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Correcting the Black–Scholes Call Formula Sign Error

Article Quant Q&A · Author: Bob
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Summary

The document examines a calculated negative value for a deep out-of-the-money call under the Black–Scholes model. The worked example substitutes zero interest rates, positive volatility, a one-year term, and a strike twice the initial underlying price into the call formula. The resulting tiny negative value raises the question of whether numerical rounding is responsible.

The accepted answer identifies the source as an algebraic sign error in the displayed expression for d1, rather than a failure of the option-pricing model. The example therefore illustrates why formula transcription should be checked before attributing an implausible price to floating-point precision. The response is brief and gives no broader numerical-analysis treatment, implementation guidance, or discussion of assumptions behind Black–Scholes; its lesson is limited to correcting this formula setup.

Key ideas

  • A negative call value in the shown example results from an incorrect sign in the d1 expression.
  • Check the transcribed pricing formula before blaming numerical rounding for an implausible result.
  • The worked example uses a deep out-of-the-money call with zero rates and positive volatility.
  • The brief correction does not address broader implementation or model-assumption issues.

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OptionsDerivatives pricing
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# Black-Scholes formula producing a negative number for a Call Option


# Black-Scholes formula producing a negative number for a Call Option












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

## Answer by Richi Wa (score 10, accepted)

https://quant.stackexchange.com/a/37651

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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