On the plot of Black-Scholes-Merton formula

The price $$C(t,S_t)$$ of a European call option is given by the famous Black-Scholes formula $$$$C(t,S_t)=S_{{t}}{\mathrm{N}}(d_{{1}})-Xe^{{-r(T-t)}}\mathrm{N}(d_{{2}})\tag{1}$$$$

$$S_t$$ (underlying asset price) is a positive stochastic process following a geometric Brownian motion, $$X$$ is a positive constant (strike price), while $$\mathrm{N}(x)$$ is the cumulative distribution function of the standard normal. $$d_2=\frac{\ln{\frac{S_t}{X}}+\left(r-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}=d_1-\sigma\sqrt{T-t}$$

Question. Is this expression supposed to match the following plot? My gut response would be no since $$(1)$$ is free to assume negative values as well; for instance, if at a certain $$t $$0

• Even if you think that (1) is free to assume negative values it won't. Proof: it is the expectation of a nonnegative payoff. Apr 24 at 16:58

The Black-Scholes call option price is the discounted expected value of the payoff $$\max(S_T-X,0$$) under the risk-neutral probability measure. The existence of such a measure is guaranteed with a market model (such as Black-Scholes) that is arbitrage-free, and a negative option price represents an (impossible) arbitrage opportunity.

However, even if you knew nothing about arbitrage pricing theory you could prove that $$C(t,S_t) > 0$$ for all $$S_t >0$$ and for all $$0 \leqslant t < T$$ directly from the formula. The inequality

$$S_t < Xe^{-r(T-t)}\frac{N(d_1)}{N(d_2)},$$

which ostensibly yields a negative option price can never hold. Note that the ratio $$N(d_1)/N(d_2)$$ is not a constant but rather depends nonlinearly on the ratio $$S_t/Xe^{-r(T-t)}$$.

Consider the price $$C(t,S) = SN(d_1) - Xe^{-r(T-t)}N(d_2)$$ treating $$S_t = S$$ as a fixed parameter. We have

$$\lim_{t \to T}d_1 = \lim_{t \to T}d_2 = \begin{cases}+\infty, & S> X\\0,&S = X\\-\infty, &S < X\end{cases}, \quad\lim_{t \to T}N(d_1) = \lim_{t \to T}N(d_2) = \begin{cases}1, & S> X\\\frac{1}{2},&S = X\\0, &S < X\end{cases}$$

and, thus, $$\lim_{t\to T}C(t,S) =\max(S-X,0) \geqslant 0$$.

Taking the partial derivative of the option price with respect to $$t$$, we find

$$\frac{\partial C}{\partial t}(t,S) = S_t N'(d_1) \frac{\partial d_1}{\partial t} - Xe^{-r(T-t)}N'(d_2)\frac{\partial d_2}{\partial t} - rXe^{-r(T-t)}N(d_2)\\= -\frac{S\sigma \phi(d_1)}{2\sqrt{T-t}}-rXe^{-r(T-t)}N(d_2)<0$$

Since the partial derivative is strictly less than $$0$$, the option price is a decreasing function of time $$t$$ for each fixed value of $$S$$.

Hence, the option price decreases to the limit as $$t$$ increases to $$T$$, that is

$$C(t,S) \downarrow \max(S-X,0) \geqslant 0 \, \text{as} \,\, t\to T,$$

and it follows that $$C(t,S) > 0$$ strictly for all $$t .

• I'm sorry but.. how did you get $\frac{S\sigma\phi(d_1)}{2\sqrt{T-t}}$ from $S_tN'(d_1)\partial_td_1-Xe^{-r(T-t)}N'(d_2)\partial_t d_2$ so easy? May 1 at 15:37
• @ric.san: It is not easy but the derivative of option price with respect to time is the well-known parameter called Theta. See here. What matters is that it is negative for a standard option. Option values decay (all other parameters held constant) as time advances towards the expiration date.
– RRL
May 1 at 16:02
• ... and by $\phi$ in that formula I mean the standard normal probability density function. $$\phi(x) = N'(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$
– RRL
May 1 at 16:07
• Note that $d_2 = d_1 - \sigma\sqrt{T-t}$ and $N'(d_2) = \frac{1}{\sqrt{2\pi}}e^{-d_1^2/2}e^{d_1\sigma\sqrt{T-t}}e^{-\sigma^2(T-t)/2}$. We also have $e^{d_1\sigma\sqrt{T-t}} = e^{\log \frac{S}{X}+ (r+\sigma^2/2)(T-t)} = \frac{S}{X} e^{r(T-t)}e^{\sigma^2(T-t)/2}$ and, hence, $Xe^{-r(T-t)}N'(d_2) = S\phi(d_1)$. That will simplify things greatly.
– RRL
May 1 at 16:57
• Thanks, I was reading Hull's book on the subject and I found this same answer. May 1 at 18:24