I want to understand the connection between the trinomial tree and the finite difference methods. As far as I understood so far is, if we transform the Black-Scholes-PDE to heat equation, the explicit finite difference methods can be written as $$V_{i-1,j}=\lambda V_{i,j+1}+(1+2\lambda) V_{i,j}+\lambda V_{i,j-1}$$ , where $i$ represent the time mesh and j the spatial mesh.

This is nothing else than a trinomial tree with probability factor $\lambda$.

Question: In finite difference method, I am using a $N\times M$ grid, with $N$ number of time mesh and $M$ number of spatial mesh. However, if I am consider only a trinomial mode, the underlying grid would be a triangle, which is not persist with the grid from finite difference method. What is the connection?

  • $\begingroup$ If you are interested only in finding the option price for one starting stock price $S_0,0$ then you need only consider a "tree" (this is how option routines usually work), but if you apply the equation to all points in a rectangular grid, you will compute the solution for all intial stock prices (and this is how PDEs are usually solved). $\endgroup$ – noob2 May 4 '17 at 19:31
  • $\begingroup$ @noob2 On a rectangular grid I will have a function of $S_0$? If I am interested in a given $S_0$, it is then better to transform the rectangular grid from some point $t$ to a triangular grid? $\endgroup$ – quallenjäger May 4 '17 at 19:35
  • $\begingroup$ @noob2 Thank you for the explanation, I understood $\endgroup$ – quallenjäger May 4 '17 at 20:16

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