I don't get why a call option's price can be independent on probability in the binomial model.Let's consider a simple case.A stock is currently worth $1. In one year it can go one of two ways: Up to $1,000,000,000 with probability p, or down to $0.5 with probability 1-p.I just don't get it. Why the hell wouldn't you want to pay more for a call option of strike $1 if p=0.99 than if it were p=0.01?Please explain to me in intuitive manner why I wouldn't want to pay more for the call option for higher p.

QuoteOriginally posted by: sburaI don't get why a call option's price can be independent on probability in the binomial model.Let's consider a simple case.A stock is currently worth $1. In one year it can go one of two ways: Up to $1,000,000,000 with probability p, or down to $0.5 with probability 1-p.I just don't get it. Why the hell wouldn't you want to pay more for a call option of strike $1 if p=0.99 than if it were p=0.01?Please explain to me in intuitive manner why I wouldn't want to pay more for the call option for higher p.Off the top of my head, the answer is that p is unique. It is not usually a dependent variable. If the arbitrage theorem applies (replicability) then p is determined by the risk-free rate. If the arbitrage theorem does not apply (no replicability) then the expected price and the p are uniquely related. Saying that the call value depends on p is the same as saying it depends on the expected price of the stock.

This doesn't explain a thing to me I'm afraid. Plus that doesn't intuitively help either. Probability is just probability. No need to write some fancy words about it in my opinion. Thanks for the reply though, and I hope for more replies!For your information, I'm reading Shreve Vol. 1, and he just says: We have a binomial model. Probability of up and p. That's it. And then comes some fancy theory about replication. But my original question is never addressed.Just imagine yourself in the situation I posted. If you know p is higher, why wouldn't you want to pay more for the call option? Just imagine you know nothing of this replication theory and are just a random guy looking at such an option in real life.

QuoteOriginally posted by: sburaThis doesn't explain a thing to me I'm afraid. Plus that doesn't intuitively help either. Probability is just probability. No need to write some fancy words about it in my opinion. Thanks for the reply though, and I hope for more replies!For your information, I'm reading Shreve Vol. 1, and he just says: We have a binomial model. Probability of up and p. That's it. And then comes some fancy theory about replication. But my original question is never addressed.Just imagine yourself in the situation I posted. If you know p is higher, why wouldn't you want to pay more for the call option? Just imagine you know nothing of this replication theory and are just a random guy looking at such an option in real life.Of course I would be willing to pay more. But p is not something you know..... What you have most information about is the risk-free rate -- and that determines p. So saying the value depends on p is just saying that it depends on the risk-free rate.As regards fancy theory about replication, there is no need to bother with most of it. Just check out the arbitrage theorem. That's the place to apply a bit of intuition.

QuoteOriginally posted by: sburaThis doesn't explain a thing to me I'm afraid. Plus that doesn't intuitively help either. Probability is just probability. No need to write some fancy words about it in my opinion. Thanks for the reply though, and I hope for more replies!For your information, I'm reading Shreve Vol. 1, and he just says: We have a binomial model. Probability of up and p. That's it. And then comes some fancy theory about replication. But my original question is never addressed.Just imagine yourself in the situation I posted. If you know p is higher, why wouldn't you want to pay more for the call option? Just imagine you know nothing of this replication theory and are just a random guy looking at such an option in real life.It is not independent. $1.000.000.000*p+$0,5*(1-p)=$1. Assuming risk-free=0This means that p will be very small.

Up to this point, and he probably never will, Shreve has not (and neither do Hull, Wilmott, Bjork in their books) mentioned any dependence between p and r. Of course p and r have been introduced but no relation between them has been. r is assumed to be known in this theory. And in no way has the author even hinted that p is some kind of a different beast that is not known.

QuoteOriginally posted by: HOOKIt is not independent. $1.000.000.000*p+$0,5*(1-p)=$1. Assuming risk-free=0This means that p will be very small.Ah hmmmm. I should have thought of that. That partially explains it I think. Thanks for this. This is some nice food for thought. But still, if the "expectation rule" applies for stock, why on earth not for call options? Keep the explanations coming. I will continue thinking about this over here.

QuoteOriginally posted by: sburaQuoteOriginally posted by: HOOKIt is not independent. $1.000.000.000*p+$0,5*(1-p)=$1. Assuming risk-free=0This means that p will be very small.Ah hmmmm. I should have thought of that. That partially explains it I think. Thanks for this. This is some nice food for thought. But still, if the "expectation rule" applies for stock, why on earth not for call options? Keep the explanations coming. I will continue thinking about this over here.Once p is determined by the formula above,The call price will be given by : Max($1.000.000.000-X;0)*p+Max($0,5-X;0)*(1-p)X=strike

You would want to pay more for a call option with a higher up probability (risk-neutral).Suppose I have p probability of price going to value vu and (1-p) going to vd. Then, for no arbitrage:v = p vu + (1-p)vdis the current forward of the asset (I'm in forward space to not have to think bout the risk-free rate for simplicity).If the call is struck at vu > k > vd (just for clarity):c = p (vu-k)If p is higher, then c is higher. Of course, if p is higher, then the STOCK PRICE must also be higher, unless you are changing vu and vd. The expectation rule applies for all financial assets: that's really a DEFINITION of what is meant by the risk-neutral measure. So long as there is no arbitrage and markets are complete, you're fine (Harrison, Kreps, Pliska, etc.)

QuoteOriginally posted by: HOOKQuoteOriginally posted by: sburaQuoteOriginally posted by: HOOKIt is not independent. $1.000.000.000*p+$0,5*(1-p)=$1. Assuming risk-free=0This means that p will be very small.Ah hmmmm. I should have thought of that. That partially explains it I think. Thanks for this. This is some nice food for thought. But still, if the "expectation rule" applies for stock, why on earth not for call options? Keep the explanations coming. I will continue thinking about this over here.Once p is determined by the formula above,The call price will be given by : Max($1.000.000.000-X;0)*p+Max($0,5-X;0)*(1-p)X=strikeI'm afraid you're not correct there, because it is well-known (as shown by fancy replication theory) that the call price has nothing to do with the (real) probability p. How to get intuitive (as opposed to finding some other valid and theoretically sound method) hold of the fact why the expectation rule doesn't hold for call option prices would be a good thing to have.

I was talking risk-neutral probabilities. With real-world probabilities, the expectation rule NEVER applies - not for stocks, options, or any other asset with any risk. Unless, of course, the marginal market participants really are risk-neutral. The stock and the option both depend on risk-neutral probabilities. The stock price may encode all the information you need about the probabilities so that the option price appears to be independent of p, but it is still affected by p in the same kind of way as the stock price.Usually, the risk-neutral probabilities are related to the real-world ones: (real) unlikely events still tend to be risk-neutrally unlikely, too. There is no NECESSARY relationship between them, though: it all depends on market-participants aggregate risk-aversion properties.Again: the risk neutral measure is precisely that measure (if it exists and is unique) in which ALL asset prices are expectations of their payoffs.

So, maybe it will just make my life easier to stop thinking about this damned (and apparantely non-existent) "expectation rule" and proceed as if I had no knowledge of such a rule? For some reasons it has been implanted into my brain that this rule must apply in the real world.Because basically many of the probabilities you have spoken of are risk-neutral probabilities, which aren't even remotely related to real probabilities if you come to think of it. They are just some theoretical numbers that help finding particular values.Hmmm I'm getting closer there mentally but some brain cells are still twitching. Ok what about this: 1) The notion of exceptation (in the real-world) is ridiculous and shouldn't even be considered. 2) Even more ridiculous is even thinking about using real expectations to price anything because the concept of p(robability) is also ridiculous except just as a helper tool to be able us to mentally grasp the binomial model.Bah. Even this is not good enough. Keep the explanation coming!

QuoteOriginally posted by: sburaI'm afraid you're not correct there, because it is well-known (as shown by fancy replication theory) that the call price has nothing to do with the (real) probability p. How to get intuitive (as opposed to finding some other valid and theoretically sound method) hold of the fact why the expectation rule doesn't hold for call option prices would be a good thing to have.Somethings are just not all that intuitive. Risk neutral pricing is one of them. At least thats the viewpoint I've settled on. Of all the finance books I've read, I think Markov and Cont, "Financial Modeling with Jump Processes" gives the best intuition on why. In the 2004 version, this would be section 9.1.1 page 296.

You should certainly not stop thinking!The meaning of the risk-neutral measure and how it relates to real-world probabilities is probably the most subtle concepts in finance: several people got Nobel prizes for it, so don't be discouraged that you find it hard to understand. I think that maybe 75% of quants don't really understand it, either. Not knowing which measure you're in is probably the most common fundamental error in math/finance.The notion of real-world probabilities is definitely not ridiculous. If I am trying to decide whether to buy a stock or not, I care about the real-world probabilities. The risk-neutral probabilities tell me only (and exactly) one thing: what does the rest of the market think. If I am trading against the market, I very definitely want to consider real-world probabilities. In that case, my valuation will depend on the real-world probabilities and MY risk aversion. By the way, this is true if you are buying a call option and not hedging it!You want to use risk-neutral probabilities if you are considering an asset as a derivative. In that case, you are not (well, in theory) taking any market risk at all. What the market encodes for you in terms of the prices is sufficient to price your derivative asset: you can synthesise the asset perfectly, so market participants agree on the price. You can think of the risk-neutral measure as just a pricing trick, if you will, though it does also tell you interesting information about risk-premia for various risk sources if you look at it carefully.These concepts are far more general than the binomial model, by the way: they apply to all models.

QuoteOriginally posted by: sburaI don't get why a call option's price can be independent on probability in the binomial model.Let's consider a simple case.A stock is currently worth $1. In one year it can go one of two ways: Up to $1,000,000,000 with probability p, or down to $0.5 with probability 1-p.I just don't get it. Why the hell wouldn't you want to pay more for a call option of strike $1 if p=0.99 than if it were p=0.01?Please explain to me in intuitive manner why I wouldn't want to pay more for the call option for higher p.Because you you can hedge a call option independly of the probability of p (taken from Shreve: Stochastic Calculus and Finance):Let S0 = 50, S(up) = 100, S(down) = 25, r = 25% and Strike = 50:Then:Call(up) = max(100-50,0) = 50Call(down) = max(25-50,0) = 0Buy 2/3 Stocks and borrow -40/3 Dollar : (100-40)/3 = 20 Net investmentThen :Up => 2/3*100 - 50/3 = (200-50)/3 = 50Down => 2/3*25 - 50/3 = 0This hedge works independly of the probability of that S is going up.The hedge costs are 20. A appropriate hedge portfolio can also determined with your example.