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options anyone ?Santa @ 2014-08-04 10:07:30
Suppose that c1, c2, c3 are the prices of european call options with strike prices K1, K2, K3.
K3>K2>K1 and K3-K2 = K2-K1

All options have the same maturity.

SHOW THAT : c2 <= {0.5(c1+c3)}

Method to be used : porfolio that is long one option with K1, long one option with K3 and short two options with K2.
Re: options anyone ? Jpoms @ 2014-08-12 08:26:55
This trade is called a Butterfly and is a neutral strategy that is profitable in a low volatility environment.

First off, saying that K3-K2 = K2-K1, all you're saying is that the option strike prices are equally spaced.

Based on the statement that K3>K2>K1, we know that C1 will have the greatest value and C3 will have the least (this being true because with call options, the lower the strike, the deeper in the money it will be).

Now lets assume the price of the underlying is P. The intrinsic value of each option would be C(n)=Max{P-K(n),0}, meaning that no option will have a negative intrinsic value. The extrinsic, or time value, should be relatively similar for each option, but again this is a very very general statement and often untrue.

Therefore based on this information...

Since C1, C2 and C3 are equally spaced, we can then do the following math...
(TV represents time value which for our purpose will remain constant across all options)

C1=(P-K1)+TV
+ C3=(P-K3)+TV
_____________
C1+C3 = 2P-K1-K3+2TV

We must then consider dividing each side in half since we are searching for the average...

.5 x C1+C3 = .5 x (2P-K1-K3+2TV)
Since we know the strike prices are equally spaced, one-half of -K1-K3 would be -K2 resulting in...
.5(C1+C3) = P-K2+TV

Therefore...
.5(C1+C3)=C2


This is a more mathematical method than the one you requested. This is how you may prove using the butterfly method...

C1 + C3 - 2*C2 >=0 (This statement is saying that by purchasing C1 and C3 and selling C2, there is some cost involved.)

then,

C1+C3>=2*C2

Solving for C2 we must divide in half both sides...

.5(C1+C3)>=C2


Much more simple that way but not as fun! :)