by **INFIDEL** » Thu Oct 25, 2007 7:22 am

OK. I've got it. Since in getting my thoughts together I've typed in both the hedging argument and the risk-neutral argument, I might as well post them. So here they are. Note that the hedging argument seems long-winded, while the r.n. argument is nice and short.

A stock is worth 100 today. Interest rates are 0. Tomorrow S= 85, 95, 105, 115.

HEDGING ARGUMENT

Buy δ stocks today and sell one option struck at 100. The hedging portfolio today is worth δ 100 - Opt, and tomorrow will be worth any one of

δ 115 - 15 --- (1)

δ 105 - 5 --- (2)

δ 95 --- (3)

δ 85 --- (4)

Now look for possible values of δ that eliminate risk between pairs of states. There are (4 choose 2) = 6 pairs.

A) Putting (1)=(2) gives δ=1, with possible portfolio values (100, 100, 95, 85) tomorrow, so today it must be worth at least 85, so

100 - Opt ≥ 85 => Opt ≤ 15.

B) Putting (1)=(3) gives δ=3/4, with possible portfolio values (71.25, 73.75, 71.25, 63.75) tomorrow, so today it must be worth at least 63.75, so

75 - Opt ≥ 63.75 => Opt ≤ 11.25.

C) Putting (1)=(4) or (2)=(3) gives δ=1/2, with possible portfolio values (42.5, 47.5, 47.5, 42.5) tomorrow, so today it must be worth at least 42.5, so

50 - Opt ≥ 42.5 => Opt ≤ 7.5.

D) Putting (2)=(4) gives δ=1/4, with possible portfolio values (13.75, 21.25, 23.75, 21.25) tomorrow, so today it must be worth at least 13.75, so

25 - Opt ≥ 13.75 => Opt ≤ 11.25.

E) Putting (3)=(4) gives δ=0, which is not a hedge.

This gives an upper bound on the option price of 7.5.

To get the lower bound, choose the max in each of the 4-tuples above, reverse (i.e. replace "at least" with "at most") the no-arbitrage argument that links tomorrow with today, and find the lower bound for Opt in each case. Then choose the min of these. It turns out that putting 50 - Opt ≤ 47.5 in (C) gives the optimal Opt ≥ 2.5 .

RISK NEUTRAL ARGUMENT

Let the probabilities of S= 115, 105, 95, 85 be p1,...,p4 respectively. Then

Opt = 15 p1 + 5 p2 ------ (*)

It's clear from these prices that the risk neutral probability of S ending above 100 equals the probability of ending below 100. Therefore

p1 + p2 = 1/2

=> p2 = 1/2 - p1, with 0 ≤ p1 ≤ 1/2.

Substituting into (*) gives

5/2 ≤ Opt ≤ 15/2.

Short and sweet.