I was asked this question in an interview.

There is an option as follows. It monitors the prices of two stocks A and B, and pays the difference in their prices at time $T$, if stock A has been higher than stock B all through till $T$. There is nothing paid if stock A has fallen lesser than stock B at any time before $T$.

How do we price this option?

I gave an answer by modeling the difference as a Brownian motion, and computing the probability that the zero-hitting time for the BM to be greater than $T$. However the interviewer said there was a simpler method based on option pricing.

Can anyone help me?

  • $\begingroup$ What was the interview for (ie what level)? The answer they're looking for will depend on this. $\endgroup$ – will Oct 3 '17 at 7:19
  • $\begingroup$ @will: quant associate position... top investment bank. $\endgroup$ – helloman Oct 3 '17 at 7:42

Well "based on option pricing" is a little vague, but the desired solution is probably to use one of the stocks (say stock $B$) as your numeraire. If you're unfamiliar the intuitive idea is that imagine instead of money, people used stock $B$ as currency / measure of wealth, etc. Then the "value" of stock $A$ (i.e. the number of shares of stock $B$ it is worth) is $S_A/S_B$ and you have a fixed strike / Barrier of $1.$ So you price a barrier option with initial price $S_{A,0}/S_{B,0}$, barrier / strike $1$ and volatility $$ \sqrt{\sigma_A^2 + \sigma_B^2-2\rho\sigma_A\sigma_B}$$ (i.e. the volatility of $S_A/S_B$) and that gives the price of the option in units of $S_B,$ so just multiply by $S_B$ to get the price in dollars.

  • $\begingroup$ I think in the setting of an interview, this is more to test of you "just know the pile of formulae from haug", or if you know them and understand their shortcomings. This formula is not helpful, since you're never going to correctly price this kind of option, since you know know either of the vols or the covariance... $\endgroup$ – will Oct 3 '17 at 7:18
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    $\begingroup$ @Quantuple The seond part of your comment was what i was trying to get at - and i think this is probably the reason it's asked in an interview, as there are multiple levels you can go into. Like asking people how to price a quanto option - you can just say change the drift, but in reality it is not that simple, and showing you're aware of that i think makes you look much better than just quoting some text book response that's only really used for quick and dirty hacks / sanity checks. $\endgroup$ – will Oct 3 '17 at 8:54

I suspect you could replicate this trade by buying $A$ and selling short $B$ (additionally, when the price of $A$ touches $B$ at any time before $T$, you liquidate the position for $0$ payoff). If so, today's price of this option is just $S_A-S_B$.

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    $\begingroup$ If the derivative had an American exercise feature, this would be right (in the zero dividend case, anyway), but since it doesn't it is worth less than the underlying spread. $\endgroup$ – spaceisdarkgreen Oct 3 '17 at 22:18
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    $\begingroup$ @spaceisdarkgreen The knockout IS American style isn't it ? So the solution is simple then. $\endgroup$ – dm63 Oct 4 '17 at 9:51
  • $\begingroup$ @dm63 That's not what I get from reading the question (it just says it pays the value of the spread at time $T$.) That said, it's entirely possible that it was American in the interviewer's actual question... would explain the 'simple solution'. $\endgroup$ – spaceisdarkgreen Oct 4 '17 at 12:18
  • $\begingroup$ The simple solution applies if the spread is paid at time T, but the knockout is American, I believe. $\endgroup$ – dm63 Oct 4 '17 at 15:04
  • $\begingroup$ @dm6 (you gotta @ me) I don't know what you mean by the knockout being American. The solution above would apply to a derivative that can be cashed in for the value of the spread at any time before expiration (i.e. American exercise) and becomes worthless if it ever hits zero. If you only get to cash out at $T$ you lose the option value of cashing out before that simply buying the spread would provide. It's easy to see such a derivative's value goes to down as vol or $T$ get large since prob of spread hitting zero before $T$ increases $\endgroup$ – spaceisdarkgreen Oct 4 '17 at 15:22

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