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Let's consider a simple European option in the Black-Scholes framework.

What is it about the maths of $SN(d_1) - KN(d_2)$ that makes its value always greater than $S-K$, when $S>K$? (I assume zero interest rate throughout).

By time value I mean the difference between the value of the option and the intrinsic value, where intrinsic value is $\max(S-K,0)$.

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  • $\begingroup$ Your question is not clear. The note in the brackets is not even correct. For example, what is SNd1 - KNd2? it is not Black Scholes formula. $\endgroup$ – Gordon Nov 1 '15 at 0:32
  • $\begingroup$ What do you mean time value? $\endgroup$ – Gordon Nov 1 '15 at 0:54
  • $\begingroup$ Fixed that. See the new version above. $\endgroup$ – quis est ille Nov 1 '15 at 8:47
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We consider the case $S\leq K$ only. In this case, the intrinsic value is zero. Note that, \begin{align*} \frac{\partial C}{\partial S} = N(d_1) >0. \end{align*} That is, $C$ is a strictly increasing function of the spot level $S$. Moreover, \begin{align*} \lim_{S\rightarrow 0} d_{1, 2} = -\infty. \end{align*} Then, \begin{align*} \lim_{S\rightarrow 0} C = 0. \end{align*} Therefore, $C>0$ holds.

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  • $\begingroup$ Very good. (Although this relies on knowing that the delta is $N(d_1)$, of course). $\endgroup$ – quis est ille Nov 1 '15 at 16:43
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I think I have part of it. Assume zero interest rate and T = 1. Then the call price C is

C  =  S.N(d1) – K.N(d2)

where S is underlying price, K is strike, and

d1 = ln(S/K)/V+V/2
d2 = d1 – V/2

d1 and d2 roughly represent the moneyness in terms of standard deviation, including the term V/2 which is added in d1, and subtracted in d2. Nd1 and Nd2 represent the moneyness in terms of probability. Note that the deeper in the money, the closer the probability gets to 1. Now when S > K, it is easy to show that time value must be positive. Let X = 1-Nd1, and Y = 1-Nd2. Then

C = S(1-x) – K(1-Y) = S-K +Y-X = intrinsic value + Y – X

Since d1 is always a bit greater than d2 because of the V/2 term, it follows that Nd1 is closer to 1, and so Y>X.

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  • $\begingroup$ The case of S<K is trickier. It follows that SNd1/KNd2 must be greater than 1 if time value is to be positive, but why is that? If I increase K then I decrease the ratio S/K, so it follows that the ratio Nd1/Nd2 must increase by at least as much. But I can’t see how that follows from the formula for d1 and d2. It must be from some property of the cumulative normal distribution, but don’t know. $\endgroup$ – quis est ille Nov 1 '15 at 8:52
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    $\begingroup$ Are you sure your formulas for d1 and d2 are correct? I think you got confused by what V means... Is it volatility $\sigma$ or variance $\sigma^2$? Anyway the reasoning that $N(\cdot)$ is strictly increasing and $d_2 < d_1$ is right, but you should still correct the formulas. $\endgroup$ – SRKX Nov 2 '15 at 4:59
  • $\begingroup$ I think the formula is correct. Please note I distributed the V (vol, not variance), which normally appears outside the bracket. I prefer this formulation as it makes clear the term ln(S/K)/V has units of standard deviation. $\endgroup$ – quis est ille Nov 2 '15 at 10:15
  • $\begingroup$ oh yeah you're right I missed one term. Sorry. $\endgroup$ – SRKX Nov 2 '15 at 10:24
  • $\begingroup$ That's all right. Thanks for the help. I couldn't find any textbook that explained this. $\endgroup$ – quis est ille Nov 2 '15 at 16:54

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