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An American call option permits buying the underlying at strike price $X$ at any chosen stopping time from the current date $t$ through maturity $T$. Under standard complete-market pricing and integrability hypotheses,The conditional expectation uses information available at time $t$, and the optimization allows only future exercise times. A European call option restricts exercise to $T$. The no early exercise of a call without dividends result applies with nonnegative interest under its stated assumptions; dividends or negative rates can change that conclusion. The discounted value is the Snell envelope of discounted exercise rewards, as in American option.

An American option permits exercise at a chosen stopping time no later than maturity. Let $G_t$ be its exercise financial payoff and $B_t>0$ the bank account. Under a specified complete-market pricing model and the usual optimal-stopping integrability conditions, its value isThe Snell envelope of the discounted exercise process is $V_t/B_t$, not the unscaled currency price $V_t$. The distinction matters away from a date at which the bank account is normalized to one. At each exercise opportunity, one compares immediate payoff with the best conditional continuation value. A European option allows only the terminal stopping time, whereas an American call option or put can have a genuine early-exercise region depending on dividends, rates and payoff shape.

A financial option granting the right to buy its underlying financial asset at a specified strike price $K$. Its exercise payoff on one stock share is the positive part $(S-K{)}^{+}$. A European call option permits exercise at expiry, while an American call option also permits earlier exercise. A forward-start call option determines the strike at a future fixing date.

For the perpetual American call option model and a fixed exercise threshold $B>K$, this is the discounted expectation of $B-K$ at its first hit, with zero payoff if the hit never occurs. The stationary ordinary differential equation and boundary conditions $C_B(0)=0$, $C_B(B)=B-K$ prove the formula. At fixed $S<B$, differentiation gives ${\partial }_B{C}_B=S^p{B}^{-p-1}[pK-(p-1)B]$, so the unique maximizing threshold is $pK/(p-1)$. If a larger threshold is imposed, $C_B^{\prime }(B-)>1$ and its value lies below immediate exercise payoff just below $B$. This is allowed for a committed policy, but violates the exercise lower bound if quoted for an unrestricted American call option.

A European call option has nonnegative financial payoff, so its price is nonnegative under no-arbitrage. A second arbitrage comparison buys one call and a zero-coupon bond paying $K$, and shorts one share. Its terminal value is $(S_T-K{)}^{+}+K-S_T=(K-S_T{)}^{+}\ge 0$, so its initial cost cannot be negative. ConsequentlyThe holder of an American call option can choose to wait until expiry, so $C_t^a\ge C_t^e$. Since $r>0$ and $t<T$, an in-the-money call has $C_t^a\ge S_t-K{e}^{-r(T-t)}>S_t-K$: immediate exercise sacrifices value. An out-of-the-money exercise pays zero and cannot improve the holder's position. This proves no early exercise of a call without dividends: waiting is always at least as good, and is strictly better for positive immediate exercise payoff. In a degenerate market with zero continuation value an out-of-the-money holder can be indifferent; the Black-Scholes model with positive volatility gives strictly positive continuation value before expiry.

An American option allows exercise at a stopping time $\tau \in [t,T]$. For its exercise financial payoff $H$, the pricing problem isThe discounted value is the Snell envelope of discounted exercise payoffs. Define the pricing operatorIn the continuation region the Black-Scholes equation holds, $\mathcal{L}V=0$. In the exercise region $V=H$, while the ability to wait and the supermartingale property give $\mathcal{L}V\le 0$. Together these conditions are the obstacle problemThus an unknown exercise boundary must be found alongside the value. At a regular boundary in the nondegenerate Itô diffusion, value matching is $V=H$, and smooth pasting is $V_S=H^{\prime }$ where the financial payoff is differentiable. A finite difference method or a binomial options pricing model steps backward taking the larger of continuation and immediate exercise values at each node.

For a non-dividend-paying American call option with $r\ge 0$, no early exercise of a call without dividends follows directly from the European lower boundAt every candidate exercise time, keeping the European call has value at least the immediate exercise financial payoff, so early exercise cannot improve the value; the American and European prices coincide. The first inequality follows from Jensen inequality or put-call parity and nonnegative put value. With $r>0$ an American put option can benefit from early receipt of its strike at sufficiently low stock prices, so its exercise boundary is generally nontrivial and its value is at least the European put value. Dividends can make call exercise worthwhile, and negative interest invalidates the stated no-early-exercise call argument.

An American call option with no expiry. For geometric Brownian motion with drift $r-y$ under a risk-neutral measure, positive dividend yield $y$, $r>0$, and volatility $\sigma >0$, let $p>1$ solve ${\sigma }^{2}p(p-1)/2+(r-y)p-r=0$. Its value is $(S_{\ast }-K)(S/S_{\ast }{)}^p$ below $S_{\ast }=pK/(p-1)$ and $S-K$ above. The power solves the stationary Black-Scholes equation with continuous stock dividends; value matching and smooth fit determine the threshold.

The candidate dominates the positive part $(S-K{)}^{+}$ because its continuation branch is convex and tangent to $S-K$ at $S_{\ast }$. In the exercise region its discounted-generator residual is $rK-yS\le 0$: the root equation gives $y{S}_{\ast }=K[r+{\sigma }^{2}p/2]>rK$. Applying Itô's formula to the discounted value, localized if necessary, gives an upper bound on every exercise policy. Stopping at the first hit of $S_{\ast }$ attains it; on paths not yet hitting, the bounded discounted continuation value tends to zero. This supplies the optimal stopping verification. As $y\downarrow 0$, $p\downarrow 1$, the threshold diverges and the value tends to $S$.