Consider the Black-Scholes equation for a European Call Option, \begin{equation} \begin{cases}\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + r\frac{\partial V}{\partial S} -rV = 0, \ &\text{for} \ (S,t)\in\mathbb{R}^+\times[0,T] \\ V(S,T) = \max(S-K,0), &\text{for} \ S\in\mathbb{R}^+ \\ V(0,t) = 0, &\text{for} \ t\in[0,T] \\ V(S,t) = S - Ke^{-r(T-t)}, &\text{as} \ S\rightarrow \infty, t\in[0,T] \end{cases} \end{equation} where $\sigma$ is the volatility of the underlying (the stock), $r$ is the interest rate, $K$ is the strike price, $T$ is the maturity time of the option, $S$ is the current stock price, and $V(S,t)$ is the value of the option.

I have few questions regarding the model. To start, why does the Black-Scholes model use a final condition at $t = T$, rather than using an initial condition, and why does it solve backward in time? For example, let us say we would like to find the value of a European Call option with a maturity time, $T=1$ year, and a strike price of $K = 10$. From my understanding, Black-Scholes should solve the value of $V(S,t)$, for all $t\in[0,T)$, for the current stock price $S$. My confusion is as to why can we let $S$ tend towards infinity if we do not know the future stock price from the get-go, and hence how would we know the value of $V(S,1 \ \text{year}) = \max(S-10,0), \text{for} \ S\in\mathbb{R}^+$? If we solve the equation numerically, we must set an upper bound to our stock price array, but how do we know this upper bound? Thus is there a formal way we may solve for this upper bound? If so how and why? If we solve the Black-Scholes equation, we will see that it solves backward in time. Hence we solve for the value of $V(S,t), \text{for} \ t\in[0,T)$. To me, this seems almost useless since European options may only be exercised at maturity time. Therefore why do we care to solve for the value of $V(S,t), \text{for} \ t\in[0,T)$?

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    $\begingroup$ You are asking at least six questions. Prioritize your concerns and limit this to one question or two questions. Otherwise, you are unlikely to receive much attention to try and explain the entire theory. With that said, the PDE is state-space problem for which the unknown value $V(S,t)$ is known at expiry $t=T$. Moreover, the behavior of $V(S,t)$ is also known at the boundaries $S=0$ and $S\to \infty$ at $t=0$. And as far as a "numerical boundary" for $S\to \infty$, that is something you need to empirically examine and will depend on $T$. $\endgroup$
    – Mark Viola
    Commented Jun 17, 2020 at 20:47
  • $\begingroup$ Please don’t crosspost questions across the network. $\endgroup$
    – Bob Jansen
    Commented Jun 17, 2020 at 20:48

1 Answer 1


Why would you know the value of the option at $T$? Because at that time there is no uncertainty anymore about its value. If at time $T$ the stock price has a value $S$, then the value of the Europen call option is by definition $\max(S-K,0)$. Because at that point you can hedge the option by selling the stock and receiving $K$ for it, effectively paying $S-K$ if the person having the option exercises it. If it doesn't exercise the option, you don't have to hedge it hence you pay $0$.

But at any time before $T$, you don't know what the value of the stock will be at $T$, hence an uncertainty exists that has to be factored into the value of the option, and that's what the Black-Scholes equation does, by weighing the price by the probabilities of all the possible paths toward a final stock price from the current stock price at $t$.


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