Deep-In-the-Money Boundary Condition for a Black–Scholes Call
Summary
The document explains the large-stock-price boundary condition for a European call under the Black–Scholes–Merton PDE. At very high stock prices, exercise at maturity is effectively certain, so the call’s value approaches the stock price minus the present value of the strike. This gives the limiting relation between call value, underlying price, and discounted exercise payment.
The explanation relies on intuition and an arbitrage argument: reserve the discounted strike amount now to fund exercise later, while the call behaves increasingly like ownership of the stock as it moves deep in the money. The document names this relationship the deep-in-the-money bound. It does not provide a formal proof or discuss assumptions such as dividends, transaction costs, or other market frictions, so the explanation should be read in the stated European-call setting.
Key ideas
- A European call’s value approaches the stock price less the discounted strike as the stock price becomes very large.
- At high stock prices, exercise at maturity is treated as virtually certain.
- The boundary condition is motivated by an arbitrage argument and called the deep-in-the-money bound.
- The explanation is intuitive and does not give a formal derivation.
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# About the boundary conditions of the Black-Scholes-Merton PDE
# About the boundary conditions of the Black-Scholes-Merton PDE
I have a question about the solution of the Black-Scholes PDE for the European call option when I read the book Stochastic Calculus for Finance II of Steven E.Shreve.
Let $c(t,x)$ be the value of the European call option at time $t$ if the stock price at that time is $S(t)=x$. Then, $c(t,x)$ satisfies the following equation: $$c_t(t,x) + r x c_x(t,x)+ \frac{1}{2} \sigma^2 x^2 c_{xx}(t,x)=rc(t,x) \text{ for all $t\in [0, T), x\geq 0$},$$ and $c(T, x)=(x-K)^{+}.$
To resolve the above equation, one needs boundary conditions at $x=0$ and $x= +\infty$. For $x=0$. It's easy to derive that $c(t,0)=0$ for all $t \in [0,T]$.
As $x=+\infty$, I do not know understand how the author finds out (w/o a detailed explanation)(c.f. page 158 in that book) that $$\lim_{x \rightarrow +\infty} c(t,x)- (x- e^{-r (T-t) }K) =0 \text{ for all $t \in [0, T]$}?$$
## Answer by nbbo2 (score 1)
https://quant.stackexchange.com/a/19058
In words the equation says that when the stock price is very high, the value of the call is (approximately) equal to the stock price minus the PV of the exercise. It is fairly intuitive: if K=100 and x is 1000, the stock is so much above K that exercise is for all practical purposes certain; the call today is worth x minus PV(k), since you can set aside PV(K) today today to exercise at maturity. So it is an arbitrage argument.
This equation is sometimes called the deep-in-the-money bound for the value of a European call.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.