European Call Option With Infinite Maturity

Worked Solution

How to Think About It: Don't reach for the Black-Scholes formula first — probe the *limit* with economics. A call is the right to buy at $K$ forever. Two forces race as $T\to\infty$: (i) the present value of the strike payment, $Ke^{-rT}$, vanishes because $r>0$ — the strike becomes free; (ii) the *deliverable* stock is worth $Se^{-qT}$ today, because a call holder receives price appreciation but not the dividends the stock leaks out at rate $q$. The whole answer hinges on which of these two survives. This is a limit/extreme-case probe: replace the messy formula with 'what does each term do at the boundary?' The naive trap is to say 'longer maturity = more optionality = more value' and expect the call to keep rising — false once dividends bleed the underlying dry.

Quick Estimate: Use $C = Se^{-qT}\Phi(d_1) - Ke^{-rT}\Phi(d_2)$ and price each piece as $T$ grows. - The drift inside $d_1$ is $(r-q+\sigma^2/2)T$, which grows like $T$ while the denominator grows like $\sqrt T$, so $d_1\to+\infty$ whenever $r-q+\sigma^2/2>0$ — true here — giving $\Phi(d_1)\to 1$ and likewise $\Phi(d_2)\to 1$. So both $\Phi$ factors saturate to $1$; the limit is decided entirely by the two exponential prefactors. - No dividends ($q=0$): prefactor on the stock is $Se^{0}=S$ (doesn't decay), prefactor on the strike is $Ke^{-rT}\to 0$. So $C\to S\cdot 1 - 0 = S$. Sanity plug: $S=100,K=90,r=3\%,\sigma=20\%$ gives $C\approx 96.3$ at $T=100$ — already $96\%$ of the way to $S=100$, and $\approx 100.0$ by $T=500$. Lands on $S$. - With dividends ($q>0$): now the stock prefactor $Se^{-qT}\to 0$ *and* $Ke^{-rT}\to 0$ — both legs decay, so $C\to 0$. Plug $q=2\%$: the same option is $\approx 27$ at $T=10$ but has collapsed to $\approx 11$ at $T=100$ and $\approx 0$ by $T=500$. It rises then falls — the dividend drag eventually wins.

Approach: Take $T\to\infty$ in the (dividend-adjusted) Black-Scholes formula, tracking each $\Phi$ factor and each discount prefactor separately.

Formal Solution:

(a) No dividends ($q=0$). The price is $$C = S\,\Phi(d_1) - Ke^{-rT}\,\Phi(d_2),\qquad d_{1}=\frac{\ln(S/K)+(r+\sigma^2/2)T}{\sigma\sqrt T},\ \ d_2=d_1-\sigma\sqrt T.$$ As $T\to\infty$: the numerator of $d_1$ grows $\propto T$, denominator $\propto\sqrt T$, so $d_1\to+\infty$ and $\Phi(d_1)\to 1$. Also $d_2=d_1-\sigma\sqrt T$; since $d_1$ grows like $\sigma\sqrt T\cdot(\tfrac{r+\sigma^2/2}{\sigma^2})\sqrt T \to$ faster than $\sigma\sqrt T$, $d_2\to+\infty$ too, so $\Phi(d_2)\to 1$. And $Ke^{-rT}\to 0$ since $r>0$. Hence $$\boxed{\lim_{T\to\infty}C = S\cdot 1 - 0\cdot 1 = S.}$$ Economically: over an infinite horizon the strike is free (its PV is zero) and the stock is almost surely far in the money, so the call *is* the stock.

(b) Continuous dividend yield $q>0$. Now $$C = Se^{-qT}\,\Phi(d_1) - Ke^{-rT}\,\Phi(d_2),\qquad d_1=\frac{\ln(S/K)+(r-q+\sigma^2/2)T}{\sigma\sqrt T}.$$ Both $\Phi$ factors still tend to a finite limit in $[0,1]$, but the prefactors both vanish: $Se^{-qT}\to 0$ and $Ke^{-rT}\to 0$. Therefore $$\boxed{\lim_{T\to\infty}C = 0.}$$ The forgone dividends, in present value, accumulate to the entire stock price over an infinite horizon, wiping out the option. (This is exactly why American calls on dividend payers can be worth early-exercising — the European version is heading to zero.)

Answer: With no dividends, $C\to S$. With continuous dividend yield $q>0$, $C\to 0$.