Showing Black–Scholes Call Value Converges to Payoff at Maturity
Summary
The document examines how a Black–Scholes call price approaches its terminal payoff as time to maturity shrinks to zero. It rewrites the price into a term involving the discounted intrinsic-value difference and a second term expressed as an integral of the normal density. Since the density is bounded and the integration interval contracts, that second contribution vanishes near maturity.
The remaining term is analyzed by comparing the terminal asset price with the strike. Above the strike, the normal cumulative probability tends to one and the call value approaches intrinsic value; below it, the probability tends to zero and the option value approaches zero. At the strike, the answer argues the limit is also zero. The reasoning relies on the Black–Scholes setup and continuous asset-price paths; it provides a pointwise argument rather than a broader treatment of continuity under other models or path behavior.
Key ideas
- The call price can be rewritten as a leading term plus an integral contribution that vanishes as maturity approaches.
- The shrinking integration interval and bounded normal density make the integral term tend to zero.
- Above the strike, the call value converges to the positive intrinsic payoff.
- Below the strike, the call value converges to zero.
- At the strike, the document concludes that the limiting call value is zero.
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Full text
# Continuity of Black-Scholes formula
# Continuity of Black-Scholes formula
How to proof B&S pricing formula is continuous in time $t$ (or it is not?).
The general pricing formula is $$ C_t = e^{-r(T-t)} \mathbb{E}^*[(S_T-K)^+ | \mathcal{F}_t] \hspace{1cm} 0\leq t\leq T $$ Then for time at maturity $t=T$ $$ C_T = \mathbb{E}^*[(S_T-K)^+ | \mathcal{F}_T] = (S_T-K)^+ $$ which is logic. For other anterior time $t<T$, the integration calculation give $$ C_t = S_t \mathcal{N}(d_+) - e^{-r(T-t)} K \mathcal{N}(d_-) $$ with $$ d\pm = \frac{\text{ln}\frac{S_t}{K} + (r\pm\frac{\sigma^2}{2})(T-t) }{\sigma\sqrt{T-t}} $$ Since $S_t$ is modelled as geometric brownian motion, it has to be continuous in $t$. I see that everything in the B&S formula is continuous in $t$. But I can not proof the continuity $$ C_t \rightarrow C_T \hspace{1cm} \text{when } t\rightarrow T $$
## Answer by David Durrleman (score 4, accepted)
https://quant.stackexchange.com/a/16131
First, let's remember that by property of $\mathcal{N}$ being a cumulative distribution function, we have that $$\lim_{x \to -\infty}\mathcal{N}(x) = 0$$ and $$\lim_{x \to \infty}\mathcal{N}(x) = 1$$.
Now, let's see how the Black-Scholes price behaves as times goes to maturity. First we have that
$$ d\pm = \left(\frac{\text{ln}\frac{S_t}{K}}{\sigma\sqrt{T-t}} + (r\pm\frac{\sigma^2}{2})\sqrt{T-t}\right) $$
and
$$C_t = \left(S_t-e^{-r(T-t)}K\right)\cdot\mathcal{N}(d+) + e^{-r(T-t)}K\cdot\left(\mathcal{N}(d+)-\mathcal{N}(d-)\right)$$
which can be further rewritten as
$$C_t = \left(S_t-e^{-r(T-t)}K\right)\cdot\mathcal{N}(d+) + \frac{e^{-r(T-t)}K}{\sqrt{2\pi}}\int_{d-}^{(d-)+\sigma^2\sqrt{T-t}} e^{-\frac{z^2}{2}}\, dz$$
First, noticing that $z \mapsto e^{-\frac{z^2}{2}}$ is bounded over $\mathbb{R}$, we can deduce that the second term goes to $0$ as $t$ goes to $T$ because the integration interval becomes arbitrarily small.
So the convergence of $C_t$ is determined by that of the first term which has different limits depending on $S_T/K$:
- if $S_T > K$, then $\ln\frac{S_T}{K} > 0$ and $\lim d+ = \infty$, so $\lim C_t = \left(S_T - K\right)\cdot\mathcal{N}(\infty) = S_T - K$
- if $S_T < K$, then $\ln\frac{S_T}{K} < 0$ and $\lim d+ = -\infty$, so $\lim C_t = \left(S_T - K\right)\cdot\mathcal{N}(-\infty) = 0$
- if $S_T = K$, then $S_t$ goes to $K$, and remembering that $\mathcal{N}$ is bounded by $0$ and $1$, we have $\lim C_t = 0$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.