April 2008: Q9

Jensen

There are a couple of ways that I can think of for solving this.

Method 1

The question boils down to a lower bound for a derivative price. Ch 9 suggests that we use no arbitrage methods for this sort of problem. So the trick is to think, what if the inequality is the other way around and then show that the assumption of no arbitrage (which is given in the exam question) is contradicted. So what if, instead:

x.C(K1)+(1-x).C(K2) < C(x.K1+(1-x).K2)
where K1<K2 and x is between 0 and 1?

In order to make a profit from this arrangement you need to buy the cheap side and sell the expensive side now (at time t say). So buy x call options with strike K1, buy (1-x) call options with strike K2 and sell 1 call option with strike K3=x.K1+(1-x).K2 at time t. Because of the way the inequality is set up, this should give us a profit at time t.

But what happens when the options get to expiry (at time T say)? We can show that the payoff on our investment strategy is always greater than or equal to 0. What is the payoff on expiry? It is:

x.max(ST-K1,0) + (1-x).max(ST-K2,0) - max(ST-(x.K1+(1-x).K2), 0)

You then need to try all the different possible combinations of ST relative to
K1, K2 and K3. For example, if ST>K1, ST>K2 and therefore ST>K3, the payoff is x(ST-K1) + (1-x)(ST-K2) - {ST-(x.K1+(1-x).K2)} = 0.

Keep going along these lines, with different combinations.

Because the payoff at time T is always >=0, yet we made a profit at time t, we have shown there is an arbitrage opportunity. Therefore, the assumption of no arbitrage is contradicted and it cannot be true that:

x.C(K1)+(1-x).C(K2) < C(x.K1+(1-x).K2)

Finally, it must therefore be true that:

x.C(K1)+(1-x).C(K2) >= C(x.K1+(1-x).K2)

Method 2

Write each of the option prices using the risk-neutral pricing formula:

C(K1) = exp(-r(T-t))E[max(ST-K1,0)|Ft]
C(K2) = exp(-r(T-t))E[max(ST-K2,0)|Ft]
C(K3) = exp(-r(T-t))E[max(ST-(x.K1+(1-x).K2),0)|Ft]

Then x.C(K1)+(1-x).C(K2)
= x.exp(-r(T-t))E[max(ST-K1,0)|Ft] + (1-x)exp(-r(T-t))E[max(ST-K2,0)|Ft]
=exp(-r(T-t))E[max(x.ST-x.K1,0) + max((1-x).ST-(1-x)K2, 0)|Ft]

You can then show that max(A,0)+max(B,0) >= max (A+B, 0) by considering possible combinations of A and B.

The result then follows.

Phew ...

Did you do anything like this in the exam?

 

No arbitrage

Jensen,

More generally if an equality or equality should hold by the principle of no arbitrage then we can prove so by contradiction.

Let's say that A >= B. Prove it.

Your Starting point is to assume B > A. This means B is expensive and A is cheap. Buy A, sell B, profit at time 0. Then get out your deckchair, sit back and relax. You will find in all scenarios that your portfolio has a value >=0. Do the maths and you will find this, (that's where A >= B comes from!)

THis is arbitrage, contradiction, THE END.

That's the rough wordy version of Anna's elegant solution. It gives you a starting point for all questions of this sort, I imagine that where to start was the biggest problem on this qn!

John