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This is in response to a number of private emails I've received, where the
question comes up of how to calculate option equivalents correctly. If
deltas don't work, which method to use?
Perhaps some of you may find it worthwhile to learn of a little-known method
using vectors with two coordinates. It's a bit tedious to explain, but once
you get the hang of it, you'll find the method quite easy to handle.
Here are the conventions we use. As you will see, it will be sufficient to
refer to the graphs of options at expiration. The *sign* of a vector
coordinate means "up" for positive, "horizontal" for zero, and "down" for
negative, always in reference to expiration graphs. The *value* of a
coordinate means the number of options.
(1,0) means long 1 call. Why? The expiration graph of a long call goes up
(1, a positive number) to the right of the strike, and is horizontal (0) to
the left of the strike. The first coordinate depicts what goes on to the
right of the strike price, the second coordinate, what happens to the left
of the strike price.
(2,0) means two long calls. The signs are the same as with 1 long call, the
values are doubled. So you see how this works?
How about 1 long put? (0,1), of course! Horizontal (0) to the right of the
strike, positive to the left. And ten long puts? Well, you just multiply the
vector by 10, so we'd have (0,10).
Similarly, a long straddle would be (1,1). Up on both sides of the strike
price!
For short options, you simply multiply those vectors by -1. Thus, (-2,0)
would be two short calls, and (0,-10) means 10 short puts. You can easily
see that the expiration graphs are again depicted correctly by the signs of
the coordinates.
We also need to describe the underlying (stocks or futures) in our little
calculus. Here the convention will be that 1 means either 100 shares of
stock, or 1 futures contract, respectively. The coordinates will now
describe what goes on to the right or left of the *current* price, since of
course there is no strike price for the underlying. The meaning of the signs
will remain the same.
Thus, (1, -1) means long 100 shares of stock (or long 1 futures contract).
Again, the signs are correct - the position goes up to the right, and down
to the left of the current price.
What would mean "short 3000 shares" in vector language? Answer: (-30,30).
And now we can finally do what we wanted to do in the first place - build
new, equivalent positions for a given option position. How? By simply adding
the vectors.
For example, suppose we have 10 long calls and 10 short puts. What is the
equivalent? We simply write down the vectors and add them:
(10, 0) + (0, -10) = (10, -10)
So you see the result is "long 10 units in the underlying": either long 1000
shares of stock, or long 10 futures contracts, as the case may be. Which
proves that the original option position was 10 synthetic longs.
It is now easy to find correct equivalents by solving simple vector
equations. Take, for example, the problem that started this thread in the
first place - how to replace a 10x10 straddle with a correct number of calls
(x) and a position the size of y in the underlying. So we add x calls and y
units of the underlying, and want the result to equal 10 long straddles,
like this:
(x, 0) + (y, -y) = (10, 10)
When you solve this equation, you get x = 20, y = -10, which means long 20
calls, short 1000 shares or 10 futures contracts, the result that you have
also seen come up in my previous posts.
Regards,
Michael Suesserott
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