Sept 06 Q8 (v)

Im having some trouble understanding part (v) of this question. The hint in the question suggests the make of the replication of the management fee is a basic fee plus 2 call options and 2 special options. The management fee is KS_1 = 0.1% when S_1<S_0
= 1% when S_1 > U
= 0.5% otherwise.

The solution suggests that the replication is possible by 0.1% of Stock price & 0.004 call options (S_0 = 100) & 0.005 call optons (strike price U) & 0.004S_0 special options and 0.005U special options.

My questions is and i hope that some one out the knows the answer. How have these numbers been derived???

Thanks

 

Does this help?

http://www.acted.co.uk/forums/showthread.php?t=1592

http://www.acted.co.uk/forums/showthread.php?t=2136

 

Those links are useful thanks but it doesnt indicate where the numbers come from, how are these selected?. How are the students expected to know what values make up to the 0.5. I must be missing something

fee = 0.001*S + 0.004*call (100) + 0.005 call (U) + 0.4 special [100,infinity) + 0.05U special [U,infinity)

Does anyone Know??

Thanks

 

I'm afraid this requires quite a bit of thought and/or inspiration.
I think of it as a puzzle where you twist and move pieces until they fit (similar to what you need to do to identify arbitrage opportunities).
But in case the inspiration doesn't hit you, here are some thought processes which may help.

First some notation
Let C(K) be a one year standard european call option on S with strike price of K,
ie C(K) pays S1-K if S1>K and nothing otherwise
Let I(a,b) be the special option in the question that pays 1 if S is between a and b, and nothing otherwise.

we need something that pays F,
F=0.001 S1 if S1<S0
F=0.005 S1 if S0<=S1<=U
F=0.01 S1 if U<S1

Now looking at the hint, since we don't know any better and it's probably the way we supposed to do it.

we supposedly can use a basic fee, 2 call options and 2 special options.
Now S0, and U are the key (known) points where the fees change, so lets take them to be the strike prices in our call options. (If we choose some other strike price it will cause the fee to change at other points which don't help)

Let's see what happens if we take just the normal fee plus two call options.
v,w,x are arbitrary amounts. Let the special options be A and B

F=v S1 + w C(S0) + x C(U) + A + B
Call options pay out when S1 is greater than a certain amount so start with the smallest value of S1 and work up (adding additional elements of the fee as we go along).

If S1<S0 we want 0.001 S1
Both C(S0) and C(U) are 0 and ignore A and B, so take v=0.001
It is ok to ignore A and B here once we make sure that whatever we make them to be is zero when S1<S0.

F= 0.001 S1 + wC(S0) +x C(U) + A + B...(2)
= 0.001 S1


If S0<=S1<=U, we need F=0.005 S1
C(S0) = S1-S0;
C(U) = 0;
Substitute in (2) we get
0.005 S1 = 0.001 S1 +w (S1-S0) +A + B ...(3)
Ignore B once we make sure that B is zero when S1<=U

0.004 S1 = w S1 - w S0 + A
Looks like w might be S1, S1 is a variable which is unknown.
S0 is known however, so to get rid of the w S0 we need to add w S0.
Adding more S1 wouldn't always exactly cancel.
We thus need a fixed payment of w S0 if S0<=S1 and 0 if S1<S0
but this is the special option, ie w S0 I(S0,infinity)

so w=0.004 and A=0.004 S0 I(S0,infinity)

F= 0.001 S1 + 0.004 C(S0) + x C(U) + 0.004 S0 I(S0,U) + B ..(4)

Now last case with S1>U
C(S0) = S1-S0;
C(U) = S1 - U;
I(S0,infinity) =1;

Need F = 0.01 S1
Sub in (4)
0.01 S1 = 0.001 S1 + 0.004 (S1-S0) +x (S1-U) + 0.004 S0 + B
0.005 S1 = x S1 - x U +B
As before scratch off the x U by making B pay x U
B = x U I(U,infinity)
and take x = 0.005

Finally F = 0.001 S1 + 0.004 C(S0) + 0.005 C(U) + 0.004 S0 I(S0,Infinity) + 0.005 U I(U,Infinity)

S0 is 100 so you can then get their expression.

And having done the first line, you can then start the "real" question, lol.