I have been thinking about structuring a trade to get exposure to the forward volatility. For example, let's say SPY ATM 1 month IV is 20 vol and SPY ATM 2 month volatility is 30 vol. Then the forward vol would be SQRT(30^2 * 2 - 20^2 * 1) or approx 37.5 vol. I want vega exposure to this forward vol. All I can think of is a gamma neutral calendar spread. Any other ideas?
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2$\begingroup$ As long as you are aware of the difference between forward vol (as you defined it) and forward starting vol (which is the vol of forward start options), your idea is fine. But what is ATM today will not be ATM tomorrow of course. So your trade will need to be rebalanced. $\endgroup$– FridoCommented Apr 25, 2023 at 7:39
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$\begingroup$ Thank you @Frido! It helps knowing someone else has the same opinion. Curious to know what differences you consider to have in mind! $\endgroup$– kwantifyCommented Apr 25, 2023 at 16:01
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$\begingroup$ Whilst waiting for someone else's opinion, what do you mean exactly with 'what differences you consider to have in mind'? $\endgroup$– FridoCommented Apr 25, 2023 at 17:23
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$\begingroup$ Just thinking about scenarios where the gamma neutral calendar spread won’t give Vega exposure to the forward vol. I understand that the gamma neutral spread gives exposure to forward starting vol and closely tracks the forward volatility. I guess I meant to ask, what scenarios would the trade structure not closely track forward volatility as I’ve put it? $\endgroup$– kwantifyCommented Apr 25, 2023 at 17:54
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$\begingroup$ Ok, I think I might have to explain more clearly the difference between forward vol (as you defined it) and forward starting vol. Give me a day or two to formulate an answer, in which I'll also try to include the structure for the trade. $\endgroup$– FridoCommented Apr 25, 2023 at 18:23
1 Answer
Let $I(K_1)$ be the IV of a vanilla option with strike $K_1$ and maturity $T_1$ and similarly $I(K_2)$ corresponds to strike $K_2$ and maturity date $T_2 > T_1$.
What I'd suggest you try to trade is not $\sqrt{I^2(K_2)T_2 - I^2(K_1)T_1}$, but the difference in total implied variance $I^2(K_2)T_2 - I^2(K_1)T_1$ instead. So basically what you want to trade is the change in the difference in total implied variance: $$ \mathrm d[I^2(K_2)T_2 - I^2(K_1)T_1] $$
Since $I^2(K_2)T_2 = (I(K_2)\sqrt{T_2})^2$ and similarly for $I^2(K_1)T_1$, $$ \mathrm dI(K_i)\sqrt{T_i} \approx \frac{1}{2I(K_i)\sqrt{T_i}} \,\mathrm dI^2(K_i)T_i $$
Now for options close to the ATM strike the vanna and volga of the option is quite small (although nonzero, but I won't go into that now). So if $K_1,K_2$ both close to ATM the market change of the options can be written as \begin{align} \frac{1}{\Gamma^{BS}(K_i)S_0^2} \left[ \mathrm dC^{BS}(K_i) - \Delta^{BS}(K_i) \mathrm dS_0 \right] &\approx I(K_i) \sqrt{T_i} \, \mathrm dI(K_i) \sqrt{T_i} + \frac12 \sigma_0^2 \, \mathrm dt\\ &\approx \frac12 \mathrm dI^2(K_i)T_i + \frac12 \sigma_0^2 \, \mathrm dt \end{align} where $\Gamma^{BS}$ is Black-Scholes gamma and $\Delta^{BS}$ is the Black-Scholes delta. I am assuming you know what the BS greeks are (including vega, and the relationship between vega and gamma).
It should be pretty clear now what the notionals are of the delta-hedged calendar spread to have a 1-day p/l equal to $\mathrm dI^2(K_2)T_2 - \mathrm dI^2(K_1)T_1$.
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1$\begingroup$ Thank you @Frido! It’s crystal clear to me now $\endgroup$– kwantifyCommented Apr 26, 2023 at 13:30
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$\begingroup$ Hello @Frido, could you please clarify the approx $\frac{1}{\Gamma S^2}[dC - \Delta dS]\approx I(K)\sqrt{(T)} dI(K)\sqrt{(T)} + \frac{1}{2}\sigma^2$? I understand instead that $\frac{1}{ \Gamma S^2}[dC - \Delta dS] = \Theta + \frac{1}{2}\sigma^2$ does this means that $ \Theta \approx I(K)\sqrt{(T)} dI(K)\sqrt{(T)}$? How do you see that? $\endgroup$– Buddy_Commented May 13, 2023 at 14:44
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$\begingroup$ @Buddy_ Vega is usually defined as the change wrt to implied vol $I$, and indeed in the usual Taylor expansion you get theta. But time in the BS formula always occurs in combination with $I$, namely $I\sqrt\tau$. This is a dimensionless quantity, and you might as well define vega as the sensitivity wrt $I\sqrt\tau$ with $\tau = T-t$. If you carry this out you'll obtain the expression I wrote down. $\endgroup$– FridoCommented May 13, 2023 at 14:46