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We can calculate the expected stock return (under the measure $Q$) from at-the-money ($K=S_t$) option prices as:

$$E\left(\frac{S_T-S_t}{S_t}\right)=\frac{e^{rT}}{S_t}(C_t-P_t)$$

The result is mainly based on the fact that $$(S_T-S_t)^+-(S_t-S_T)^+=S_T-S_t$$ and $C_t=e^{-rT}E((S_T-K)^+)$.

I am looking for an expression of stock return using options with $K\neq S_t$.

My first approach was to equate the two sides and determine the difference:

$$(S_T-K)^+-(K-S_T)^+\stackrel{!}{=} S_T-S_t+\left[(S_T-S_t)^+-(S_t-S_T)^+-((S_T-K)^+-(K-S_T)^+)\right]$$

Maybe it would be possible to rearrange this term to get a sum of option payoffs plus a deterministic part (i.e. a bond).

Please let me know if you find a solution.

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1 Answer 1

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Note that \begin{align*} \frac{S_T-S_t}{S_t} &= \frac{S_T-K +K-S_t}{S_t}\\ &=\frac{(S_T-K)^+-(K-S_T)^+ +K-S_t}{S_t}. \end{align*} Then, \begin{align*} E\left(\frac{S_T-S_t}{S_t} \mid \mathcal{F}_t \right) &= \frac{e^{rT}}{S_t}(C_t-P_t)+ \frac{K-S_t}{S_t}. \end{align*} where \begin{align*} C_t &= e^{-rT} E\left((S_T-K)^+ \mid \mathcal{F}_t \right),\\ P_t &= e^{-rT} E\left((K-S_T)^+ \mid \mathcal{F}_t \right). \end{align*}

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  • $\begingroup$ Thank you, can you explain how you come from the first to second step, i.e. why exactly the term $+K-S_t$? $\endgroup$
    – emcor
    Jun 17, 2016 at 14:47
  • $\begingroup$ @emcor: Do you mean this step: $S_T-S_t =S_T-K+K-S_t$? $\endgroup$
    – Gordon
    Jun 17, 2016 at 15:00
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    $\begingroup$ +1 for the answer. @emcor, this is merely an application of call-put parity. Since $E_t((S_T-S_t)/S_t) = (F(t,T) - S_t)/S_t$ and by call-put parity the fair forward price is $F(t,T) = K + (C(t,T)-P(t,T))e^{rT}, \forall K$. Just plug it in and you're done. $\endgroup$
    – Quantuple
    Jun 17, 2016 at 19:24

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