I have a question regarding the intrinsic value of an European option. I use the following notations: $S_t$ price of the non dividend paying stock at time $t<T$, $T$ is the maturity, $r$ risk-free interest rate p.a., $K$ strike price, $\sigma$ volatility p.a. The Black-Scholes formula for the value of an European call option at time $t$ is given by: \begin{align} C_t&=S_t \Phi(d_1)-Ke^{-r(T-t)}\Phi(d_2) \\ d_1&=\frac{\ln\left(\frac{S_t}{K}\right)+(r+\frac{\sigma^2}{2})(T-t)}{\sigma \sqrt{T-t}} \\ d_2&=\frac{\ln\left(\frac{S_t}{K}\right)+(r-\frac{\sigma^2}{2})(T-t)}{\sigma \sqrt{T-t}} \end{align} In the textbook from Hull I find the following definition of the intrinsic value:

The intrinsic value of an option is defined as the maximum of zero and the value the option would have if it were exercised immediately. For a call option, the intrinsic value is therefore $\max\left\{S_t-K;0 \right\}$...

Now I wonder why it is defined that way. I can only exercise the option at the expiration date. Intuitively I would have defined it as $\max\left\{S_t-Ke^{-r(T-t)};0\right\}$, which is the lower bound for the price of an European call option. Also consider the case $S_t \rightarrow \infty$ when holding all other parameter fixed. In this case $C_t-(S_t-K)$ converges to $K-Ke^{-r(T-t)}$ which is the time value of the option. On the other hand $C_t-(S_t-Ke^{-r(T-t)})$ converges to zero.

I think that as the value of the share increases, the time value should actually go to zero, since it is virtually certain that the option will be exercised. However, if the intrinsic value is defined as in Hull then this is not the case.

I am grateful for any answers and thanks in advance !


1 Answer 1


Intrinsic value, by definition, is the value of the option if it were to be exercised today, so there is no time value involved, and no consideration as to if the option could actually be exercised today. If the underlying is at \$50, then a call option with a strike of \$40 has an intrinsic value of \$10 by definition - if I exercise the option today, I buy a stock that's worth \$50 for \$40 for a \$10 instant profit.

I think you are trying to apply the theoretical value of the option today and not the profit if it were to me exercised today.

What your algebra shows is that prior to expiry, the time value is not floored at zero but at the difference between the present and future value of the strike.

  • $\begingroup$ Thanks for your answer. What confuses me is the interpretation of the time value of the option when the intrinsic value is defined this way. Example: Assume the strike price is \$50 and the stock price is \$1000. Intrinsic value = \$950. Choosing some arbitrary values for the other parameters I calculate with BS a value of \$951.48 for the option. So the time value is \$1.48. Now assume that the stock price is \$1.000.000. Intrinsic value \$999.950, BS gives a value of \$999.951.48 so in this case the time value is also \$1.48. So the time value does not converge to zero. $\endgroup$
    – Count
    Mar 10, 2021 at 9:16
  • 1
    $\begingroup$ Right - the time value prior to expiry is floored at the difference between the present and future value of the strike. (K(1-e^[-r(T-t)] as you show). As you get closer to expiry, the time value does converge to zero. $\endgroup$
    – D Stanley
    Mar 10, 2021 at 14:39
  • $\begingroup$ This makes sense. Thanks for your help ! $\endgroup$
    – Count
    Mar 10, 2021 at 16:19

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