# Black-Scholes公式的連續性

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 $t ## 最佳答案 首先，讓我們記住，通過$ \ mathcal {N} $作為累積分布函數的屬性，我們得到$$\ lim_ {x \ to - \ infty} \ mathcal {N}（x）= 0$$和$ $\ lim_ {x \ to \ infty} \ mathcal {N}（x）= 1 $$。 現在，讓我們看看Black-Scholes價格隨著時間的推移如何表現。首先，我們有$$ d \ pm = \ left（\ frac {\ text {ln} \ frac {S_t} {K}} {\ sigma \ sqrt {Tt}} +（r \ pm \ frac {\ sigma ^ 2} {2}） \ SQRT {Tt的} \右）  C_t = \ left（S_t-e ^ { - r（Tt）} K \ right）\ cdot \ mathcal {N}（d +）+ e ^ { - r（Tt）} K \ cdot \ left（\ mathcal {N}（d +） - \ mathcal {N}（d - ）\右）$$可以進一步改寫為$$ C_t = \ left（S_t-e ^ { - r（Tt）} K \ right）\ cdot \ mathcal {N}（d +）+ \ frac {e ^ { - r（Tt）} K} {\ sqrt {2 \ pi}} \ int_ {d - } ^ {（d - ）+ \ sigma ^ 2 \ sqrt {Tt}} e ^ { - \ frac {z ^ 2} {2}} \，dz$$首先，註意到$ z \ mapsto e ^ { - \ frac {z ^ 2} {2}} $限制在$ \ mathbb {R} $之上，我們可以推斷出第二個期限為$ 0  因為積分間隔變得任意小，所以轉到$T$。

• if $S_T > K$, then $\ln\frac{S_T}{K} > 0$ 和 $\lim d+ = \infty$, so $\lim C_t = \left(S_T - K\right)\cdot\mathcal{N}(\infty) = S_T - K$
• if $S_T < K$, then $\ln\frac{S_T}{K} < 0$ 和 $\lim d+ = -\infty$, so $\lim C_t = \left(S_T - K\right)\cdot\mathcal{N}(-\infty) = 0$
• if $S_T = K$, then $S_t$ goes to $K$, 和 remembering that $\mathcal{N}$ is bounded by $0$ 和 $1$, we have $\lim C_t = 0$