I seen two variations of the Black-Scholes PDE with either $+{\frac {\partial V}{\partial t}}$ or $-{\frac {\partial V}{\partial t}}$, and wanted to ask why that is?

a) https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_equation#Solving_the_PDE $${\frac {\partial V}{\partial t}}+{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+rS{\frac {\partial V}{\partial S}}-rV=0$$

b) https://www.quantstart.com/articles/C-Explicit-Euler-Finite-Difference-Method-for-Black-Scholes $$-{\frac {\partial V}{\partial t}}+{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}+rS{\frac {\partial V}{\partial S}}-rV=0$$

Please see below derivation of Black-Scholes PDE:

\begin{align*} \Pi &= C(S,t)+\Delta S\\ d\Pi &=dC(S,t)+\Delta dS\\ &= C_S dS+C_t dt + \frac{1}{2}C_{SS}d[S]+ \frac{1}{2}C_{tt}d[t]+ C_{St}d[S,t]+\Delta dS\\ &= C_S(S\mu dt+S\sigma dW)+C_tdt+\frac{1}{2}C_{SS}S^2\sigma^2dt+\Delta (S\mu dt+S\sigma dW)\\ &= (\Delta S\sigma+C_SS\sigma)dW+(C_SS\mu+C_t+\frac{1}{2}C_{SS}S^2\sigma^2+\Delta S\mu)dt \end{align*} Hedge: \begin{align*} d\Pi&\stackrel{!}{=}\Pi rdt\Rightarrow \Delta = -C_S:\\ &\Pi rdt= (C_t+\frac{1}{2}C_{SS}S^2\sigma^2)dt\\ &\Leftrightarrow(C-C_S S) rdt= (C_t+\frac{1}{2}C_{SS}S^2\sigma^2)dt \end{align*} $$\Rightarrow C_t+\frac{1}{2}C_{SS}S^2\sigma^2-r(C-C_S S)= 0$$

  • 1
    $\begingroup$ It depends whether you measure t forward (starts at 0 and increases to T) or backwards (time to maturity dwindles down to zero as maturity approaches). In the former case you have a minus sign in front of DV/dt. $\endgroup$ – Alex C Dec 8 '19 at 3:31
  • $\begingroup$ @AlexC I added a derivation of the PDE. Please show where the sign change would be observed? $\endgroup$ – emcor Dec 8 '19 at 3:41

Your derivation is right and as Alex said, the only difference between the equations is how you measure time. There are two possibilities

  • Time going forward, $t\in[0,T]$,
  • Time going backwards, $\tau\in[T,0]$, i.e. $\tau=T-t$.

And clearly, after the change of variables, $\frac{\partial V}{\partial t}=-\frac{\partial V}{\partial\tau}$. Amongst others, this change of variables is employed to transform the Black-Scholes PDE into the heat (diffusion) equation.

Note that using $t\in[0,T]$ means that the corresponding PDE needs to be solved subject to a terminal payoff condition $V(t=T,S_t)=\varphi(S_T)$. On the other hand, using the time-to-maturity $\tau$, the PDE is solved subject to an initial condition $V(\tau=0,S_\tau)=\varphi(S_T)$.

  • $\begingroup$ @AlexC said "In the former case you have a minus sign in front of DV/dt" but it seems to be the opposite? $\endgroup$ – emcor Dec 8 '19 at 11:58
  • 1
    $\begingroup$ Yes, sorry, it is the opposite. I could not edit the comment after 5 minutes. $\endgroup$ – Alex C Dec 8 '19 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.