I want to convert the payoff of an Asian and a lookback Call option with prices in their corresponding with returns. Example: for an European Call $\varphi(S_T)=(S_T-K)^+$, so knowing that $S_T=S_0(1+r_t)^T=S_0\prod_{i=1}^{T}(1+r_i)=S_0R_T$ I can write that $(S_T-K)^+=(S_0R_T-K)^+$.

If an Asian Call payoff is $\varphi(S_T)=(\frac{1}{T}\sum_{t=1}^{T}S_t-K)^+$ and a lookback Call payoff is $\varphi(S_T)=(S_{\operatorname{max}}-K)^+$, how do I obtain respectively

  • $(\frac{1}{T}\sum_{t=1}^{T}S_t-K)^+=(\frac{S_0}{T}\sum_{t=1}^{T}R_t-K)^+$,

  • $(S_{\operatorname{max}}-K)^+=(S_0R_{\operatorname{max}}-K)^+$ for $R_{\operatorname{max}}:=\underset{1 \leq t \leq T}{\operatorname{max}}\begin{Bmatrix} R_t \end{Bmatrix}$?

Thanks in advance.

  • $\begingroup$ Hi, what do you mean with "how do I obtain respectively": What do you want to obtain? A pricing formula or payoff formula? That would be helpful. In any case, it is most often quite helpful to transform to log-returns, i.e. $S_T=S_0e^{x_T}$. $\endgroup$ Nov 30, 2020 at 14:10
  • $\begingroup$ @Kermittfrog Thanks for your answer! I have to verify, mathematically, the equalities to in points 1 and 2, but I couldn't do it until now. PS: I know that is more helpful using log-returns $\operatorname{log}(\frac{S_{t+1}}{S_t})=\operatorname{log}(1+r_t)$ but the author of book, to prove the equality in the example for European options, uses the compound interest regime and not that continuos. $\endgroup$
    – user51121
    Nov 30, 2020 at 14:19
  • 1
    $\begingroup$ Note that, your return at time $t$ is basically defined by $\frac{S_t}{S_0}$. Then the conversion is straightforward. $\endgroup$
    – Gordon
    Nov 30, 2020 at 14:28
  • $\begingroup$ @Gordon Omg... I was getting lost in a glass of water. Thanks! $\endgroup$
    – user51121
    Nov 30, 2020 at 14:42


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.