Why the Barone-Adesi–Whaley Approximation Drops a Boundary Term

# why we drop the last term in the Barone-Adesi Whaley formula


# why we drop the last term in the Barone-Adesi Whaley formula












In this paper Efficient Analytic Approximation of American Option Values

in the first several lines of page 306, the author dropped the last term in equation 11, he explained that when $T\to 0$, we have $f_K\to 0$, but I don't know why.

Can anyone explain it?

$\epsilon$ is the early exercise premium, and $T$ is the time to maturity. Define $K(T):=1-e^{-rT}$

write $\epsilon$ as $\epsilon(S,K)=K(T)f(S,K)$

the equation 11 is $$S^2f_{SS}+NSf_S-(M/K)f-(1-K)Mf_K=0$$

## Answer by Quantuple (score 4, accepted)

https://quant.stackexchange.com/a/33122

Below is a hand-wavy way to reach the above result. I suspect there is a more elegant way to show it though.

The early exercise premium is defined as the difference between the American and European option prices $$ \epsilon(S,T) := C(S,T)-c(S,T) \tag{0}$$ In the paper it is further rewritten as $$ \epsilon(S,T) = K(T) f(S,K(T)) = \epsilon^*(S,K(T)) \tag{1} $$ for some function $K: T \to 1-e^{-rT}$ and where we have let $$\epsilon^* : (S,K) \to K f(S,K)$$

From the above expression $$ f(S,K) = \frac{\epsilon^*(S,K)}{K} $$ hence $$ f_K(S,K) = -\frac{\epsilon^*(S,K)}{K^2} + \frac{\epsilon^*_K(S,K)}{K} $$ We would like to compute $$ \lim_{T \to 0} f_K(S,K(T)) = \lim_{K \to 0} f_K(S,K) = \lim_{K \to 0} -\frac{\epsilon^*(S,K)}{K^2} + \frac{\epsilon^*_K(S,K)}{K} \tag{2} $$

To help us in our computations, we note that:

> As the time to maturity tends towards 0 an American option becomes strictly equivalent to its European counterpart (i.e. same price + same Greeks), hence we notably have: $$\lim_{T \to 0} \epsilon(S,T) = 0 \tag{A}$$ $$\lim_{T \to 0} \epsilon_T(S,T) = 0 \tag{B}$$ $$\lim_{T \to 0} \epsilon_{TT}(S,T) = 0 \tag{C}$$

Using (A) and noting that $$ \lim_{T \to 0} \epsilon(S,T) = 0 \iff \lim_{K \to 0} \epsilon^*(S,K) = 0 $$ we can rewrite (2) using l'Hôpital rule as

$$ \lim_{K \to 0} f_K(S,K) = \lim_{K \to 0} -\frac{\epsilon^*_K(S,K)}{2K} + \frac{\epsilon^*_K(S,K)}{K} = \lim_{K \to 0} \frac{\epsilon^*_K(S,K)}{2K} \tag{3} $$

From (1) $$\epsilon_T(S,T) = \epsilon^*_K K_T = \epsilon_K^*(S,K(T)) re^{-rT} $$ Using (B) along with the above equation yields $$ \lim_{T\to 0} \epsilon_T = 0 \iff \lim_{T \to 0} \epsilon_K^*(S,K(T)) r = 0 $$ hence $$ \lim_{K \to 0} \epsilon_K^*(S,K) = 0 \tag{4} $$ Plugging this into (2) and using l'Hôpital rule gives $$ \lim_{T \to 0} f_K(S,K(T)) = \lim_{K \to 0} \frac{\epsilon^*_{KK}(S,K)}{2} $$

From (1) $$ \epsilon_{TT}(S,T) = \epsilon^*_{KK} K_T^2 + \epsilon_K^* K_{TT} $$ Simarly to what we did earlier, using (C) along with (3) will lead to $$ \lim_{K \to 0} \epsilon^*_{KK} = 0 \tag{5} $$ and plugging that back into our expression for $\lim_{T\to 0} f_K$ finally yields $$ \lim_{T\to 0} f_K = \frac{1}{2} \lim_{K \to 0}\epsilon^*_{KK}(S,K) = 0$$