newbie question about options

[Paluka]

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it is never optimal to exercise the American call option early.



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This is absolutely false.

[jason1990]

I thought it was clear from the context of the thread what I was talking about. Apparently it wasn't. I am talking about an American call option without a continuous payout. The link you have provided is for an American call option with a continuous payout.

If you would like to see a proof of the fact I mentioned, you may see Proposition 14.3 on page 245 of "Stochastic Calculus and Financial Applications" by J. Michael Steele.

[jason1990]

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it is never optimal to exercise the American call option early.



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This is absolutely false.

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Earlier in this section of the thread, Rotating Rabbit pointed out that

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An american call is worth the same as a european call provided there are no dividends to be paid.

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This is the context in which I was making my comment and in this context, it is absolutely true.

(By the way, I'm simply making a mathematical claim here for the benefit of those that find it interesting. I am not making any claims about the "real world" of finance, if that's what you're referring to.)

For a rigorous formulation of my claim and a proof of it, you can see the reference I gave in my previous post.

[crazy canuck]

I'm not talking about real world option prices either.

Have you read the link I posted?

First sentence:

We now discuss an approximation to the option price of an American option on a commodity having a continuous payout


So according to your claim the American option price should be the same as European option price.

I don't care whatever book it is in, your claim is wrong.

[jason1990]

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Have you read the link I posted?

First sentence:

We now discuss an approximation to the option price of an American option on a commodity having a continuous payout

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Did you read the post I posted?

Second sentence:

I am talking about an American call option WITHOUT a continuous payout.

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So according to your claim the American option price should be the same as European option price.

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For the kind of American call option I'm talking about, yes this is true. The prices are the same.

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I don't care whatever book it is in, your claim is wrong.

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I gave you the book reference so you could read the proof for yourself. As a professional mathematician whose research area is stochastic analysis and mathematical finance, I can assure you that what I'm claiming is the absolute truth. If you choose not to believe it or choose not to try to better understand exactly what it is I'm claiming, then I suppose that's your business.

[crazy canuck]

You said in your first post:


h(x)=max{x-K,0}.

This function is convex and has h(0)=0. These two properties are all one needs in order to conclude that it is never optimal to exercise the option before expiry. Notice that the corresponding function for the put is still convex, but it is not 0 at 0.



No, as you said yourself these properties are not enough.

Yes, there are options for which there is no early exercise premium (like probably the example in your book), but in GENERAL the claim is not true.

In fact even if an option has non continuous payout, early exercise can be optimal. For example find a simple example of binomial option pricing.

Maybe you forgot to mention continuous stock price and some other conditions too?

[jason1990]

Whatever. This is just silly now. My original post was a reply to Rotating Rabbit, who was explicitly talking about an American call option without any payoff rate (continuous or otherwise) other than the terminal payoff, in a Black-Scholes model using geometric Brownian motion.

You obviously either didn't read the thread carefully enough, or you don't have enough interest in it to discern the context for yourself. Frankly, this is not worth my time.

If you have any genuine interest in learning more about the theorem I mentioned about options of this kind that are defined as convex functions of the stock price, then you'll have to PM me, because I'm not going to bother with this thread anymore.

[Rotating Rabbit]

Hi Jason. Indeed he's not read my answer properly, I clearly stated that there were exceptions (e.g. dividend payments continuous or otherwise etc).

When you say the value of the call option (at s(t) = x):

h(x) = max {x - K, 0}

I presume you mean the payoff (ie not including cost) as opposed to the value.

If you dont mind, I'm very interested to know the gist of why max{s(t)-K, 0} being convex and passing the origin implies its an increasing function of t (and why max{K-S(t), 0} i.e. the put with coordinate (s(0),K) , is different)?

[jason1990]

Hi Rabbit. Thanks for the PM. So here's the idea. Our stock is simple geometric Brownian motion S(t). Our bond is simple exponential growth b(t). Assume h is convex with h(0)=0. Then for all p between 0 and 1, we have

(*) h(px) <= ph(x) + (1 - p)h(0) = ph(x).

Now, let M(t)=h(S(t))/b(t), let F_t be the Brownian filtration, and let Q be the equivalent martingale measure under which the discounted stock price D(t)=S(t)/b(t) is a martingale. Then

E_Q[M(t+s) | F_t] = E_Q[h(S(t+s))/b(t+s) | F_t]
= b(t)^{-1} E_Q[h(S(t+s))b(t)/b(t+s) | F_t]
>= b(t)^{-1} E_Q[h(S(t+s)b(t)/b(t+s)) | F_t]

by (*). Since h is convex, Jensen's inequality gives

E_Q[h(S(t+s)b(t)/b(t+s)) | F_t]
>= h(E_Q[S(t+s)b(t)/b(t+s) | F_t])
= h(b(t)E_Q[S(t+s)/b(t+s) | F_t])
= h(b(t)E_Q[D(t+s) | F_t])
= h(b(t)D(t))
= h(S(t)),

where we have used the fact that D is a Q-martingale. Putting these two facts together gives

E_Q[M(t+s) | F_t] >= b(t)^{-1}h(S(t)) = M(t),

so M is a Q-submartingale. Hence, by optional sampling, if T is expiry and tau is any stopping time with tau<=T, then

(**) E_Q[h(S(T))/b(T)] >= E_Q[h(S(tau))/b(tau)].

By the martingale pricing formula, the price of this option is

sup E_Q[h(S(tau))/b(tau)],

where the sup is taken over all stopping times tau<=T; and the optimal exercise time is the stopping time which achieves this supremum. Therefore, by (**), we see that the optimal exercise time is simply T, the time of expiry.

In terms of arbitrage, let's see if I remember this correctly. As the seller of the option, suppose I use the same hedging portfolio I would use if it were a European option. Then my initial investment is

V(0) = E_Q[h(S(T))b(0)/b(T)]

and my wealth at time t will be

V(t) = E_Q[h(S(T))b(t)/b(T) | F_t]
= b(t)E_Q[M(T) | F_t].

If the buyer exercises at time tau, then my wealth will be

V(tau) = b(tau)E_Q[M(T) | F_tau]
>= b(tau)M(tau) = h(S(tau)).

So I will have enough to cover him, plus a possible surplus.

Edit: There is a b(0) missing in at least one place above. So let's just assume b(0)=1.

[Rotating Rabbit]

Well thankyou for this Jason. I'm not fluent enough in martingales to have been able to derive this myself. I'd previously been trying to think of a way of adapting the security/bond hedging european-call pricing argument to find a price. I note your edit I would normally assume b(0) is 1 anyway unless its someone being particularly annoying [img]/images/graemlins/smile.gif[/img]

Its interesting to me that weak-convexity is so powerful even when the functions dont even have continuous first derivatives.

There's probably a paper in here somewhere for someone looking at concave utility functions and american calls conbined, if it hasnt already been examined in detail.