Forward-Start Option Delta and Volatility Exposure

# Show that a forward starting option has 0 delta, and no sensitivity to volatility until the strike is determined


# Show that a forward starting option has 0 delta, and no sensitivity to volatility until the strike is determined












I need to show that the payoff: $([(S_{T2}-S_{T1})/S_{T1}]-k)^+$

a. Has 0 delta

b. Has no sensitivity to quadratic variation of the underlying till $T_1$

Additionally, I would like to know for what payoff of the kind $f(S_{T1},S_{T2})$ do these results hold? Particularly, I would intuitively assume that a payoff of the kind $(S_{T2}-KS_{T1})^+$ also respects these rule (b).

My thoughts:

For a lognormal model, (a) and (b) are immediate by just substituting the closed form solution to the SDE. Beyond that, I'm not able to say anything.

## Answer by user34971 (score 3, accepted)

https://quant.stackexchange.com/a/67750

I'll rewrite the first payoff in the more common form $(S_T/S_{t^*} - k)_+$, where $t^*$ is the forward start date, $T$ the expiry date, and today is $t$. So $t < t^* < T$.

I'll assume a pure stochastic volatility model (quite important to specify the model).

Then the forward start option price today is \begin{align} E_t \left[ \left( \frac{S_T}{S_{t^*}} - k \right)_+\right] &= E_t \left[ E_{t^*} \left[ \left. \left( \frac{S_T}{S_{t^*}} - k \right)_+ \right| \mathcal{F}_{t^*}\right]\right] \\ &=E_t \left[ \frac{1}{S_{t^*}} E_{t^*} \left[ \left. \left(S_T - k S_{t^*} \right)_+ \right| \mathcal{F}_{t^*}\right]\right] \\ &= E_t \left[ \frac{1}{S_{t^*}} BS(S_{t^*},t^*, kS_{t^*},T, I^* ) \right] \end{align} Since we are working with a pure SV model, first of all the implied volatility $I^*$ at time $t^*$ is a funtion of moneyness $S_{t^*}/(kS_{t^*}) = 1/k$. Furthermore, the Black-Scholes call price function is homogeneous of degree 1 in spot and strike, meaning that $$ BS(S_{t^*},t^*, kS_{t^*},T, I^*(k) ) = S_{t^*} BS(1,t^*, k,T, I^*(k) ) $$ Thus, the price of the forward start option today does not depend on the future value of the spot at $t^*$: \begin{align} E_t \left[ \left( \frac{S_T}{S_{t^*}} - k \right)_+\right] &= E_t \left[ BS(1,t^*, k,T, I^*(k) ) \right] \\ &= BS(1,t,k,I^{FS}(k)) \end{align} where $I^{FS}(k)$ is the (definition of) forward start implied volatility. It can be shown that the forward start implied volatility depends not only on $k$ but also on the future volatility over the interval $[t^*,T]$. But since $E_t \left[ BS(1,t^*, k,T, I^*(k) ) \right]$ does not depend on spot it is clear that the forward start IV does not depend on spot price.

Now for the second forward start option with payoff $\left(S_T - kS_{t^*} \right)_+$: repeating the conditioning argument above, and again using the homogeneity of the BS model you will obtain \begin{align} E_t \left[ \left( S_T- kS_{t^*}\right)_+\right] &= E_t \left[ S_{t^*} BS(1,t^*, k,T, I^*(k) ) \right] \end{align} Now there is a dependence on the future spot within the expectation, and in general, because the implied volatility $I^*(k)$ depends on volatility which is correlated with the spot price, the expectation cannot be evaluated easily. However, by change of numeraire (i.e. under the share measure), you can write \begin{align} E_t \left[ \left( S_T- kS_{t^*}\right)_+\right] &= S_t E^{\mathbb S}_t \left[ BS(1,t^*, k,T, I^*(k) ) \right] \end{align} In any case, the second payoff you specificied has delta and also exposure to (future) volatility.

The derivations / arguments above are valid for pure SV models. In (S)LV models, since the implied volatility is no longer a simple function of moneyness (and volatility), even for the first payoff you wrote there will be delta.

As for the statement/question do these options have exposure to quadratic variation up to strike date? I would say yes unless the quadratic variation over $[t^*,T]$ is independent of the quadratic variation over $[t,t^*]$ which I don't think is the case in general.