# Is negative forward variance an arbitrage?

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.

• 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/… Jun 9, 2023 at 19:14
• @Frido I added the missing details below. Jun 9, 2023 at 21:15

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 $$t, $$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$$