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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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