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

$S_t$ (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 $t<T$ $$0<S_t< Xe^{-r(T-t)}\frac{\mathrm{N}(d_2)}{\mathrm{N}(d_1)} $$

enter image description here

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    $\begingroup$ Even if you think that (1) is free to assume negative values it won't. Proof: it is the expectation of a nonnegative payoff. $\endgroup$
    – Kurt G.
    Commented Apr 24, 2023 at 16:58

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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 $t <T$.

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  • $\begingroup$ 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? $\endgroup$
    – ric.san
    Commented May 1, 2023 at 15:37
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    $\begingroup$ @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. $\endgroup$
    – RRL
    Commented May 1, 2023 at 16:02
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    $\begingroup$ ... and by $\phi$ in that formula I mean the standard normal probability density function. $$\phi(x) = N'(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$$ $\endgroup$
    – RRL
    Commented May 1, 2023 at 16:07
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    $\begingroup$ 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. $\endgroup$
    – RRL
    Commented May 1, 2023 at 16:57
  • $\begingroup$ Thanks, I was reading Hull's book on the subject and I found this same answer. $\endgroup$
    – ric.san
    Commented May 1, 2023 at 18:24

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