Why an American Call Can Have Higher Delta Than a European Call

# Comparison of the American and European call deltas


# Comparison of the American and European call deltas












Suppose the interest rate is zero. A stock with price $S(t)$ at time $t$ pays only one dividend at time $t_1$ such that $S(s_+)=S(t_1^-)q$ where $q\in[0,1]$ is a constant. Consider a European call and an American call with the same strike $K$ and expiry $T>t_1>0$ at time $0$. Define $$C_{\text{E}}(S_t,t):=\mathbf E\big[(S(T)-K)_+\big|S(t)=S_t\big]$$ Then $$C_{\text{E}}(S_0,0)=\mathbf E\big[(S(T)-K)_+\big|S(0)=S_0\big]=\mathbf E\big[C_{\text{E}}(S_{t_1^-},t)\big|S(0)=S_0\big].$$ The only time this American option may exercise early is time $s$. So $$C_{\text{A}}(S_0,0)=\mathbf E\big[\max\left(S(t_1^-)-K,\,C_{\text E}(S_{t_1^-},t_1^-)\right)\big|S(0)=S_0\big].$$

Is the following comparison of the American and the European call deltas true? $$\frac{\partial}{\partial S}C_{\text{A}}(S,0)\ge\frac{\partial}{\partial S}C_{\text{E}}(S,0).$$

## Answer by Hans (score 3)

https://quant.stackexchange.com/a/75026

This is true, $\frac{\partial}{\partial S}C_\text A(S,0)\in [0,1]$ and that the difference is bound above.

Proof: Suppose $S\ge 0$ follows a local volatility process $$dS(t) = \sigma(S(t),t)S(t)\,dB(t)$$ where $B(t)$ is the standard Brownian motion. We can extend this dynamics later to account for a stochastic volatility process.

Because $C_\text E(S,t_1^-)=C_\text E(qS,t_1^+)$, $\frac{\partial}{\partial S}C_\text E(S,t_1^-)=q\frac{\partial}{\partial x}C_\text E(x=qS,t_1^+)<q<1,\,\forall t<T$, $S-K=C_{\text E}(S,t_1^-)$ has a unique solution for $S$. We denote it by $S_c$. Because the value of a call, either American or European, such as $C_\text E(S,t)$, is convext with respect to stock $S$, $\min_{S\ge S_c}\frac{\partial}{\partial S}C_\text E(S,t_1^-)=\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-)$.

Let $\Theta$ be the Heaviside step function. At time $t<s$, both calls, and their difference $C=C_{\text A}-C_{\text E}$ due to the linearity thereof, satisfy $$\bigg[\frac{\partial}{\partial t}+\frac12(\sigma S)^2\frac{\partial^2}{\partial S^2}\bigg]C(S,t)=0,$$ $$C(S,t=t_1^-)=\big(S-K-C_\text E(S,t_1)\big)\Theta(S-S_c).$$ Take partial derivative of the above partial differential equation over $S$ and $C_1:=\frac{\partial}{\partial S}C$. We have $$\bigg[\frac{\partial}{\partial t}+uS\frac{\partial}{\partial S} +\frac12(\sigma S)^2\frac{\partial^2}{\partial S^2}\bigg]C_1(S,t)=0, \tag1\label{1}$$ $$C_1(S,t=t_1^-)=\Big(1-\frac{\partial}{\partial S}C_\text E(S,t_1)\Big)\Theta(S-S_c),$$ where $u:=\frac1{2S}\frac{\partial}{\partial S}(\sigma S)^2$. It can be shown by repeating the same methodology here that $$0\le C_1(S,t_1^-)\le1-\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-).$$ By the strong maximum principle of elliptical and parabolic partial differential equation (c.f. also the Wikipedia article), $$0=\inf_S C_1(S,t_1^-)\le C_1(S,t)\le\sup_S(S,t_1^-)=1-\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-), \quad\forall t<t_1. \tag2\label2$$

An alternative to the last step is the Feynman-Kac formula.

The solution to Equation $\eqref1$ is $$C_1(S,t)=\mathbf E\big[C_1(S(t_1^-),t_1^-)\big|S(t_1^-)=S\big]$$ under the probability measure such that $S(t)$ is a new Ito process $$dS=uSdt+\sigma S\,dB.$$ Equation $\eqref2$ is then obvious.

To recapitulate, we have proved -- more than what is sought by the question -- that $$0\le\frac{\partial}{\partial S}\big(C_{\text{A}}(S,t)-C_{\text{E}}(S,t)\big)\le1-\frac{\partial}{\partial S}C_\text E(S=S_c,t_1^-), \quad\forall t<t_1,$$ which is the desired result for the difference.

Apply the same method, we can show \begin{align*} & C_\text A(S,T) = C_\text E(S,T) = (S-K)_+ \\ \implies & \frac{\partial}{\partial S}C_\text A(S,T) = \frac{\partial}{\partial S}C_\text E(S,T) = \Theta(S-K)\in [0,1] \\ \implies & \frac{\partial}{\partial S}C_\text A(S,t_1^+) = \frac{\partial}{\partial S}C_\text E(S,t_1^+) \in [0,1] \end{align*} and thus $$\frac{\partial}{\partial S}C_\text A(S,t)\in [0,1], \quad\forall t<t_1.$$

$$\tag*{$\square$}$$