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 $$ C_t \rightarrow C_T \hspace{1cm} \text{when } t\rightarrow T $$


1 Answer 1


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

$$ d\pm = \left(\frac{\text{ln}\frac{S_t}{K}}{\sigma\sqrt{T-t}} + (r\pm\frac{\sigma^2}{2})\sqrt{T-t}\right) $$


$$C_t = \left(S_t-e^{-r(T-t)}K\right)\cdot\mathcal{N}(d+) + e^{-r(T-t)}K\cdot\left(\mathcal{N}(d+)-\mathcal{N}(d-)\right)$$

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 $\lim C_t = 0$
  • $\begingroup$ I largely rewrote my proof because the first version wasn't very rigorous (especially in the case where $S_T = K$). I hope the improved, correct version is still clear enough. $\endgroup$ Jan 8, 2015 at 18:51

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