Reading Black–Scholes Call Price Surfaces Across Spot and Time
Summary
The document discusses how to interpret a plotted European call value as the underlying price and time to maturity vary. The response first flags a likely implementation or scaling problem: with the stated strike and plotted values, the displayed prices appear inconsistent with a standard call valuation. It gives a corrected illustrative value under a different interpretation of the underlying price, underscoring that plot axes and units need checking before drawing conclusions.
The qualitative surface behavior is then explained. For a fixed maturity, a call generally gains value as the underlying price rises because finishing in the money becomes more likely. As maturity approaches, a call’s value tends toward its increasingly certain expiry payoff; volatility also raises option value by widening possible outcomes. These observations are conceptual, not a validation of the original plot. Exact behavior depends on model inputs and conventions, and the discussion does not derive the Black–Scholes equation or assess model assumptions.
Key ideas
- A surface plot reports the option value for each combination of underlying price and time, so axis units must be checked.
- The response identifies a likely inconsistency between the stated inputs and plotted call values.
- At fixed maturity, a call generally becomes more valuable as the underlying price rises.
- As expiry approaches, the call value converges toward the payoff implied by the increasingly certain terminal price.
- Higher volatility generally increases option value by expanding the range of possible outcomes.
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# Implications of Black Scholes Plot
# Implications of Black Scholes Plot
I'm pretty new to finances, but I'm heavily into scientific computation. For my scientific computations class, I need to have at least a basic understanding of finances for the presentation I'm going to give.
Given the Black-Scholes equation, \begin{equation*} \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2\frac{\partial^2 C}{\partial S^2} + rS \frac{\partial C}{\partial S} - rC = 0 \end{equation*} I was able to plot the following graph where the volatility $\sigma = 0.08$, risk-free interest $r = 0.05$, strike price K = 10, C is the payoff, S is the current market price, and t is the time to expiry.
However, I am at loss at how I can interpret this. Can anyone give me any insight about the meaning of this graph and its implications? Any help would be highly appreciated.
## Answer by bcf (score 3, accepted)
https://quant.stackexchange.com/a/22073
Something is off in your plot. The value of a call should be very near zero with a strike price $10$ for the stock prices and times you have plotted. At first I thought you may have plotted "moneyness" defined as $S/K$ instead of $S$, but then your values are too low for that. May want to check your implementation.
Besides that, the plot is telling you just what any other two-variable plot is: the value of the function for a given pair of variables. In this case, e.g. the (supposed, but not correct) price of a European call option at which $S = 1.4$ and $t = 5$ is about 2.5 (peak of your graph). (Note the correct price, assuming the stock price is actually 14, is about 4.48 - see here and plug in "8" for volatility and "5" for interest rate, leave dividends blank).
The behavior of you plot does seem correct, though. You can observe that, for fixed $t$, the price of the option increases as the stock price increases. This makes sense, since it is increasingly more likely to expire with a positive value. Also, for fixed $S$ and decreasing $t$, meaning as we approach maturity (note this is actually moving forward in time), the call becomes worth less and less, since its value at expiration is become more and more certain (for fixed $S$). This is always true for calls, but in some extreme situations puts do increase in the value as a function of $-t$.
## Answer by SmallChess (score 1)
https://quant.stackexchange.com/a/22077
I'd want to add what @bcf has already mentioned:
To answer your comment, as your option approach to maturity, your likelihood of your final position being different to your current value decrease. This is related to volatility. The more volatility your option has, the more expensive it is. What does this have to do with investors? An investor with a long term expectation would go for a longer maturity, but the investor must be prepared to pay for more for the option. A short-term investor might not want a short-term option (because it doesn't have enough risk). The short term investor might prefer a forward contract where no option premium is required (easier accounting).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.