0
$\begingroup$

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.

$\endgroup$
0
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks! Can you provide me a reference/explanation for "Implied volatility is a function of only moneyness under pure SV model"? $\endgroup$
    – Arshdeep
    Sep 6 at 12:07
  • $\begingroup$ Take a look at Makr Joshi's paper "Log type models, homogeneity of option prices and convexity". The point is that for these models it can be shown that the implied vol is a function of moneyness. If you can't derive it as a consequence of homogeneity feel free to post a new question. $\endgroup$ Sep 7 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.