## Chapter 3

This forum is to discuss the book "the concepts and practice of mathematical finance" by Mark Joshi.

### Chapter 3

In exercise 3.4, let the hedging portfolio be

Pi = delta S - Opt.

Now look at the stock values 115 and 105 at time t=T, and set the value of the respective hedging portfolios equal:

delta 115 - 15 = delta 105 - 5 => delta = 1 => portfolio value Pi(T) = 100.

By no arbitrage, Pi(0) = 100 too. This gives

Pi(0) = 1 x 100 - Opt = 100 => Opt = 0.

i.e. the minimum option value is 0. The solutions give 5/2. Why?
INFIDEL

Posts: 62
Joined: Sun Aug 26, 2007 5:57 am

the portfolio consisting of 0.5 stocks and -47.5 bonds sub-replicates.
mj

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Joined: Fri Jul 27, 2007 7:21 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.
INFIDEL

Posts: 62
Joined: Sun Aug 26, 2007 5:57 am

the risk neutral argument is actually a bit more subtle than that in that there are other risk-neutral measure in which p1+p2 does not equal 0.5

there are three ways to do this problem
1) get an upper bound on the set of arbitrage-free prices using super and sub replication, use RN evaluation to get a lower bound

2) classify all super and sub replicating portfolios

3) classify all risk-neutral measures

approach 1 which is essentially what you've done is the easiest in this problem, but observe that you really need both the RN and replication arguments to make it work
mj

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Joined: Fri Jul 27, 2007 7:21 am

Deleted. Double post.
Last edited by INFIDEL on Thu Oct 25, 2007 6:07 pm, edited 1 time in total.
INFIDEL

Posts: 62
Joined: Sun Aug 26, 2007 5:57 am

mj wrote:the risk neutral argument is actually a bit more subtle than that in that there are other risk-neutral measure in which p1+p2 does not equal 0.5

Aha. To prevent arbitrage, we have the much broader condition than I had, of strict inequalities 0 < p1+p2 < 1. [Else if the sum equals one, set up a portfolio consisting of 1 share of stock (value 100) and -100 riskless bonds. The stock is guaranteed to rise above 100, with no chance of falling below 100, while the bonds stay constant (since r=0). i.e. we have a guaranteed profit and an arbitrage. Similarly if p1+p2 = 0.]

So there's definitely more work to be done in my post above.

mj wrote:but observe that you really need both the RN and replication arguments to make it work

Do you mean "replication" as in replicating the value of the option in all possible states using stocks and bonds? Or setting up a hedging portfolio whose value is the same in all world states?

It's just that this is the terminology you seemed to settle on in the book.
INFIDEL

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Joined: Sun Aug 26, 2007 5:57 am

What i mean was that sub and super replication arguments generally tell you that some prices are arbitrageable.

Risk-neutral arguments tell you that some prices are not arbitrageable.

To be sure that you have the set of non arbitrageable prices, you either need to use both types of arguments, or be very careful that you really have considered all possible portfolios or measures.
mj

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Joined: Fri Jul 27, 2007 7:21 am

How exactly do you apply the RN argument to obtain the lower bound? Starting with probabilities a,b,c and d for stock values of 115,105,95 and 85, I can state the following :

1) 115a + 105b + 95c +85d = 100
2) a +b + c + d = 1
3) 15a + 5b < 7.5 (obtained using the hedging arguments)

Whats the next step?

In the 3 state example solved in the text, the probability of the stock having the highest value is equated with that of the stock having its lowest value (say p) and the third intermediate state then has the probability 1-2p. Is there any particular reason why this is done, that is equating the probabilites of the two extreme states?
The tools of a mathematician are pencil, paper and a waste paper basket. The waste paper basket distinguishes him from a philosopher.
f0X_in_s0X

Posts: 86
Joined: Tue Dec 04, 2007 5:47 pm

The three state problems are simpler in any case. I just solved problem 3.7, wherein the RN probabilities are obtained quite simply by solving the resulting equations (which are just eqns of lines). In the 4 state case, they are equations of planes... :(

1) 115a + 105b + 95c +85 (1-a-b-c) = 100 or 30a + 20b + 10c = 15
2) a + b + c < 1
3) a > 0, b > 0 and c >0

Solving these eqns would help us characterise all the RN probabilities, and then we wouldn't need the hedging arguments. But using (2) in (1) would only give us a relationship in two variables (substituting c = 1-a-b). That wont be very helpful I guess.

Any other ideas? Or am I missing something?
The tools of a mathematician are pencil, paper and a waste paper basket. The waste paper basket distinguishes him from a philosopher.
f0X_in_s0X

Posts: 86
Joined: Tue Dec 04, 2007 5:47 pm

well if you want to characterize the set of all risk-neutral probabilities and their implied prices you would have to parametrize the 2 dimensional solution set of probabilities and then work out the the prices they imply.

However, this is too fiddly to be appealing.

So alternatively, work out a subset and use this to get a set of non-arbitrageable prices, N. (as INFIDEL did)

Use sub and super replication to get a set of arbitrageable prices, A.

If N union A is everything you are done.
mj

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Joined: Fri Jul 27, 2007 7:21 am

Thanks, I get it. The best way I think then is to use Infidel's hedging arguments. RN is messy here.

I was having trouble with this statement of Infidel's solution

Then choose the min of these. It turns out that putting 50 - Opt ≤ 47.5 in (C) gives the optimal Opt ≥ 2.5 .

What Infidel means I guess is to take the max (and not min). Or in other words the intersection of all the sets we get when we bound the stock-option portfolio.
The tools of a mathematician are pencil, paper and a waste paper basket. The waste paper basket distinguishes him from a philosopher.
f0X_in_s0X

Posts: 86
Joined: Tue Dec 04, 2007 5:47 pm

f0X_in_s0X wrote:What Infidel means I guess is to take the max (and not min). Or in other words the intersection of all the sets we get when we bound the stock-option portfolio.

Yes!
INFIDEL

Posts: 62
Joined: Sun Aug 26, 2007 5:57 am

I apologize in advance if I am missing something obvious. In the second part of problem 3.4, why is Opt(K=110) nonzero?

The payoff of the call option struck at K=100 can be

15
5
0
0

depending on whether the price of the stock tomorrow is 115, 105, 95, or 85, while the payoff of the call struck at K=110 is

5
0
0
0

Portfolios i) of two Opt(K=110) and 5\$ in riskless bonds and ii) Opt(K=100) have exactly the same payoffs, and therefore must be worth the same:

2 Opt(110) + 5\$ = Opt(K=100).

Since Opt(K=100)=5\$, one has Opt(110)=0.

I would very much appreciate if someone would point out a mistake in the argument above.
dual

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what are your two portfolios worth when K =95?
mj

Posts: 1380
Joined: Fri Jul 27, 2007 7:21 am

Thanks! It was indeed a silly error on my part.
dual

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Joined: Sat Mar 01, 2008 11:46 pm

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