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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.

thisgraph

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.

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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 $-t$.

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  • $\begingroup$ Hi, thanks for you reply. I can't seem to access the link you've provided. Nonetheless, you've provided some very valuable insight. A couple of clarification though - as we approach T (time to expire) the value of the option decreases. What are the implications of this in terms of the choices we ought to make as investors? Since for European call option, we can't exercise our options any earlier than time T, what exactly is the value of this graph? $\endgroup$ – meraxes Dec 3 '15 at 3:22
  • $\begingroup$ @meraxes It sounds like you're just getting started learning quant finance? I recommend you pick up a good book on the subject, and one of the most popular is Hull - Options, Futures, and Other Derivatives. Another option is to pick one from the Wilmott empire, but you'll have to mind his ego. $\endgroup$ – bcf Dec 3 '15 at 3:31
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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).

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    $\begingroup$ The drift is unnecessary when pricing options - it is replaced by the risk-free rate. $\endgroup$ – bcf Dec 4 '15 at 12:58
  • $\begingroup$ @bcf Oh yeah... Sorry I forgot. Edited. $\endgroup$ – SmallChess Dec 4 '15 at 12:58

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