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Assume you have two stocks $S$ and $P$ so that at initial time $t = 0$: $S_0 > P_0$.

You bought an option which pays off $S_T - P_T$ as long as $S_t > P_t$ through the time $0 < t < T$.

What would the price of such option be?

*I am looking for a non-arbitrage argument avoiding any specific distribution assumptions (log-normal, normal etc) if possible.

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I solved it the following way, just want make sure I'm not missing something obvious.

Set up a portfolio $PF$ consisting of long $S$ and short $P$ at time $t = 0$. Choose arbitrary time $0 < t < T$. If $S_t > P_t$ then $PF_t = S_t - P_t$ which coincides with the value of the option. If $S_t$ hits $P_t$ from above, then dissolve the portfolio by selling $S$ and buying $P$. Again both the portfolio $PF$ and the option have the same value 0 in this case.

So we have a self-financing portfolio which has the same payoff at time $T$ as the option. So the option value at $t=0$ must be the same as the portfolio value in the absence of arbitrage, i.e. option value is $S_0 - P_0$.

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    $\begingroup$ Interesting approach I wonder if you get the same result as @Gordon though. However, the is the portfolio self-financing there, I mean you do have to pay for $S_0 - P_0$ it does not change the result though? $\endgroup$ – SRKX Nov 23 '15 at 1:14
  • $\begingroup$ I agree with this argument. But at the start, you need to short the option, short the stock $P$, and buy the stock $S$. Then you have a zero cost for your portfolio, assuming that the option price is $S_0-P_0$. $\endgroup$ – Gordon Nov 23 '15 at 15:13
  • $\begingroup$ You may need to merge your two accounts. $\endgroup$ – Gordon Nov 23 '15 at 15:20
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The option payoff at maturity $T$ is defined by \begin{align*} (S_T-P_T)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1}. \end{align*} Let $Q$ be the risk-neutral probability measure and $E$ be the corresponding expectation operator. Let $Q_p$ be a probability measure defined by \begin{align*} \frac{dQ_p}{dQ}\big|_t = \frac{P_t}{e^{rt} P_0}. \end{align*} Moreover, let $E_p$ be the corresponding expectation operator. Then the option value can be computed by \begin{align*} e^{-rT}E\left((S_T-P_T)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1} \right) &= e^{-rT}E_p\left(\left(\frac{dQ_p}{dQ}\big|_T\right)^{-1}(S_T-P_T)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1} \right)\\ &=P_0 E_p\left(\left(\frac{S_T}{P_T}-1\right)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1} \right), \end{align*} which can be treated as a down-and-out barrier call option, assuming that $S_t/P_t$ is log-normally distributed under the measure $Q_p$.

Note that, under $Q_p$, the process $\{S_t/P_t \mid t \geq 0\}$ is a martingale, that is, we can treat $S_t/P_t$ as an asset process with zero interest and zero dividend. Using the down-and-out barrier call option formula in John Hull, we obtain that \begin{align*} E_p\left(\left(\frac{S_T}{P_T}-1\right)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1} \right) = \frac{S_0}{P_0}-1. \end{align*} That is, the option price is $S_0-P_0$.

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  • $\begingroup$ I know you did most of the job here, but I'd really love to see the example go on until the close-form assuming a lognormal. But I guess that's just long a tedious algebra to get the value of the expectation given the joint distribution of $S$ and $P$ right? Or maybe you define $Z_t=\frac{S_t}{P_t}$ which is also a GBM and which should help already. $\endgroup$ – SRKX Nov 22 '15 at 11:45
  • $\begingroup$ @SRKX: You are right. Since $Z_t$ is a GBM, I tried to leverage on existing result. $\endgroup$ – Gordon Nov 22 '15 at 14:15
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The option payoff is equivalent to $Z_{\tau \wedge T}-1$ where $\tau=\inf\{t | Z_t = 1\}$ provided that $Z_t$ is assumed to be continuous. Since $Z_t=S_t/P_t$ is a martingale under $Q_P$, we have $E_P[Z_{\tau \wedge T}]=Z_0$ and the option value is $P_0 (Z_0 - 1)=S_0-P_0$ regardless of the model.

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  • $\begingroup$ Excellent! I have not thought about this. $\endgroup$ – Gordon Nov 23 '15 at 17:57

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