Why Black–Scholes Uses a Terminal Payoff and Solves Backward

# May someone please explain the intuition behind the Black-Scholes Equation?


# May someone please explain the intuition behind the Black-Scholes Equation?












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.

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? From my understanding, Black-Scholes should solve the value of $V(S,t)$, for all $t\in[0,T)$, for the current stock price $S$. Hence how would we know the value of $V(S,T) = \max(S-T,0), \text{for} \ S\in\mathbb{R}^+$? Moreover, why do we care to solve for $V(S,t), \text{for} \ t<T$ if a European option may only be exercised at the maturity time $t=T$?

## Answer by Magic is in the chain (score 2, accepted)

https://quant.stackexchange.com/a/55001

In terms of the settings, we know the current stock price, we have assumed that the stock price dynamics follow Geometric Brownian motion (GBM), we know the parameters of this process (volatility etc), and we know the characteristics of the options (option type, maturity). In practice, we know the current price of the option as well, but we pretend we don’t, or you can say we want the model to re-produce this price, so knowing the price does not matter! With this context, here are some notes:

As you correctly pointed out, the European options payoff is at maturity: $\max \left(S_T-K,0\right)$ for a call option and $\max \left(K-S_T,0\right)$ for a put option. Hence if we know the stock price at maturity, we know the payoff, and we will know how much the option is worth at maturity. But we need to find out how much is this option worth today so that we can determine the fair price when buying or selling. There are two ways to go about it:

- One can simulate the value of the stock at maturity (using the assumed GBM dynamics), and then average the payoff as per the relevant probability distribution, and then discount it to today to get the price. The reason we have to simulate the price at maturity is because the option payoff depends on the stock price at maturity, and we can simulate the stock price using the dynamics (GBM) we assumed.

- An equivalent way is to approach the problem in terms of deterministic PDE, and solve it using numeric methods. This equivalence between the stochastic approach and the PDE is a consequence of a more general result, but we can set it aside for now. The simple reasoning goes as follows. The terms of the option contracts gives us the terminal condition (the payoff at maturity), so one can work backward. If we assume we know the stock price at maturity (knowing it can be between 0 and 1 million is enough!), we can calculate the value of the option at maturity. Using these prices at maturity, we can calculate the values at the preceding step (the PDE you have takes care of the probability/weights of moves from one time step to the next assuming the step size is very small).

Now focusing on approach no. 2 above, we know the stock price at maturity can be anything from zero to infinity, but then the probability is usually concentrated in a relatively small region, so the range is not as wide as one might think. But the numeric method won't know this. So alternatively, if you have say a call option with a strike of 100, then if the stock price turns out to be 1 trillion, does the strike of 100 matter in such situations? And the stock price can not go below zero, and the option does not pay when the stock price is below K, so you can safely assume the following boundary conditions.

- For large S, $V\left(t,S\right) \approx S$

- For very small S,$V\left(t,S\right) \approx 0$

Similar considerations give the boundary conditions for the put option. Of course I simplified a lot of technicalities in the above notes, but hope it is intuitive.

## Answer by Aksakal almost surely binary (score 2)

https://quant.stackexchange.com/a/54998

> Hence how would we know the value of V(S,T)=max(S−T,0),for S∈R+?

You know the value at time T as a function of S: it is simply the pay out, which is $\max(S-K,0)$, where $K$ is a strike.

> Moreover, why do we care to solve for V(S,t),for t<T if a European option may only be exercised at the maturity time t=T?

No, we're not interested in value at time T. It's trivial as I shown above. We're interested in value right now $V(S,0)$ or in future before maturity $T.