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#1
07-06-2017, 03:00 PM
 mistersunnyd Member SOA Join Date: Aug 2016 Studying for how to find a job Posts: 175
Pricing options with average payoffs

Test is coming up, and I just want to understand some things intuitively before I go in. One thing that has always confounded me is why an option is priced with average payoffs instead of average stock prices.

Let's say I use MC valuation to find 4 stock prices in one year. Why can't I average the stock prices and then price the option? Why must I find the individual payoffs and then average them for the stock price?
#2
07-06-2017, 03:25 PM
 The Disreputable Dog Member CAS Join Date: Dec 2011 Studying for prelims College: Somersby School Favorite beer: Worldwide Stout Posts: 996

This is just an off-the-cuff answer, but suppose you have a call. The call pays if the stock value is above some strike (say K=50). The average stock price may be 49, in which case the payoff is zero. However, within that average of 49, some simulated values of the stock could/would have been higher than 50. Those would have resulted in payoffs, which would then get blended into the value of the call.

E.g., stock prices of 55, 49, and 43 would average to 49. But payoffs of 5, 0, and 0 would average to 1.67.

The former shows no value to the call. The latter shows an average payoff of 1.67.
#3
07-06-2017, 03:36 PM
 mistersunnyd Member SOA Join Date: Aug 2016 Studying for how to find a job Posts: 175

But let's say you you picked four random numbers [0,1] and got their z-scores and then computed their stock price, wouldn't each simulated stock price have the same probability of being the real stock price? Thus, even if the average of the simulated stock prices yields a payoff of 0, why is that incorrect? After all, these four numbers are randomly chosen, and they just so happened to average out at the strike price.
#4
07-06-2017, 03:47 PM
 InTheDark SOA Join Date: May 2017 College: University of Waterloo Posts: 26

Try understand this:
E((X+1)^2) is not eq to (E(X)+1)^2, so for nonlinear function f,
E(f(X)) is not eq to f(E(X)).
Imagine f as the converter from stock price to option payoff.
#5
07-06-2017, 04:07 PM
 kodak11111 Member SOA Join Date: Nov 2016 Location: Winnipeg, Canada Studying for ERM College: U of Manitoba, 2019 Posts: 88

Quote:
 Originally Posted by InTheDark Try understand this: E((X+1)^2) is not eq to (E(X)+1)^2, so for nonlinear function f, E(f(X)) is not eq to f(E(X)). Imagine f as the converter from stock price to option payoff.
In other words, the payoff for a call is only when the stock price is above the strike price. If it paid off everywhere then you could average the stock prices. But since it doesn't, we can't to average the stock prices.

Also it makes sense to only care about payoffs because that what we actually get. The stock price only determines the payoff.
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#6
07-06-2017, 04:09 PM
 mistersunnyd Member SOA Join Date: Aug 2016 Studying for how to find a job Posts: 175

Quote:
 Originally Posted by kodak11111 In other words, the payoff for a call is only when the stock price is above the strike price. If it paid off everywhere then you could average the stock prices.
Ahh ok that makes sense. Thank you

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