\partial V_t^{T_1}}{\partial \widetilde{V}_t^{T_1}}=\frac{1}{T_1-t}\frac{\partial P_1}{\partial V_t^{T_1}}>0 $$ and similarly for $T_2$ superscript. But since the forward variance is negative, we have $\widetilde{V}_t^{T_1}>\widetilde{V}_t^{T_2}$ and since the PV of the option is an increasing function of this variable, this means that the PV of the call option of maturity $T_2$ is smaller than the PV of the call option of maturity $T_1$. So $P_1>P_2$. This can be shown to be a calendar arbitrage as explained here: Check for arbitrage - European calls with same strike price, different duration and price . More specifically, under no-arbitrage we would have: $$ P_2=e^{-(T_2-t)r_d}\mathbb{E}_{\mathbb{Q}}\left[\left(S_{T_2}-F_{t,T_2}\right)^{+}|\mathcal{F}_t\right]=e^{-(T_2-t)r_d}\mathbb{E}_{\mathbb{Q}}\left[\mathbb{E}_{\mathbb{Q}}\left[\left(S_{T_2}-F_{t,T_2}\right)^{+}|\mathcal{F}_{T_1}\right]|\mathcal{F}_t\right]\geq\\ \geq e^{-(T_2-t)r_d}\mathbb{E}_{\mathbb{Q}}\left[\left(\mathbb{E}_{\mathbb{Q}}\left[S_{T_2}|\mathcal{F}_{T_1}\right]-F_{t,T_2}\right)^{+}|\mathcal{F}_t\right]=\\ =e^{-(T_2-t)r_d}\mathbb{E}_{\mathbb{Q}}\left[\left(\underbrace{S_{T_1}e^{(r_d-r_f)(T_2-T_1)}}_{F_{T_1,T_2}}-F_{t,T_2}\right)^{+}|\mathcal{F}_t\right]=\\ =e^{-(T_1-t)r_d}e^{(T_1-T_2)r_d}\mathbb{E}_{\mathbb{Q}}\left[\left(\underbrace{S_{T_1}e^{(r_d-r_f)(T_2-T_1)}}_{F_{T_1,T_2}}-F_{t,T_2}\right)^{+}|\mathcal{F}_t\right]=e^{-(T_2-T_1)r_f}P_1 $$

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  • $\begingroup$ Yes seems correct to me, and for this particular derivation it's good you looked at vega as the sensitivity wrt to de-annualized variance and not to annualized variance. $\endgroup$
    – Frido
    Jun 10, 2023 at 6:56

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