European vs American derivative securities, interesting question

Question

Let us denote by $c^A(t, S(t))$ the price, at time $t$ of a certain American-style derivative security, whose instrinsic value, at time $t$ is denoted by $V(t)$.From the no-arbitrage principle, we know that, at each time $t\in [0,T]$, we must have $c^A(t, S(t))\geq V(t)$.

Let us now denote by $c^E(t, S(t))$ the price of the European counterpart of the same derivative, having payoff $V(T)$ exercisable only at time $T$. In this case we know that the relation $c^E(t, S(t))\geq V(t)$, for $t\in [0,T]$ in general does not hold anymore: for example if the European derivative security is a put option.

Now, my question is: suppose that, for a given European security, we actually have the relation $c^E(t, S(t))\geq V(t)$, for all $t\in [0,T]$. Can we say that in this case the American derivative security is equivalent to its European counterpart? That is, can we conclude that $c^E(t, S(t))=c^A(t, S(t))$, for all $t\in [0,T]$?

This is certainly the case for call options, where we have indeed that $c^E(t, S(t))\geq V(t)$, $t\in [0,T]$. But what about for other types of derivatives?

Answer 1

Consider two options with maturity $T$ that only differ in their exercise styles, one being European (holder can only exercise at $T $), the other American (holder exercises when it's best for him/her). These options need not necessarily be vanilla options.

Let us further denote by $I (S_t) $ the intrinsic value of these contigent claims at time $t $, i.e. the value that the holder would get by exercising at $t$.

Your question then translates to

Assuming that $\forall t \in [0,T]$, the following inequality holds $$ V^E(t,S_t) \geq I(S_t) \tag{A} $$ does that imply the following equality $$ V^A(t,S_t) = V^E(t,S_t) \tag{B} $$

Proof $(B) \Rightarrow (A)$

The proof is straightforward. Indeed, as you've stated in your question the $t$-value of an American option is always greater than the intrinsic value at time $t$: $V^A(t,S_t) \geq I(S_t)$ while by $(B)$ $V^A(t,S_t)=V^E(t,S_t)$. The former inequality comes from the fact that immediate exercise is merely one of the many stopping strategies that the holder of an American option could resolve to and he/she is expected to pick the one which maximises his/her gains.

Proof $(A) \Rightarrow (B)$

Starting from the definition of the American option \begin{align} V^A(t,S_t) &= \text{sup}_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} I(S_\tau) \right] \tag{1} \end{align} where $\tau$ represents a family of stopping times with values in $[t,T]$.

Assume that $(A)$ holds i.e. $I(S_t) \leq V^E(t,S_t), \forall t \in [0,T]$. In that case from equation $(1)$, we can write that \begin{align} V^A(t,S_t) &\leq \text{sup}_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} V^E(\tau,S_\tau) \right] \tag{2} \\ \end{align} by linearity of the expectation operator.

Now, noting that in the absence of arbitrage $\frac{V^E(t,S_t)}{B_t}$ should emerge as a $\mathbb{Q}$-martingale (remember that $V^E(t,S_t)$ is a tradable asset), with $B_t$ the $t$-value of the risk-free money market account, the optimal sampling theorem gives: \begin{align} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} V^E(\tau,S_\tau) \right] &= e^{rt} \mathbb{E}_t^\mathbb{Q} \left[ \frac{V^E(\tau,S_\tau)}{B_\tau} \right] \\ &= e^{rt}\frac{V^E(t,S_t)}{B_t} \\ &= V^E(t,S_t) \end{align} hence $(2)$ becomes \begin{align} V^A(t,S_t) &\leq \text{sup}_{\tau \in [t,T]} V^E(t,S_t) = V^E(t,S_t) \tag{I1} \\ \end{align}

On the other hand we have that $$ V^A(t,S_t) \geq V^E(t,S_t) \tag{I2} $$ i.e. the $t$-value of an American option is always greater than the $t$-value of its European counterpart. This is because European exercise at expiry is merely one of the many stopping strategies that the holder of an American option could resolve to and he/she is expected to pick the one which maximises his/her gains.

Combining inequalities $(I1)$ and $(I2)$ then trivially yields: $$ V^A(t,S_t) = V^E(t,S_t) $$

Answer 2

The European call price converges to zero as the underlying price converges to zero, reflecting the vanishing probability of exercise.At the other extreme, the European call price converges to the discounted value of the difference between the asset price and the strike.For intuition note that the probability of exercise converges to one as $S_t$ becomes large and therefore that the call value converges to the present value of $S_T - K$. As the call price is non-negative and the call payoff dominates $S_T - K$, we also have the lower bound $$\color{green}{C_E(S,t)\ge (S_t e^{-q\tau}-K e^{-r\tau})^+}$$

Now let we Consider an American-style call option with exercise price $K > 0$ and maturity date $T$, written on an underlying asset whose price $S$ satisfies the stochastic differential equation (under the risk neutral measure) $$dS_t=(r-q)S_tdt+\sigma S_t dW_t\mathbb{^{Q}}$$Here $r$, $q$ and $\sigma$ are constant parameters: the price is a geometric Brownian motion (GBM) process. As exercise cannot be optimal when $S<K$ it has become customary to write the option payoff in the form $V=(S-K)^+$ Our first objective is to characterize the structure of the exercise region and its boundary. Given that the environment is Markovian, the state of nature is completely described by the asset price S and the pair $(S, t)$ contains all the information required for pricing and decision-making purposes. Let $C (S, t)$ be the option price at the point $(S, t)$. The immediate exercise region, denoted by $D$, is the set of pairs $(S, t)$ at which immediate exercise is an optimal policy.That is $$\color{green}{D=\{(S,t)\in\mathbb{R}^+\times[0,T]: C(S,t)=(S-K)^+\}}$$ Its complement $$\color{green}{ D^c=\{(S,t)\in\mathbb{R}^+\times[0,T]: C(S,t)>(S-K)^+\}}$$ is the continuation region, i.e., the set of prices-dates at which immediate exercise is sub-optimal. Now we describes elementary properties of the exercise region. We can say

• $C_A(S,t)$ is continuous on $\mathbb{R}^+\times[0,T]$.
• $C_A(.,t)$ is non-decreasing and convex on $[0,T]$ for all $t\in[0,T]$.
• $C_A (S,.)$ is non-increasing on $[0, T]$ for all $S \in \mathbb{R}^+$.
• $0\le \frac{\partial C(S,t)}{\partial S}\le 1$ on $\mathbb{R}^+\times[0,T]$
• $\frac{\partial C(S,t)}{\partial S}= 1$ for every $(S,t)\in D^c$ kim(1990), carr, Jarrow and Myneni (1992) show for $t\in[0,T]$ $$\color{green}{C_A(S,t)=C_E(S,t)+\pi(S,t,B(.))}$$ where $\pi(S,t,B(.))$ is the early exercise premium is given by $$\pi=\int_{t}^{T}(q S_t e^{-\delta(s-t)}N(d_1(S_t,B_s,s-t))-rKe^{-r(s-t)}N(d_2(S_t,B_s,s-t))ds$$ such that $$d_1=\frac{\ln\left(\frac{S_t}{B_s}\right)+(r-q+\frac{1}{2}\sigma^2)(t-s)}{\sigma\sqrt{s-t}}$$ $$d_2=d_1-\sqrt{s-t}$$ The immediate exercise boundary $B$ solves the recursive non-linear integral equation $$B_t-K=C_E(B_t,t)+\pi(B_t,t,B(.))$$ for $t \in [0, T)$, subject to the boundary condition $$B_{T^{-}}=\max\{K,\frac{r}{q}K\}$$ and $$B_T=K\le B_{T^{-}}.$$

Finally because American options offer more exercise opportunities they therefore have higher values than the European options. In the absence of dividends($q=0$) the American call price is the same as the European one, since the optimal exercise policy for such an option is not to exercise.