4
$\begingroup$

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.

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
3
$\begingroup$

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 $S_0 - P_0$.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ 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
    Commented Nov 23, 2015 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
    Commented Nov 23, 2015 at 15:13
  • $\begingroup$ You may need to merge your two accounts. $\endgroup$
    – Gordon
    Commented Nov 23, 2015 at 15:20
1
$\begingroup$

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 $S_0-P_0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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
    Commented Nov 22, 2015 at 11:45
  • $\begingroup$ @SRKX: You are right. Since $Z_t$ is a GBM, I tried to leverage on existing result. $\endgroup$
    – Gordon
    Commented Nov 22, 2015 at 14:15
1
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Excellent! I have not thought about this. $\endgroup$
    – Gordon
    Commented Nov 23, 2015 at 17:57

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.