[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.