Why European Call Options Have Positive Time Value
Summary
The discussion explains why a European call’s Black–Scholes value exceeds intrinsic value when rates are assumed to be zero. For an out-of-the-money or at-the-money call, intrinsic value is zero, while the option price increases strictly with the underlying price and approaches zero as the underlying price approaches zero. This supports a positive option value for positive spot levels.
For an in-the-money call, another explanation rewrites the Black–Scholes price as intrinsic value plus a residual term. Since the model’s d1 exceeds d2, the corresponding cumulative normal probabilities differ in a way that leaves a positive residual. The exchange gives an intuitive mathematical argument, not a broader treatment of discounting, dividends, early exercise, or options outside the stated European zero-rate framework.
Key ideas
- A call’s intrinsic value is zero when spot is at or below strike.
- In the Black–Scholes model, call value increases with the underlying spot price.
- The call value approaches zero as spot approaches zero.
- For an in-the-money call, the difference between model value and intrinsic value remains positive because d1 exceeds d2.
- The explanations assume a European call and zero interest rates.
Tags
Full text
# Why is the time value of an option mathematically always positive?
# Why is the time value of an option mathematically always positive?
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 $\max(S-K,0)$.
## Answer by Gordon (score 6, accepted)
https://quant.stackexchange.com/a/21527
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.
## Answer by quis est ille (score 0)
https://quant.stackexchange.com/a/21525
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.