Was Fermat's Theorem Really Proven?

[PairTheBoard]

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I'm not sure how serious you are, but: do you understand the strategy I outline? you don't need to have black scholes and differential equations to get it. if all you're getting out of it is a "smell" I would suggest, if I may be so bold, that you work harder on the analyticcs. get a pen and paper and work through several scenarios of stock price movement and trades.

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I believe I understand the strategy. I've studied a rigorous proof of it and sat in lecture where the proof was given. As I understand it, the hedging strategy is done by the seller of the option to insure he doesn't lose a ton of money on it. If he's sold a call and the stock rises he basically takes a gradually increasing position in the stock to offset potential loses on the call he's sold. From the seller's point of view he doesn't care if it goes up as long as it doesn't go up too much - in which case he must hedge and protect, but potentially lose his profit on the sale of the call in doing so. He would much rather the stock goes down because he then automatically keeps the profit on the sale of the call. The Sales Price for the option is dictated by BS to be such as to probablistically offset the costs involved in hedging against a rising stock price. It's a "fair" price because the buyer of the call has the same EV as the seller of the call. That's entirely different from saying that either the buyer of the call or the seller of the call doesn't care whether the stock goes up or down. That smells fishy because it's not true. While the EV of the seller and buyer are equal their results are different depending on the size and direction of the stock price movement.

The reason BS was such a breakthrough result is that in computing this "fair" price for the option, which gives both buyer and seller equal EV, it was able to take the Stock "Trend" - assuming there is such a thing - out of the final calculation. The final calculation of the "fair" option price according to BS does not depend on the "Trend" except in how the Trend effects Volatility. This was an astonishing result by BS and had a huge freeing up effect on the options market. The BS mathematics has undergone tremendous scrutiny so it's almost certainly correct. However I still can't believe it.

PairTheBoard

[mosta]

I'm not extremely advanced with this material (or with math in general). If you have studied the material technically and considered my explanation and still have doubts, I'm probably not sophisticated enough to address your concerns. As far as Ito's lemma and the dropping out of the stochastic part of the differential equation goes, I have an at best indirect understanding of the proof in specifics. However I am comfortable with teh claim that the trading strategy wherein the options provide protection for stock scalping, and the value of the option is the expected amount of scalp based on motion irrespective of direction--I forgot how to finish this sentence--I'm comfortable with that model and stock direction does not play into it. now in even simpler terms with no calculus and all, just diagram a PNL graph of a delta neutral call and then do a delta neutral put. they are exactly the same in all respects except interest and dividends. same delta, gamma, vega, theta. if you want to be more analytical look at the expressions for those partial derivatives. they're all the same (except the interest difference has a slight impact on the theta expression). but I find the graph more convincing. I mean all you need to konw is that the gamma is the same. that essentially tells you that it's the same option. and if you use the hedging strategy that I outline, it works out exactly the same if you buy a call and sell 50 shares or buy a put and buy 50 shares. you have 50 more shares to trade on either stock price direction, with either kind of option.

[jason1990]

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The reason BS was such a breakthrough result is that in computing this "fair" price for the option, which gives both buyer and seller equal EV, it was able to take the Stock "Trend" - assuming there is such a thing - out of the final calculation. The final calculation of the "fair" option price according to BS does not depend on the "Trend" except in how the Trend effects Volatility. This was an astonishing result by BS and had a huge freeing up effect on the options market. The BS mathematics has undergone tremendous scrutiny so it's almost certainly correct. However I still can't believe it.

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It's not about EV at all. It's about the fact that the option is not the only game in town. You can trade in the stock directly and take either side of the wager.

I'm sure you've seen it, but I'll repeat a canonical example. Today the stock is $100. Tomorrow it will be $200 with probability p and $50 with probability 1-p. I want to sell you an option to buy the stock tomorrow for $170. How much should you pay me?

How much will the option be worth tomorrow? It'll be worth $30 is the stock goes up and $0 is it goes down. Can you achieve that same result without going through me? Sure. Take $10 out of your pocket and borrow $10 from your uncle. Buy 1/5 of a share. If the stock goes up, sell it for $40, pay back your uncle, and you're left with $30. If it goes down, sell it for $10, pay back your uncle, and you're left with $0. So you can achieve the same result without my option and it cost you $10 out of your pocket. So you'd be a fool to pay me any more than $10 for this option. (In fact, this is exactly what I will be doing to make sure I have what I need to pay you tomorrow.)

But if I'm selling it for $9, then you've got a nice opportunity. Borrow 1/5 of a share from your uncle and sell it for $20. Buy my option. You've got $11 left. If the stock goes up, borrow $170 from your uncle, use the option, sell 4/5 share for $160, pay back your uncle his $170 plus his 1/5 share and you're left with $1. If the stock goes down, borrow $50 from your uncle, buy a share, sell 4/5 share for $40, pay back your uncle his $50 plus his 1/5 share and you're left with $1. Where does your free $1 come from? It comes from me. So I'd be a fool to sell the option for anything less than $10.

So the price is $10 and it has nothing to do with p. It therefore has nothing to do with EV. For example, if p=0.25, then your EV when you buy the option is $-2.50 (= $30*p - $10) and my EV when I sell it is $2.50. The option is not priced according to EV. It's priced according to the fact that it's nothing more than a shortcut. When I sell you the option, I'm just doing you a service. I'm making your life easier so that you don't have to go through the rigamarole of borrowing from your uncle, etc.

Now, it's an interesting fact that if you artificially set p=1/3, so that the stock has no "trend" and then price the option according to this fake EV, then you come to the right price. For no other value of p will you come to the correct price when you use the EV method. Now that may be something to be mystified by.

But the fact that the true EV doesn't matter is not mysterious. The option is a derivative. It's outcome is simply a shortcut for some other procedure. That other procedure is of the form, "if this happens, I'll do that; if that happens, I'll do this, etc." The procedure doesn't at all depend on the specific value of any of the probabilities of those events. So the startup cost for that procedure doesn't either. And that startup cost is exactly what you should pay for the option. It would be an entirely different story if the buyer of the option was unable to trade directly in the stock.

[wmspringer]

Note that, while in science a theory is basically a "best guess", in math a theorem is something which has been proved; if the logic of the proof is sound, the theorem can never be disproved.

Edit: Really I should say, if the logic AND the axioms are sound.

[wmspringer]

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Personally, I believe the proof. I watched the Nova special about Andrew Wiles three or four times--I still have it. I didn't understand most of it, but it was fascinating none-the-less. Early on he thought he solved it, but there was a problem. That problem was solved. You should view this tape. You might be able to find it at your local library. Nova repeats their programs from time to time as well.

Will anyone ever be able to come up with such a "simple" proof, and perhaps more importantly--was Fermat telling the truth?

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I've seen the special and it was interesting, although I prefer the book that goes with it (Fermat's Enigma by Simon Singh)

The interesting thing with Fermat is, I understand that most if not all (I forget?) of his notes turned out to be correct. Most likely he made an error or overlooked something, but wouldn't it be amazing if there really was a proof using 17th century mathematics?

[wmspringer]

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My thoughts exactly...
and those colored maps are giving me bad flash backs of my combinatorics class last quarter [img]/images/graemlins/frown.gif[/img]

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Heh, that was actually my specialty...I had a paper out the other year on the four color theorem [img]/images/graemlins/grin.gif[/img]

[PairTheBoard]

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The theory seems to imply that a put in such a stock is worth just as much as a call. Makes no sense.

PairTheBoard

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yes, put-call parity is one of the first things you learn as an options trader. all that put-call parity amounts to is that buying a (at the money) call on 100 shares of stock and hedging it by selling 50 shares of stock short against it amounts to entirely the same thing (except for a minor wrinkle with basis exposure) as buying the (same strike) put on 100 shares and hedging it by buying 50 shares of stock. same risk exposure for stock price movement, time decay, and implied volatility change (different for interest rate and dividend change but those are usually pretty minor).

the way to understand BS is to understand the hedging/trading strategy that they use to derive the option value. suppose a stock is trading at $60 and I buy 1 contract of the 60 call for $1. one strategy would be to hold the calls and hope stock goes up over $61, where my profit starts. here's the BS strategy. (remember a "contract" is for a call on 100 shares--it's like 100 calls you might say.) when I buy the call at the same time I can sell 50 shares $60. (remember stock is currently trading at the strike price, 60.) now the game changes. now I don't need stock to go up, rather I can profit on an up move or a down move. consider the two scenarios:

a. stock goes up. what's my strategy here? I have the right to buy 100 shares at 60 a share (my call contract). I paid $1 for this. so to break even I need to sell 100 shares at an average price of $61. I already sold 50 shares at $60. so now if I sell 50 more shares at $62, I break even. or to be more cautious, what I might do is sell some at $60.50, $61, $61.50, $62, $63, etc a little at a time. once I'm short 100 shares, I could call it a day. hopefully I've worked my price average up over $61. I cover those short shares with my call contract and close everything out. my profit is the differency between my average sale price starting with teh first 50 shares I sold at $60, and $61 (60 for the exercise price of the call plus $1 for what the call cost me).

b. but what if stock goes down? well now the call is not looking so useful anymore. but I can still make money. I can buy my 50 short shares back. I break even if I can make a $1 buying the shares back. I sold 50 shares at $60, so if I can buy 50 shares at an average price of $58, I break even. hopefully I can buy it better than that. now my call is not something I'm thinking about using (I wouldn't exercise the right to buy at $60 when stock is under $60), but I made money on my short stock hedge.

now take another step. suppose you bought the calls at the money and sold the 50 shares, and then the stock kept running up and you kept selling a little at a time until you had sold 50 more shares. in a sense you're done. but actually you're not: now suppose stock drops back down suddenly to $60. what do you do? start the whole process over! sell 50 shares again. if stock runs up again, sell more and more until you've sold 50 more. if stock drops in price, buy more and more until you've bought back the 50. you can do this indefinitely. you can scalp stock against your options, because they stop you out from having any loss on an extreme price move. in fact yo make money on an extreme price move in either direction. one more refinement: you don't have to wait till stock as at strike and sell exactly 50 shares. whenever stock moves up you sell some and whenever it moves down you buy some. your stock position will vary between 0 and short 100 shares. if you can scalp on a continuous basis, then you can use calculus and differential equations to model this strategy.

and that is how we value the option. it doesn't matter whether stock goes up or down. I will scalp stock either way. I don't care which _way_ it moves, but I do care how much it moves. I need to make the price of the option in by trading stock against my options as it moves up or down. the more it moves the more I can trade stock and make little profits putting my hedge on and off (or holding my hedge and trying to catch a single big move-same thing either way, it scales). this is why the standard deviation of stock price movement is all we need to konw to value an option--because we have a money making strategy to trade against the option that is indifferent to which way stock moves. but it does need stock to move. time passing is lost opportunity for the stock to move and for me to trade against it. so as time passes options drop in value. you're biggest risk when you buy the calls and sell the 50 shares is that the stock price might not move at all.

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ok, I broke down and actually read your explanation here. I'm getting an inkling you might be right. But let me seek clarification. I believe you're saying you do equally well with the Call-Hedge combo as with a Put-Hedge combo regardless of how the stock moves. Let's say your Call-Hedge Combo has you all-in with 100 total shares sold at say $80/sh and your Put-Hedge combo has you all-in with 100 total shares bought at $40/sh. Due to put-call parity you paid the same price for them. Now suppose due to the extreme underlying upward "Trend" in the stock - assuming there could be such a thing - the stock triples in price straight to $180/sh. Your Call-Hedge strategy has made you something like $10/sh on 50 shares. Your Put-Hedge strategy has made you what? As the stock moves up you gradually sell your 50 shares until it hits $80/sh and you make something like $10/sh on 50 shares.

By god I think you're right. The Put-Hedge is just as profitable as the Call-Hedge even with a Tripling-Trend. That's amazing. Why didn't my phd advisor explain it to me that way instead of just smiling and winking? That's not how I remember the BS proof going but it makes sense. I've been waiting years for somebody to explain this to me in a satisfactory way and I think you've done it. Finally I can declare an inner peace treaty with Black-Scholes.

Kudos to mosta

and Kudos to the Collective Intelligence of 2+2


PairTheBoard

[PairTheBoard]

Nice explanation. You guys have convinced me. Thanks.

PairTheBoard

[mosta]

[ QUOTE ]

saying you do equally well with the Call-Hedge combo as with a Put-Hedge combo regardless of how the stock moves. Let's say your Call-Hedge Combo has you all-in with 100 total shares sold at say $80/sh and your Put-Hedge combo has you all-in with 100 total shares bought at $40/sh. Due to put-call parity you paid the same price for them. Now suppose due to the extreme underlying upward "Trend" in the stock - assuming there could be such a thing - the stock triples in price straight to $180/sh. Your Call-Hedge strategy has made you something like $10/sh on 50 shares. Your Put-Hedge strategy has made you what? As the stock moves up you gradually sell your 50 shares until it hits $80/sh and you make something like $10/sh on 50 shares.


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A. long the 60 call for $1, short 50 shares at $60.
let's do this as discrete one time period to expiration model.

1. stock goes to $80. sell 50 shares. net sold 100 shares at average price of $70. exercise call to buy 100 shares for $60. all positions closed. profit = $20 on 50 shares and $0 on 50 shares, minus $1 on call contract.

2. stock goes to $40. buy 50 shares. net zero stock position (sold 50, bought 50). call expires worthles. so no position. profit = $20 on 50 shares, minus $1 on call contract.

B. long the 60 put for $1, long 50 shares at $60.

1. stock goes to $80. sell 50 shares. net zero stock position (bought 50, sold 50). put expires worthless. so no position. profit = $20 on 50 shares, minus $1 on put contract.

2. stock goes to $40. buy 50 shares. net bought 100 shares at average price of $50. exercise put option to sell 100 shares at $60. all positions closed. profit = $20 on 50 shares and $0 on 50 shares, minus $1 on put contract.

now you said stock goes to $180. we add one time period. it matters whether it went to $180 from $40 or from $80. consider each of the four scenarios above as a starting point.

A.1. stock goes from $80 to $180. you have sold 100 shares (but you haven't exercised your call yet and it hasn't expired because there's a time period left). your position is neutral. profit/loss is zero.

B.1. stock goes from $80 to $180. you sold the 50 shares you bought already. your stock position is zero. your put is way out of the money. your position is neutral. profit/loss is zero.

A.2. stock goes from $40 to $180. you had zero stock at $40 because you bought back your short 50 shares. now on this gap move to $180 you will sell 100 shares at $180. your call expires and is exercised so you cover your short 100 shares buying for $60. you make $120 on 100 shares in addition to the scalp you made in time period one.

B.2. stock $40 to $180. you were long 100 shares total at $40 against your puts. on this gap move to $180 you will sell those 100 shares at $180. you have zero stock position and your put expires worthless, out of the money. you make $120 on 100 shares in addition to the scalp you made in the first time period.


Here's how it works from the trader's perspective, analytically. you buy a call or a put with a given strike and expiration. let's say the 60 strike again, but this time stock is at $70 when you do the trade. you look at your sheets to get the delta of the option. the call delta is 65. the put delta is 35.

A. buy ITM call for $11 ($10 parity plus $1 intrinsic/time value). sell 65 shares. now you are delta neutral. you have 35 shares to sell as stock goes up and 65 to buy as stock goes down.

B. buy OTM put for $1 ($0 parity, plus same $1 intrinsic/time value). buy 35 shares. now you are delta neutral. you have 35 shares to sell as stock goes up and 65 to buy as stock goes down.

the next morning stock is up $5, to $75. you look at your sheets to see what your new delta is. the call and the put had the same gamma, let's say 2 (deltas per $1 of stock move). so in both cases your delta on your sheets is now long 10. so you say, I better sell 10 shares.

A. you now have sold a total of 75 shares. you have 25 more to sell as stock goes up and 75 to buy as it goes down. you are delta neutral again. and your gamma is now a less (stock farther from strike). your gamma is like 1.5 now.

B. you bought 35 shares before and now you sold back 10. you have 25 more to sell as stock goes up and 75 to buy as it goes down. delta neutral. gamma 1.5.

both are the same. the gamma (convexity) ratchets your delta up or down as stock moves. that signals you to trade stock and improve your average price or scalp. as stock moves away from strike you run out of gamma, so run out of stock to trade and end up fully hedged--and then you hope that stock reverses and goes back to strike so you get gamma again and can scalp again.

graphically you know how a call or a put is a curve inside a hockey stick (laid flat on it's shaft). when you trade stock you're applying a linear transformation to that graph which means that you rotate it. so whichever way you started--the call was pointing to the right, the put was pointing to the left--after the stock hedge the graphs have rotated to be exactly the same bowl. delta neutral means the level point on the bowl is where stock is. and the graph curves up in either direction. that curve is your gamma. the curve up is also your profit as stock moves. as stock moves up the bowl, that means you are getting a delta. so you trade stock to get neutral and rotate the bowl flat again, so it's symmetric. the stock trades that rotate the bowl flat lock in little bits of profit by either improving your average price as you compelete your hedge or scalping as you take off your hedge.

[mosta]

jason's probabilistic argument and my trading/hedging strategy are two different ways to see put-call parity. here's a third one: consider the combination of long call and short put on the same strike. what is that position? if stock is above strike at expiration you exercise the call. if stock is below strike at expiration you get assigned on your put. so regardless, the combo amounts to a contract to buy the stock at the strike price at expiration, whether stock is above or below. what's that? a forward contract. what's the value of a forward? parity plus basis. there is no time value/intrinsic value to a forward. thus all the time value/optionality of the long call and short put must cancel out--must be the same.

suppose stock is at strike. the call is trading $1 and the put is trading $2. (suppose the specialist is bearish and doesn't know how to price options and thinks the put should be worth more because he "knows" stock is going down.) what do you do? buy the call, sell the put, sell 100 shares, and never look at the position again. you pay $1, collect $2, sell stock at strike, and it back at strike at expiration either by exercising your call or being assigned on your put depending on whether stock has gone up or down. riskless $1 profit/arbitrage. (okay there are interest and dividend factors/risks, and that is why call and put are different in that regard.) you have a forward hedged/arb'd against the underlying. you have no risks, not for stock price or time (yes for interest and dividend...).

you can also do it where stock is not at strike. then there will be parity in one optoin. but the value above parity must be the same or you get an arb.