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I would expect that the Black Scholes model should always give a value for a call option, $c$, to be at least $0$. However, I am seeing some cases where that is not the case. Here is the Black-Scholes model for a call option.
\begin{eqnarray*} c &=& S_0 N(d_1) - Ke^{-rT}N(d_2) \\ d_1 &=& \frac{ \ln{ \frac{S_0}{K} } - (r + \frac{\sigma^2}{2})T }{\sigma \sqrt{T}} \\ d_2 &=& d_1 - \sigma \sqrt{T} \\ \end{eqnarray*} Now please consider the special case of $\sigma = 0.1$, $T = 1$, $r = 0$ and $K = 2S_0$. We have: \begin{eqnarray*} c &=& S_0 N(d_1) - Ke^{-(0)(1)}N(d_2) = S_0 N(d_1) - 2S_0N(d_2) \\ d_1 &=& \frac{ \ln{( \frac{S_0}{2S_0} )} - (0 + \frac{\sigma^2}{2})(1) }{\sigma \sqrt{1}} \\ d_1 &=& \frac{ \ln{( \frac{1}{2} )} - (0 + \frac{\sigma^2}{2})(1) }{ 0.1 } \\ d_1 &=& 10 \ln{( \frac{1}{2} )} - 10\Big( \frac{.01}{2} \Big) \\ d_1 &=& 10 \ln{( \frac{1}{2} )} - \frac{.1}{2} \\ d_1 &=& -6.9814718 \\ N(d_1) &=& 0.000000000001461 \\ d_2 &=& -6.9814718 - 0.01 \sqrt{1} = -6.9914718 \\ N(d_2) &=& 0.000000000001360 \\ c &=& S_0 (0.000000000001461) - 2S_0 (0.000000000001360) \\ c &=& (-1.259E-12) S_0 \\ \end{eqnarray*} Why am I getting a negative number? Is it round off error?
Thanks,
Bob

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Your $d_1$ is wrong - the minus sign is wrong. It should be: $$ d_1 = \frac{\ln(S_0/K) + (r+\sigma^2/2)/T}{\sigma \sqrt{T}}. $$ See e.g. here.

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    $\begingroup$ re: Your d1 is wrong - the minus sign is wrong. It should be: d1=(ln(S0/K)+σ2/2)/σ. Where is the T factor? Isn't the equation $$\frac{\ln(S_0/K)+(r+\sigma^2/2)T}{\sigma\sqrt T}$$ $\endgroup$
    – David
    Jun 5, 2023 at 13:50
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    $\begingroup$ @David I think you're right, please provide an edit with the proper $\LaTeX$ formatting. Of course, in this example we have $T=1$ and it doesn't matter but it might be confusing to readers. $\endgroup$
    – Bob Jansen
    Jun 5, 2023 at 14:26
  • $\begingroup$ Thanks @David I edited my answer so that it is fully clear., $\endgroup$
    – Richi Wa
    Jun 15, 2023 at 6:35

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