• Doing last step -- un-change of variable, where in my case I have $$k = -\frac{2r}{\sigma^{2}},$$ $$v(\tau, x) = u(\tau, x) \cdot \exp\left(-\frac{1}{4}(k+1)^{2} \tau - \frac{1}{2}(k-1)x\right),$$ $$x = \ln\left(\frac{S}{F}\right),$$ $$\tau = \frac{\sigma^{2}}{2}(T-t),$$ $$C = F \cdot v(\tau,x),$$ getting $$C(t,S) = S \Phi(z_{1}) - F e^{-r(T-t)} \Phi(z_{2}),$$ where $$z_{1} = \frac{\ln(\frac{S}{F}) + (\color{red}{-r} + \frac{1}{2}\sigma^{2})(T-t)}{\sigma \sqrt{T-t}}$$ and $$z_{2} = \frac{\ln(\frac{S}{F}) + (\color{red}{-r} - \frac{1}{2}\sigma^{2})(T-t)}{\sigma \sqrt{T-t}}.$$

I can't figure out why I am getting a $\color{Red}{\text{negative rate}}$; any suggestions or hunches are much appreciated.

  • 3
    $\begingroup$ It is hard to understand your question. For example, from where your un-change of variables should be applied. Do not assume everyone have your context. $\endgroup$ – Gordon Apr 16 at 18:05

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