Was Fermat's Theorem Really Proven?

[jason1990]

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Without doing the math I'm guessing that in that example, with stock price of $130, a put with strike of $90 would be priced the same as a call with strike of $170.

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A call with strike $170 is worth $30 if up, $0 if down. Hedge it by taking $16 from pocket, borrowing $10, and buying 1/5 share for $26. Your 1/5 share is worth $40 if up, $10 if down, and you owe $10. So the call has price $16.

A put with strike of $90 is worth $0 if up, $40 if down. Hedge it by taking $18.67 from pocket, borrowing 4/15 share, and selling it for $34.67. You have $53.33. If up, this buys you exactly 4/15 share to settle your debt. If down, 4/15 share costs $13.33, leaving you with $40. So the put has price $18.67.

Put-call parity only applies when comparing calls and puts with the same strike price.

[jason1990]

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Without doing the math I'm guessing that in that example, with stock price of $130, a put with strike of $90 would be priced the same as a call with strike of $170.

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A call with strike $170 is worth $30 if up, $0 if down. Hedge it by taking $16 from pocket, borrowing $10, and buying 1/5 share for $26. Your 1/5 share is worth $40 if up, $10 if down, and you owe $10. So the call has price $16.

A put with strike of $90 is worth $0 if up, $40 if down. Hedge it by taking $18.67 from pocket, borrowing 4/15 share, and selling it for $34.67. You have $53.33. If up, this buys you exactly 4/15 share to settle your debt. If down, 4/15 share costs $13.33, leaving you with $40. So the put has price $18.67.

Put-call parity only applies when comparing calls and puts with the same strike price.

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In fact, if I've done my algebra right, then if the current stock price is S and the strike price is K, then the call and put prices are

call:
4S/3 - 200/3 - KS/150 + K/3

put:
S/3 - 200/3 - KS/150 + 4K/3

Of course, that's only true in this simple model, whereas put-call parity is true in general.

[PairTheBoard]

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Without doing the math I'm guessing that in that example, with stock price of $130, a put with strike of $90 would be priced the same as a call with strike of $170.

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A call with strike $170 is worth $30 if up, $0 if down. Hedge it by taking $16 from pocket, borrowing $10, and buying 1/5 share for $26. Your 1/5 share is worth $40 if up, $10 if down, and you owe $10. So the call has price $16.

A put with strike of $90 is worth $0 if up, $40 if down. Hedge it by taking $18.67 from pocket, borrowing 4/15 share, and selling it for $34.67. You have $53.33. If up, this buys you exactly 4/15 share to settle your debt. If down, 4/15 share costs $13.33, leaving you with $40. So the put has price $18.67.

Put-call parity only applies when comparing calls and puts with the same strike price.

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In fact, if I've done my algebra right, then if the current stock price is S and the strike price is K, then the call and put prices are

call:
4S/3 - 200/3 - KS/150 + K/3

put:
S/3 - 200/3 - KS/150 + 4K/3

Of course, that's only true in this simple model, whereas put-call parity is true in general.

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jason1990 --
"Put-call parity only applies when comparing calls and puts with the same strike price. "

?

Just to be clear. In the real market where the Black-Scholes theoretical modeling after brownian motion applies, if the stock price is $60 would a Call with Strike 62 be in parity - ie. equally valued - with a Put with strike $58? Or not?

PairTheBoard

[jason1990]

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Just to be clear. In the real market where the Black-Scholes theoretical modeling after brownian motion applies, if the stock price is $60 would a Call with Strike 62 be in parity - ie. equally valued - with a Put with strike $58? Or not?

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Not. In the simplest setting, we'll take a unit bond (so interest is zero) and a unit volatility. Also assume our expiration time is 1. Then a call with strike price K is valued at

60*F(ln(60/K) + 1/2) - K*F(ln(60/K) - 1/2),

where F is the cdf of a normal. Take K=62 and get 22.37. Take K=58 and get 23.60.

So the call with strike price 62 has value 22.37. By put-call parity, the put with strike price 58 has value

23.60 + 58 - 60 = 21.60.

[PairTheBoard]

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Just to be clear. In the real market where the Black-Scholes theoretical modeling after brownian motion applies, if the stock price is $60 would a Call with Strike 62 be in parity - ie. equally valued - with a Put with strike $58? Or not?

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Not. In the simplest setting, we'll take a unit bond (so interest is zero) and a unit volatility. Also assume our expiration time is 1. Then a call with strike price K is valued at

60*F(ln(60/K) + 1/2) - K*F(ln(60/K) - 1/2),

where F is the cdf of a normal. Take K=62 and get 22.37. Take K=58 and get 23.60.

So the call with strike price 62 has value 22.37. By put-call parity, the put with strike price 58 has value

23.60 + 58 - 60 = 21.60.

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Really? I'll take your word on it, but doesn't that seem a bit strange? With a stock price of $60 and strike prices of $60 for both the put and call they are valued equally by parity. But if you move the Strike price up $2 for the call and down $2 for the put, the Call becomes worth more than the Put. Do you have an intuitive explanation for this?


So in general Put-Call parity doesn't mean what I thought but actually means:

jason1990 --
"The put-call parity can be expressed as

put - call = strike - stock."

where they have the same Strike.


I'm now thinking that what Black-Scholes did was construct the theoretical Equivalent Trading Strategy for an option and calculate the theoretical cost of implemening the strategy - thus providing a rigorous pricing method for the option. Was Black-Scholes even needed then for the Put-Call parity you express above? Or does that follow automatically from the kind of arguments mosta has been making?

This is very cool stuff.

PairTheBoard

[jason1990]

Let S(t) be the stock price at time t. We'll assume there's no interest and consider calls and puts with expiration t=1 and strike price K. Denote their values at t=0 by C(K) and P(K). At expiration, the call and put are worth, respectively,

max(S(1)-K,0) and max(K-S(1),0).

Suppose you have found a CETS and a PETS. If you do the CETS and the opposite of the PETS, then at expiration you will have

max(S(1)-K,0) - max(K-S(1),0) = S(1) - K.

But you could guarantee yourself S(1)-K at time t=1 by simply borrowing K and buying the stock now for S(0). So we have

C(K) - P(K) = S(0) - K.

For this to work, we must assume that options are priced according to the startup cost of an equivalent trading strategy. In particular, we must know that such strategies exist and are unique. But we needn't assume, for example, that the stocks are performing geometric Brownian motion.

Now, you've already observed that C(S)=P(S), where S=S(0). Let a=C'(S). We might expect a<0. That is, if we increase the strike price, the value of the call will diminish. Then

C(S + dK) = C(S) + a*dK.

But P(K)=C(K)+K-S, so P'(K)=C'(K)+1, and P'(S)=a+1. Hence,

P(S - dK) = P(S) - (a+1)*dK.

In the Black-Scholes example I gave you, a=-0.3085 and we have C(S)=P(S)=22.98. This gives

C(60 + dK) = 22.98 - 0.3085*dK
P(60 - dK) = 22.98 - 0.6915*dK.

If you take dK=2 you will get very close to the values I gave you. This also shows (mathematically, though perhaps not intuitively) why the call ends up worth more than the put.

[jason1990]

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Let a=C'(S). We might expect a<0. That is, if we increase the strike price, the value of the call will diminish. Then

C(S + dK) = C(S) + a*dK.

But P(K)=C(K)+K-S, so P'(K)=C'(K)+1, and P'(S)=a+1. Hence,

P(S - dK) = P(S) - (a+1)*dK.

In the Black-Scholes example I gave you, a=-0.3085 and we have C(S)=P(S)=22.98. This gives

C(60 + dK) = 22.98 - 0.3085*dK
P(60 - dK) = 22.98 - 0.6915*dK.

If you take dK=2 you will get very close to the values I gave you. This also shows (mathematically, though perhaps not intuitively) why the call ends up worth more than the put.

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Here's some more info to overwhelm you. [img]/images/graemlins/smile.gif[/img] In the Black-Scholes setting and in the dicrete setting (the two examples I provided) you can create an "artificial" probability distribution on the movement of the stock so that

C(K) = E[max(S(1)-K,0)].

This means that for dK>0,

C(K) - C(K+dK) = E[max(S(1)-K,0) - max(S(1)-(K+dK),0)]
= E[(S(1)-K)*1_{K<S(1)<=K+dK} + dK*1_{S(1)>K+dK}].

(The notation "1_{...}" stands for the random variable which is 1 if ... happens and 0 otherwise.) Hence,

dK*P(S(1)>K+dK) <= C(K) - C(K+dK) <= dK*P(S(1)>K).

What this means is that if C'(K) exists, it must satisfy

C'(K) = -P(S(1)>K).

(Don't forget, this is the "artificial" probability.) This verifies the intuition in the previous post that a=C'(S)<0. Also, it shows that -1<=a<=0, so that if we decrease the strike price, the value of the put diminishes. (Recall that P(S-dK)=P(S)-(a+1)*dK.) If |a|<1/2, as in the Black-Scholes example, the put diminishes faster. But if |a|>1/2, then the call would diminish faster. To get an example of this, look at the discrete example I gave and start the stock at S(0)>125.

[PairTheBoard]

jason1990 --
Let S(t) be the stock price at time t. We'll assume there's no interest and consider calls and puts with expiration t=1 and strike price K. Denote their values at t=0 by C(K) and P(K). At expiration, the call and put are worth, respectively,

max(S(1)-K,0) and max(K-S(1),0).

Suppose you have found a CETS and a PETS. If you do the CETS and the opposite of the PETS, then at expiration you will have

max(S(1)-K,0) - max(K-S(1),0) = S(1) - K.

But you could guarantee yourself S(1)-K at time t=1 by simply borrowing K and buying the stock now for S(0). So we have

C(K) - P(K) = S(0) - K.
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This is kind of tricky and cool. Let me see if I understand it. You have introduced a 3rd Trading strategy that is equivalent to the Dual Strategy of a CETS and an Opposite-PETS. The 3rd strategy has a clear start up cost of S(0)-K, which is actually negative if K>S(0). Since the 3rd strategy is equivalent to the Dual Strategy they must have the same start up cost. And you further say that the start up cost of the PETS is negative the start up cost of an Opposite-Pets. This means the start of cost of an Opposite-PETS is itself Negative. Does this make sense? A Negative Cost is really a Revenue. Since a PETS is equivalent to buying a Put, is an Opposite PETS equivalent to selling a Put? This would make sense because selling a Put produces start up Revenue rather than start up cost. Maybe the easier way to see it is that the dual strategy of a PETS and Opposite-PETS would just be a do nothing strategy, thus zero start-up cost.

ok. However, I'm still not sure why the CETS and PETS are needed to get the parity. Why can't you just say that buying a call and selling a Put is equivalent to borrowing K and buying stock at S(0) so C(K)-P(K) must equal S(0)-K?

Maybe I can answer my own question. We need to know that C(K) and P(K) equate to startup costs of their Equvalent Trading Strategies?



Your use of the Calculus derivitive is really nice and proves that the Call decreases slower than the Put as the Srikes respectively move away from the stock price. I assume the $60/sh is irrelevant and the thing would scale to any stock price. It proves it but I'm still curious as to what's "really" going on. Arguments that the stock price can go up easier than go down or can go up further than can go down don't seem to make sense because of the Parity at a common $60 strike. I'll bet there's something fairly simple going on here that I'm not seeing.

Thanks,

PairTheBoard

[mosta]

is it not just because stock prices are log-normally distributed, not normally, and therefore not symmetric?

[PairTheBoard]

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is it not just because stock prices are log-normally distributed, not normally, and therefore not symmetric?

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Sounds good. But then, why doesn't that affect the Put-Call parity when both strike prices equal the stock price?

PairTheBoard