# 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? $$\frac{\partial}{\partial S}C_{\text{A}}(S,0)\ge\frac{\partial}{\partial S}C_{\text{E}}(S,0).$$

• Never really thought about this (good question), but how about this: if you 'Europeanize' an American option I suspect its IVs will be higher than that of the European. Hence, the difference between the deltas will depend on the value of vanna. I can see many holes with this shortcut, so not saying at all that it is right. Just a thought. Mar 24 at 9:03
• @Frido: Well, "Europeanization", and thus the corresponding IV, is not a well defined concept. It is a somewhat heuristic. I actually have a proof. I just need some time to write it up.
– Hans
Mar 24 at 9:26
• Yes it is heuristic. Once you have written up your proof, it would be great to post it here as an answer. Mar 24 at 9:31
• @Frido: That is my plan. :-) The proof can be generalized to the case with continuous time dividend.
– Hans
Mar 24 at 9:35
• @Frido: Have you seen my proof below? It is not the most tidily formatted yet. I will revamp it later.
– Hans
Mar 28 at 20:17

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 $$C_\text E(S,t_1^-)=C_\text E(qS,t_1^+)$$, $$\frac{\partial}{\partial S}C_\text E(S,t_1^-)=q\frac{\partial}{\partial x}C_\text E(x=qS,t_1^+), $$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 $$t, both calls, and their difference $$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

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 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

$$\tag*{\square}$$