# Prices and returns

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 $$R_{\operatorname{max}}:=\underset{1 \leq t \leq T}{\operatorname{max}}\begin{Bmatrix} R_t \end{Bmatrix}$$?

• 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. $S_T=S_0e^{x_T}$. – Kermittfrog Nov 30 '20 at 14:10
• @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. – user51121 Nov 30 '20 at 14:19
• Note that, your return at time $t$ is basically defined by $\frac{S_t}{S_0}$. Then the conversion is straightforward. – Gordon Nov 30 '20 at 14:28