We can solve the Black Scholes PDE by numerical methods like Euler \begin{equation} \frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S}+\frac{1}{2} \sigma^2 S^2\frac{\partial^2 V}{\partial S^2}-rV=0 \end{equation} In order to do that, we need to discretize the stock space. A simple way is to set a range of potential stock prices and then uniformly divide it.

My confusion is that, in this process, the distribution of the underlying stock price, i.e., $dS_t=rS_tdt+\sigma dW$, seems to only factor in determining the range of the space grid. So the probability of getting to each space grid does not affect the PDE solution. But on the other hand, if the stock price follows another process even if the distribution has a similar range the option price should certainly change. What am I missing here?


1 Answer 1


Assume a sufficiently 'wide and dense' mesh of sampling points $(S_i,t_j), i=0..N, j=0..M$, $S_i=S_{low}+i\Delta S$, $t_j=j\Delta t$. Given the payoff encoded in $v_{i,M}$ and some boundary conditions for $j=M$, sarting from $j=M$ backwards, an explicit recursive numerical approximation scheme to the value equation then works along the lines of

$$ v_{i,j}=\alpha v_{i-1,j+1}+\beta v_{i,j+1}+\gamma v_{i+1,j+1} $$

where $\alpha,\beta,\gamma$ are some weights defined by the parameters of the underlying process, i.e. $r,\sigma$, and your choice discretization step sizes $\Delta S,\Delta t$. As you have already 'fixed' some method and (sensible) discretization, the parameters $\alpha,\beta,\gamma$ are solely driven by the underlying process - and that's what will influence the option price.

Did that help?

  • 1
    $\begingroup$ @J.Lin: I think what KermittFrog is saying ( which I only see now ) is that the value of the $v_{ij}$ are determined by the underlying process., So, if the process changes, then the $v_{i,j}$ change which means that the parameters estimates of $\alpha$, $\beta$, $\gamma$ will change. $\endgroup$
    – mark leeds
    Feb 2 at 14:23
  • $\begingroup$ Yep! Should I update my answer a bit? $\endgroup$ Feb 2 at 15:41
  • $\begingroup$ IMHO, no. I was just giving the "layman's interpretation" since yours was somewhat for the non-layman and I didn't see a response from the OP that it helped. Great answer and thanks because the question had me confused initially also. $\endgroup$
    – mark leeds
    Feb 2 at 18:46

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